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\begin{document}
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\selectlanguage{british}%

\title{Graduate Algebra}


\author{Diane Maclagan\\
Notes by Florian Bouyer}

\maketitle
\begin{center}
Copyright (C) Bouyer 2011. 
\par\end{center}

\begin{center}
Permission is granted to copy, distribute and/or modify this document 
\par\end{center}

\begin{center}
under the terms of the GNU Free Documentation License, Version 1.3 
\par\end{center}

\begin{center}
or any later version published by the Free Software Foundation; 
\par\end{center}

\begin{center}
with no Invariant Sections, no Front-Cover Texts, and no Back-Cover
Texts. 
\par\end{center}

\begin{center}
A copy of the license can be found at \href{http://www.gnu.org/licenses/fdl.html}{http://www.gnu.org/licenses/fdl.html}
\par\end{center}

\tableofcontents{}

\newpage{}


\section{Introduction}


\subsection{Groups }
\begin{defn}
A \emph{semigroup} is a non-empty set $G$ together with a binary
operation (``multiplication'') which is associative ($(ab)c=a(bc)\,\forall a,b,c\in G$)

A \emph{monoid} is a semigroup $G$ which contains an element $e\in G$
such that $ae=ea=a\,\forall a\in G$.

A \emph{group} is a monoid such that $\forall a\in G\,\exists a^{-1}$
such that $aa^{-1}=a^{-1}a=e$.\end{defn}
\begin{note*}
Many authors say ``semigroup'' for monoid. e.g. $\nn=\{0,1,\dots\}$
is called a semigroup.\end{note*}
\begin{example*}[Semigroups that are not monoids]

\begin{itemize}
\item A proper ideal in a ring under multiplication
\item $(\nn\setminus\{0\},+)$
\item $(2\zz,\times)$
\item $(M_{n}(2\zz),\times)$
\item $(\rr,\min)$
\end{itemize}
\end{example*}

\begin{example*}[Monoids that are not groups]

\begin{itemize}
\item $(\nn,+)$
\item Polynomials in 1 variable under composition
\item Rings with identity that has non-invertible elements under multiplication
\item $(\rr\cup\infty,\min)$
\end{itemize}
\end{example*}
\begin{xca*}
{}
\begin{itemize}
\item In a monoid, identities are unique
\item In a group, inverses are unique.
\end{itemize}
\end{xca*}
\begin{defn}
Let $G$ and $H$ be semigroups. A function $f:G\to H$ is a \emph{homomorphism}
\emph{of semigroups} if $f(ab)=f(a)f(b)\,\forall a,b\in G$

If it is a bijection, it is called an \emph{isomorphism}.

Let $G$ and $H$ be monoids. A \emph{monoid homomorphism} is a semigroup
homomorphism with $f(e_{G})=e_{H}$

A \emph{group homomorphism} between groups $G,H$ is a semigroup homomorphism
between the underlying semigroups.
\end{defn}
Group homomorphisms are automatically monoid homomorphisms: $f:G\to H$,
$f(e_{G})=f(e_{G}e_{G})=f(e_{G})f(e_{G})$. Multiply by $f(e_{G})^{-1}$
then we get $e_{H}=f(e_{G})^{-1}f(e_{G})=f(e_{G})^{-1}f(e_{G})f(e_{G})=e_{H}f(e_{G})=f(e_{G})$.
\begin{example*}[Important example of a group: Permutation Group]


Let $X$ be a non-empty set. Let $P(X)$ be the set of all bijection
$f:X\to X$. $P(X)$ is a group under function composition that is
$fg:X\to X$ is $f\circ g:X\to X$.
\begin{itemize}
\item This is associative because function composition is
\item The identity is $\id$ (the identity map)
\item The inverse of $f$ is $f^{-1}:X\to X$. (Which exists since $f$
is a bijection
\end{itemize}
If $|X|=n$ then $P(X)\cong S_{n}$ (the symmetric group on $n$ elements)\end{example*}
\begin{defn}
A \emph{sub\{group, monoid, semigroup\}} of a \{group, monoid, semigroup\}
$G$ is a subset $H\subset G$ that is a \{group, monoid, semigroup\}
under the operation of $G$.

Let $\phi:G\to H$ be a group homomorphism, the \emph{kernel} \emph{of
$\phi$} is $\ker\phi=\{a\in G|\phi(a)=e_{H}\}$\end{defn}
\begin{note*}
The kernel of $\phi$ is a subgroup of $G$. In fact it is \uline{normal}\emph{\uline{
}}(i.e., $\forall g\in G$, $ghg^{-1}\in H=\ker\phi$ for all $h\in H$\}\end{note*}
\begin{defn}
A group $G$ is \emph{abelian} if $ab=ba$ for all $ab\in G$\end{defn}
\begin{xca*}
Find $\phi:G\to H$ ($G,H$ monoid) that is a semigroup homomorphism
but not a monoid homomorphism

$(\rr,\times)\to(M_{2}(\rr),\times)$ by $\phi(a)\mapsto\begin{pmatrix}a & 0\\
0 & 0
\end{pmatrix}$
\end{xca*}

\subsection{Rings}
\begin{defn}
A \emph{ring }$R$ is a non-empty set $R$ together with binary operations
$+,\times$ such that
\begin{enumerate}
\item $(R,+)$ is an abelian group (write identity as $0$)
\item $a(bc)=(ab)c$ (multiplication is associative, so $(R,\times)$ is
a semigroup)
\item $a(b+c)=ab+ac$ and $(a+b)c=ac+bc$ (distributivity)
\end{enumerate}
If there is $1_{R}\in R$ such that $1_{R}a=a1_{R}=a\,\forall a\in R$
then $R$ is a ring \emph{with identity}

$R$ is \emph{commutative} if $ab=ba\,\forall a,b\in R$.

Let $R,S$ be rings. A ring \emph{homomorphism} $\phi:R\to S$ is
a function $\phi$ such that :
\begin{enumerate}
\item $\phi(r+s)=\phi(r)+\phi(s)$ (group homomorphism)
\item $\phi(rs)=\phi(r)\phi(s)$ (semigroup homomorphism)
\end{enumerate}
\end{defn}
\begin{note*}
We do \uline{not} require that if $R,S$ have identities, that
$\phi(1_{R})=1_{S}$ (e.g., $\phi(a)=0_{S}\,\forall a$ is OK)\end{note*}
\begin{defn}
Let $R$ be a ring with identity. An element $a\in R$ is \emph{left}
(respectively \emph{right}) \emph{invertible} if $\exists b\in R$
(respectively $c\in R$) such that $ba=1_{R}$ (respectively $ac=1_{R}$)

If $a$ is left and right invertible then $a$ is called \emph{invertible},
or a \emph{unit}.

A ring with identity $1_{R}\neq0_{R}$ in which every non-zero element
is a unit is a \emph{division ring}. A commutative division ring is
a \emph{field}. 

A field homomorphism is a ring homomorphism $\phi$ of the underlying
rings.\end{defn}
\begin{example*}[Useful example of a ring: Group rings]
 Let $R$ be a commutative ring with $1$. Let $G$ be a group. The
group ring $R[G]$ has entries $\left\{ \sum_{g\in G}r_{g}g:r_{g}\in R\right\} $
``formal sums'' (all but finitely many $r_{g}=0$). This is a ring
under coordinate wise addition, and multiplication is induced from
$(g_{1})(g_{2})=(g_{1}g_{2})$. 

e.g.: $R=\cc,G=\zz$ then $\cc[\zz]=\cc[t,t^{-1}]$. $\cc[\zz/3\zz]=\cc[t]/(t^{3})$\end{example*}
\begin{defn}
Let $R$ be a ring. A (\emph{left}) $R$-\emph{module} is an abelian
group $M$ (write additively) together with a function $R\times M\to M$
such that
\begin{enumerate}
\item $r(m+m')=rm+rm'$
\item $(r+s)m=rm+sm$
\item $r(sm)=(rs)m$
\end{enumerate}
\end{defn}
If $R$ is a field an $R$-module is a vector space. If $R$ has $1_{R}$
we usually ask $1_{R}m=m$ for all $m\in M$.
\begin{defn}
An $R$-module \emph{homomorphism} is a group homomorphism $\phi:M\to M'$
such that $\phi(rm)=r\phi(m)$. 

\newpage{}
\end{defn}

\section{Category Theory}
\begin{defn}
A \emph{category} is a class $\ob(\mathscr{C})$ of objects (write
$A,B,C,\dots$) together with:
\begin{enumerate}
\item a class, $\mor(\mathscr{C})$, of disjoint sets $\hom(A,B)$. one
for each pair of objects in $\ob(\mathscr{C})$. An element $f$ of
$\hom(A,B)$ is called a \emph{morphism} from $A$ to $B$. (write
$f:A\to B$)
\item For each triple $(A,B,C)$ of objects: a function $\hom(B,C)\times\hom(A,B)\to\hom(A,C)$
(write $(f,g)\mapsto f\circ g$) ``composition of morphism satisfying:

\begin{enumerate}
\item associativity: $h\circ(g\circ f)=(h\circ g)\circ f$ with $f\in\hom(A,B)$,
$g\in\hom(B,C)$ and $h\in\hom(C,D)$
\item Identity: For each $B\in$$\ob(\mathscr{C}$) there exists $1_{B}:B\to B$
such that $\forall f\in\hom(A,B)$ $1_{B}\circ f=f$ and $\forall g\in\hom(B,C)$
$g\circ1_{B}=g$
\end{enumerate}
\end{enumerate}
\end{defn}
\begin{example*}
\uline{{}}
\begin{elabeling}{00.00.0000}
\item [{Sets:}] Objects:\uline{ }the class of all sets. Morphisms $\hom(A,B)$
is the set of all functions $f:A\to B$
\item [{Groups:}] Objects: Groups. Morphisms: group homomorphism.
\item [{Semigroups:}] Object: semigroups. Morphisms: semigroup homomorphism
\item [{Monoids:}] Object: monoids. Morphisms: monoid homomorphism.
\item [{Rings:}] Objects: Rings. Morphisms: ring homomorphism
\item [{Ab:}] Objects: abelian groups. Morphisms: group homomorphism
\item [{Vect$_{k}$:}] Objects: Vector spaces over (a field) $k$. Morphisms:
linear transformations.
\item [{Top:}] Objects: Topological spaces. Morphisms: Continuous functions.
\item [{Manifolds:}] Objects: Manifolds. Morphisms: Continuous maps.
\item [{Diff:}] Objects: Differentiable manifolds. Morphisms: differentiable
maps
\item [{Point}] Let $G$ be a group. Object: one point. Morphisms: $\hom(\mathrm{pt},\mathrm{pt})=G$
(composition is multiplication)

\begin{note*}
$\forall f\in\hom(\mathrm{pt},\mathrm{pt})$ there exists $g$ such
that $f\circ g=1_{\mathrm{pt}}=g\circ f$. (This example is useful
for Groupoid)
\end{note*}
\item [{Open~Sets}] Fix a topological space $X$. The category of open
set on $X$: Objects: Open sets. Morphisms: inclusions. (i.e., $\hom(A,B)$
is empty or has size one) (This example is useful for sheaves)
\item [{$R$-module}] Fix a ring $R$. Objects: are $R$-modules. Morphisms:
$R$-module homomorphism $\phi(rm)=r\phi(m)$
\end{elabeling}
\end{example*}
\begin{defn}
In a category a morphism $f\in\hom(A,B)$ is called an \emph{equivalence}
if there exists $g\in\hom(B,A)$ such that $g\circ f=1_{A}$ and $f\circ g=1_{B}$.

If $f\in\hom(A,B)$ is an equivalence then $A$ and $B$ are said
to be \emph{equivalent.} \end{defn}
\begin{example*}
{}
\begin{elabeling}{00.00.0000}
\item [{Groups}] Equivalence is isomorphism
\item [{Top}] Equivalence is homeomorphism
\item [{Set}] Equivalence is bijection.
\end{elabeling}
\end{example*}
\begin{defn}
Let $\mathscr{C}$ be a category and $\{A_{\alpha}:\alpha\in I\}$
be a family of objects of $\mathscr{C}$. A \emph{product} for the
family is an object $P$ of $\mathscr{C}$ together with a family
of morphisms $\{\pi_{\alpha}:P\to A_{\alpha}:\alpha\in I\}$ such
that for \uline{any} object $B$ with morphisms $\phi_{\alpha}:B\to A_{\alpha}$
$\exists!\phi:B\to P$ such that $\phi_{\alpha}\circ\phi=\phi_{\alpha}\,\forall\alpha$ \end{defn}
\begin{example*}
$|I|=2$

$\xyC{3pc}\xymatrix{ & B\ar@{-->}[d]^{\exists!}\ar[ddl]_{\phi_{1}}\ar[ddr]^{\phi_{2}}\\
 & P\ar[dr]_{\pi_{2}}\ar[dl]^{\pi_{1}}\\
A_{1} &  & A_{2}
}
$
\end{example*}
\textbf{Warning:} Products don't always exists, but when they do,
we often recognize them
\begin{example*}
{}\end{example*}
\begin{elabeling}{00.00.0000}
\item [{Set:}] Products is Cartesian product. 
\item [{Groups}] Product is direct product.
\item [{Open~sets~of~$X$}] Interior $(\cap A_{\alpha})$.\end{elabeling}
\begin{lem}
If $(P,\pi_{\alpha})$ and $(Q,\psi_{\alpha})$ are both products
of the family $\{A_{\alpha},\alpha\in I\}$ then $P$ and $Q$ are
equivalent (isomorphic).\end{lem}
\begin{proof}
Since $Q$ is a product $\exists!f:P\to Q$ such that $\pi_{\alpha}=\psi_{\alpha}\circ f$.
Since $P$ is a product $\exists!g:Q\to P$ such that $\psi_{\alpha}=\pi_{\alpha}\circ g$.
So $g\circ f:P\to P$ satisfies $\pi_{\alpha}=\pi_{\alpha}\circ(g\circ f)\,\forall\alpha$.
Since $P$ is a product $\exists!h:P\to P$ such that $\pi_{\alpha}=\pi_{\alpha}\circ h$.
Since $h=1_{P}$ satisfies this, we must have $g\circ f=1_{P}$. Similarly
$f\circ g:Q\to Q$ equals $1_{Q}$. So $f$ is an equivalence.\end{proof}
\begin{defn}
An object $I$ in a category $\mathscr{C}$ is \emph{universal} (or
initial) if for all objects $C\in\ob(\mathscr{C})$ there is an unique
morphism $I\to C$. $J$ is \emph{couniversal} (or terminal) if for
all object $C$ there is a unique morphism $C\to J$.\end{defn}
\begin{example*}
{}\end{example*}
\begin{elabeling}{00.00.0000}
\item [{Sets}] $\emptyset$ initial, $\{x\}$ terminal.
\item [{Groups}] Trivial group, initial and terminal.
\item [{Open~sets}] $\emptyset$ is initial. $X$ is terminal\end{elabeling}
\begin{example*}
Pointed topological spaces: Objects: Pairs $(X,p)$ where $X$ is
a non-empty topological space, $p\in X$. Morphisms: Continuous maps
$f:(X,p)\to(Y,q)$ with $f(p)=q$. $(\{p\},p)$ is terminal and initial.\end{example*}
\begin{thm}
Any two initial (terminal) objects in a category are equivalent.\end{thm}
\begin{proof}
Let $I,J$ be two initial objects in $\mathscr{C}$. Since $I$ is
initial $\exists!f:I\to J$. Since $J$ is initial $\exists!g:J\to I$.
Since $I$ is initial, $1_{I}$ is the only morphism $I\to I$, so
$g\circ f=1_{I}$. Similarly, $f\circ g=I_{J}$ so $f$ is an equivalence.
For terminal objects the proof is the same with the arrows reversed.
\end{proof}
Why is the lemma a special case of the theorem. Let $\{A_{\alpha}:\alpha\in I\}$
be a family of objects in a category $\mathscr{C}$. Define a category
$\mathscr{E}$ whose objects are all pairs $(B,f_{\alpha}:\alpha\in I)$
where $f_{\alpha}:B\to A_{\alpha}$. The morphisms are morphisms $(B,f_{\alpha})\to(C,g_{\alpha})$
are morphisms $h:B\to C$ such that $f_{\alpha}=g_{\alpha}\circ h$. 

Check:
\begin{itemize}
\item $I_{B}:B\to B$ induces $1_{(B,f_{\alpha})}$in $\mathscr{E}$
\item Composition of morphisms is still ok (These first two checks that
$\mathscr{E}$ is a category)
\item $h$ is an equivalence in $\mathscr{E}$ implies $h$ is an equivalence
in $\mathscr{C}$. (This will help us show what we wanted)
\end{itemize}
If a product of $\{A_{\alpha}\}$ exists, it is terminal in $\mathscr{E}$.
We just showed terminal objects are unique (up to equivalence) so
products are unique (up to equivalence).
\begin{note*}
Not every category has products. (for example finite groups)\end{note*}
\begin{defn}
A \emph{coproduct} of $\{A_{\alpha}\}$ in $\mathscr{C}$ is ``a
product with the arrows reversed'', i.e., $Q$ with $\pi_{\alpha}:A_{\alpha}\to Q$
such that $\forall C$ with $\phi_{\alpha}:A_{\alpha}\to C$, $\exists!f:Q\to C$
such that $\phi_{\alpha}=f\circ\pi_{\alpha}$

$\xyC{3pc}\xymatrix{A_{1}\ar[dr]^{\pi_{1}}\ar[ddr]_{\phi_{1}} &  & A_{2}\ar[dl]_{\pi_{2}}\ar[ddl]^{\phi_{2}}\\
 & Q\ar@{-->}[d]^{\exists!}\\
 & C
}
$ \end{defn}
\begin{example*}
The coproduct of sets is a disjoint union.

For the pointed topological space we have the product is $(\prod X_{\alpha},\prod p_{\alpha})$.
The coproduct is the wedge product, that is, (in the case of the coproduct
of two object) $X\coprod Y/p\sim q$.

For abelian groups the coproduct is direct sum, i.e., $\oplus_{I}G_{\alpha}\ni(g_{\alpha}:\alpha\in I,g_{\alpha}\in G_{\alpha})$
and all but finitely many $g_{\alpha}=e_{G_{\alpha}}$. 
\end{example*}

\subsection{Functors}
\begin{defn}
Let $\mathscr{C}$ and $\mathscr{D}$ be categories. A \emph{covariant}
functor $T$ from $\mathscr{C}$ to $\mathscr{D}$ is a pair of functions
(both denoted by $T$):
\begin{enumerate}
\item An object function: $T:\ob(\mathscr{C})\to\ob(\mathscr{D})$
\item A morphism function $T:\mor(\mathscr{C})\to\mor(\mathscr{D})$ with
$f:A\to B\mapsto T(f):T(A)\to T(B)$ such that 

\begin{enumerate}
\item $T(1_{C})=1_{T(C)}\forall C\in\ob(\mathscr{C})$
\item $T(g\circ f)=T(g)\circ T(f)$ for all $f,g\in\mor(\mathscr{C})$ where
composition is defined
\end{enumerate}
\end{enumerate}
\end{defn}
\begin{example*}
{}
\begin{itemize}
\item The ``forgetful functor'' from Groups to Sets. $T(G)=$underlying
set and $T(f)=f$ (i.e. same functions, thought of as a map of sets)
\item $\hom(G,-):$Groups$\to$Sets. Let $G$ be a fixed group. Let $T$
be the functor that takes a group $H$ to the set $\hom(G,H)$. If
$f:H\to H'$ is a group homomorphism, then $T(f):T(H)\to T(H')$ is
given by $T(f)(g)=f\circ g$.


Check:
\begin{itemize}
\item $T(1_{H})(g)=1_{H}\circ g=g$ so $T(1_{H})=1_{T(H)}$
\item $T(g\circ f)(h)=(g\circ f)\circ h=g\circ(f\circ h)=T(g)(f\circ h)=T(g)(T(f)(h))=(T(g)\circ T(f))(h)$
\end{itemize}
\end{itemize}
\end{example*}
\begin{defn}
Let $\mathscr{C}$ and $\mathscr{D}$ be categories. A \emph{contravariant}
functor $T$ from $\mathscr{C}$ to $\mathscr{D}$ is a pair of functions
(both denoted by $T$):
\begin{enumerate}
\item An object function: $T:\ob(\mathscr{C})\to\ob(\mathscr{D})$
\item A morphism function $T:\mor(\mathscr{C})\to\mor(\mathscr{D})$ with
$f:A\to B\mapsto T(f):T(B)\to T(A)$ such that 

\begin{enumerate}
\item $T(1_{C})=1_{T(C)}\forall C\in\ob(\mathscr{C})$
\item $T(g\circ f)=T(f)\circ T(g)$ for all $f,g\in\mor(\mathscr{C})$ where
composition is defined
\end{enumerate}
\end{enumerate}
\end{defn}
\begin{example*}
$\hom(-,G):$Groups$\to$Sets. Let $G$ be a fixed group. Let $T$
be the functor that takes a group $H$ to the set $\hom(H,G)$. If
$f:H\to H'$ is a group homomorphism, then $T(f):T(H')\to T(H')$
is given by $T(f)(g)=g\circ f$.\end{example*}
\begin{defn}
Let $\mathscr{C}$ be a category. The \emph{opposite category} $\mathscr{C}^{\mathrm{op}}$
has object $\ob(\mathscr{C})$ and $\hom_{\mathscr{C}^{\mathrm{op}}}(A,B)=\hom_{\mathscr{C}}(B,A)$.
(``reverse the arrows'')
\end{defn}
One can see that this is a category with $g^{\mathrm{op}}\circ f^{\mathrm{op}}=(f\circ g)^{\mathrm{op}}$. 

If $T:\mathscr{C}\to\mathscr{D}$ is a contravariant functor then
$T^{\mathrm{op}}:\mathscr{C}^{\mathrm{op}}\to\mathscr{D}$ defined
by $T^{\mathrm{op}}(C)=T(C)$ and $T^{\mathrm{op}}(f)=T(f)$ is covariant.


\subsection{Some natural occurring functors}
\begin{enumerate}
\item Fundamental group (ref: Hatcher ``Algebraic Topology'')


$\pi_{1}:$ Pointed topological spaces $\to$Groups. $\pi_{1}(X,p)=$homotopy
classes of maps $f:[0,1]\to X$ such that $f(0)=f(1)=p$. This is
a group under concatenation of loops. $f\circ g:[0,1]\to X$ with
$f\circ g(t)=\begin{cases}
g(2t) & 0\leq t\leq\frac{1}{2}\\
f(2t-1) & \frac{1}{2}\leq t\leq1
\end{cases}$, $f^{-1}(t)=f(1-t)$. If $\phi:(X,x)\to(Y,y)$ is a continuous map
with $\phi(x)=y$, then we get an induced map $\pi_{1}(X,x)\to\pi_{1}(Y,y)$
by $(f:[0,1]\to X)\mapsto(\phi(f):[0,1]\to Y)$ by $\phi(f)(t)=\phi\circ f(t)$. 


Check:
\begin{enumerate}
\item This is a group homomorphism
\item $\pi_{1}(1_{(X,x)})=1_{\pi_{1}(X,x)}$
\item $\pi_{1}(\phi\circ\psi)=\pi_{1}(\phi)\circ\pi_{1}(\psi)$
\end{enumerate}

\emph{Recall:} A group is a category with one object where all morphisms
are isomorphisms (have inverses). A groupoid is a category where all
morphisms are isomorphisms.

\item Consider the category $\mathscr{U}(X)$ of open sets on $X$ with
morphisms inclusion $T:\mathscr{U}\to$Sets, $T(U)=\{$continuous
functions from $U$ to $\rr$\}. If $V\subseteq U$ then $T(V)\leftarrow T(U)$
(by restriction). Good easy exercise is to finish checking that this
is a functor. This is an example of a presheaf.
\end{enumerate}

\subsection{Natural Transformations}
\begin{defn}
Let $\mathscr{C}$ and $\mathscr{D}$ be categories and let $S$ and
$T$ be covariant functors from $\mathscr{C}$ to $\mathscr{D}$.
A\emph{ natural transformation} $\alpha$ from $S$ to $T$ is a collection
$\{\alpha_{c}:c\in\ob(\mathscr{C})\}$ in $\mor(\mathscr{D})$, where
$\alpha_{c}:S(C)\to T(C)$ such that if $f:C\to C'$ is a morphism
in $\mathscr{C}$ then 
\[
\xymatrix{S(C)\ar[r]^{\alpha_{c}}\ar[d]_{S(f)} & T(C)\ar[d]^{T(f)}\\
S(C')\ar[r]^{\alpha_{c'}} & T(C')
}
\]
commutes.\end{defn}
\begin{example*}
$\mathscr{C}=$groups, $\mathscr{D}=$sets. $S=\hom(G,-)$ and $T=\hom(H,-)$.
Let $\phi:H\to G$ be a group homomorphism. Given a group $A$, we
construct $\alpha_{A}:\hom(G,A)\to\hom(H,A)$ by $g\mapsto g\circ\phi$
(where $\phi:H\to G$). Let $f:A\to B$
\[
\xymatrix{\hom(G,A)\ar[r]^{g\mapsto g\circ\phi}\ar[d]_{g\mapsto f\circ g} & \hom(H,A)\ar[d]^{g'\mapsto f\circ g'}\\
\hom(G,B)\ar[r]^{g'\mapsto g\circ\phi} & \hom(H,B)
}
\]
\end{example*}
\begin{defn}
A natural transformation where all $\alpha_{c}$ are isomorphism is
called a \emph{natural isomorphism.}\end{defn}
\begin{example*}
Let $\mathscr{C}=\mathscr{D}=$$n$-dimensional vector space over
$k$. Let $S=\id$ and $T:V\to V^{**}$ (i.e. $T(V)=V^{**}$ if $f:V\to W,w^{*}\to k$
then $T(f):V^{**}\to W^{*}$, $T(f)(\beta)\in W^{**}$ we have $T(f)(\beta)(\psi)=\beta(\psi\circ f)\in k$

We claim $T$ and $S$ are naturally isomorphic. For $V\in$Vect$_{n}^{k}$,
let $\alpha_{v}$ be the linear transformation $V\to V^{**}$ given
by $v\mapsto\phi_{V}$ where $\phi_{V}(\psi)=\psi(v)$. Then for $f:V\to W$
\[
\xymatrix{V\ar[r]^{\alpha_{V}}\ar[d]_{S(f)=f} & V^{**}\ar[d]^{T(f)}\\
W\ar[r]^{\alpha_{W}} & W^{**}
}
\]
\begin{eqnarray*}
T(f)(\alpha_{V}(v))(\psi) & = & T(f)(\phi_{V})(\psi)\\
 & = & \phi_{V}(\psi\circ f)\\
 & = & \psi\circ f(v)\\
 & = & \phi_{f(v)}(\psi)
\end{eqnarray*}
Since $\alpha_{w}\circ f(v)=\phi_{f(v)}$ means the diagram commutes.
Since each $\alpha_{V}$ is an isomorphism (exercise that this uses
finite dimension), $T$ is naturally isomorphic to $S$.\end{example*}
\begin{defn*}
Two categories $\mathscr{C}$ and $\mathscr{D}$ are equivalent if
there are functors $f:\mathscr{C}\to\mathscr{D}$ and $g:\mathscr{D}\to\mathscr{C}$
such that $f\circ g$ is natural isomorphic to $1_{\mathscr{D}}:\mathscr{D}\to\mathscr{D}$,
$g\circ f$ is naturally isomorphic to $1_{\mathscr{C}}:\mathscr{C}\to\mathscr{C}$.

\newpage{}
\end{defn*}

\section{Free Groups}

Intuitive idea: Group formed by ``words'' in an alphabet. Multiplication
is concatenation, e.g. $F_{2}=$words in $x$ and $y$ for example
$xyx^{-1}y^{-1}x^{3}y^{2}$

\textbf{Construction}
\begin{lyxcode}
{}
\begin{elabeling}{00.00.0000}
\item [{Input:}] A set $X$ (might be infinite)
\item [{1)}] Choose a set $X^{-1}$ disjoint from $X$ with $|X^{-1}|=|X|$
and a bijection $X\to X^{-1}$, $x\mapsto x^{-1}$. Choose an element
$1\notin X\cup X^{-1}$
\item [{2)}] A \emph{word} on $X$ is a sequence $(a_{!},a_{2},\dots)$
with $a_{i}\in X\cup X^{-1}\cup\{1\}$ such that there exists $N$
such that $a_{n}=1\,\forall n>N$. $(1,1,1,\dots)$ is the empty word
and written as $1$
\item [{3)}] A word is \emph{reduced }if:

\begin{enumerate}
\item $\forall x\in X$, $x$ and $x^{-1}$ are never adjacent (i.e., if
$a_{k}=x$ then $a_{k-1},a_{k+1}\neq x^{-1}$)
\item $a_{k}=1\Rightarrow a_{i}=1\,\forall i>k$
\end{enumerate}

A non-empty reduced word has the form $(x_{1}^{\lambda_{1}},x_{2}^{\lambda_{2}},\dots,x_{n}^{\lambda_{n}},1,1,\dots)$
with $x_{i}\in X$ and $\lambda_{i}=\pm1$. Write this as $x_{1}^{\lambda_{1}}x_{2}^{\lambda_{2}}\dots x_{n}^{\lambda_{n}}$.

\item [{4)}] Our group $F(X)$ as a set is the set of reduced words. 


Naive attempt at defining multiplication: Define $(x_{1}^{\lambda_{1}}\dots x_{n}^{\lambda_{n}})(y_{1}^{\delta_{1}}\dots y_{m}^{\delta_{m}})$
to be $x_{1}^{\lambda_{1}}\dots x_{n}^{\lambda_{n}}y_{1}^{\delta_{1}}\dots y_{m}^{\delta_{m}}$


\textbf{Problem:} This product might not be reduced


\textbf{Solution:} Reduce it:


Formally: if $a=(x_{1}^{\lambda_{1}}\dots x_{m}^{\lambda_{m}})(y_{1}^{\delta_{1}}\dots y_{n}^{\delta_{n}})$
(and suppose that $m\leq n$). Let $K=\max_{0\leq k\leq n}\{k:x_{m-j}^{\lambda_{m-j}}=y_{j+1}^{-\delta_{j}+1}$
for all $0\leq j\leq k-1$\}. Then define $ab=\begin{cases}
x_{1}^{\lambda_{1}}\dots x_{m-k}^{\lambda_{m-k}}y_{k+1}^{\delta_{k+1}}\dots y_{n}^{\delta_{n}} & k<m\\
y_{m+1}^{\delta_{m+1}}\dots\delta_{n}^{\delta_{n}} & k=m<n\\
1 & k=m=n
\end{cases}$ (and analogously if $m>n$)


This define a multiplication $F(X)\times F(X)\to F(X)$. 


\textbf{Claim: }This is a group
\begin{itemize}
\item $1$ is the identity
\item The inverse of $x_{1}^{\lambda_{1}}\dots x_{m}^{\lambda_{m}}$ is
$x_{m}^{-\lambda_{m}}\dots x_{1}^{-\lambda_{1}}$
\item For associativity see Lemma \ref{lem:associative}
\end{itemize}
\end{elabeling}
\end{lyxcode}
\begin{lem}
\label{lem:associative}The multiplication $F(X)\times F(X)\to F(X)$
is associative.\end{lem}
\begin{proof}
For each $x\in X$ and $\delta=\pm1$ let $|x^{\delta}|:F(X)\to F(X)$
be the map given by:
\begin{itemize}
\item $1\mapsto x^{\delta}$
\item $x_{1}^{\delta_{1}}\dots x_{n}^{\delta_{n}}\mapsto\begin{cases}
x^{\delta}x_{1}^{\delta_{1}}\dots x_{n}^{\delta_{n}} & x^{\delta}\neq x_{1}^{-\delta_{1}}\\
x_{2}^{\delta_{2}}\dots x_{n}^{\delta_{n}} & x^{\delta}=x_{1}^{-\delta_{1}},n>1\\
1 & n=1,x^{\delta}=x^{-\delta_{1}}
\end{cases}$
\end{itemize}
Note that this map is a bijection, since $|x^{\delta}||x^{-\delta}|=1=|x^{-\delta}||x^{\delta}|$.
Let $A(X)$ be the group of all permutations of $F(X)$. Consider
the map $\phi:F(X)\to A(X)$ given by
\begin{itemize}
\item $1\mapsto1_{A(X)}$
\item $x_{1}^{\delta_{1}}\dots x_{n}^{\delta_{n}}\mapsto|x_{1}^{\delta_{1}}||x_{2}^{\delta_{2}}|\dots|x_{n}^{\delta_{n}}|$
\end{itemize}
since $ $$|x_{1}^{\delta_{1}}|\dots|x_{n}^{\delta_{n}}|:1\mapsto x_{1}^{\delta_{1}}\dots x_{n}^{\delta_{n}}$
we have $\phi$ is injective. Note that if $w_{1},w_{2}\in F(X)$,
then $\phi(w_{1}w_{2})=\phi(w_{1})\phi(w_{2})$. Since $A(X)$ is
a group, the multiplication is associative, so the multiplication
in $F(X)$ is associative.\end{proof}
\begin{example*}
{}
\begin{itemize}
\item $X=\{x\}$, $F(X)\cong\zz$ (reduced words are $1,x^{n},x^{-n}$)
\item $X=\{x,y\}$. $F(X)=$''words in $x,y,x^{-1},y^{-1}$, e.g. $xyx^{-1}y^{-1}$
is reduced. So $F(X)$ is not abelian as $xyx^{-1}y^{-1}\neq1$ so
$xy\neq yx$. 
\end{itemize}
\end{example*}
\begin{note*}
There is an inclusion $i:X\to F(X)$.\end{note*}
\begin{lem}
If $G$ is a group and $f:X\to G$ is a map of sets then $\exists!$
homomorphism $\overline{f}:F(X)\to G$ such that $\overline{f}i=f$.\end{lem}
\begin{proof}
Define $\overline{f}(1)=e\in G$. If $x_{1}^{\delta_{1}}\dots x_{n}^{\delta_{n}}$
is a non-empty reduced word on $X$, set $\overline{f}(x_{1}^{\delta_{1}}\dots x_{n}^{\delta_{n}})=f(x_{1})^{\delta_{1}}\dots f(x_{n})^{\delta_{n}}$.
Then $\overline{f}$ is a group homomorphism with $\overline{f}i=f$
by construction and it is unique by the homomorphism requirement.
\end{proof}
This says the free group is a free object in the category of group.
If $\mathscr{C}$ is a concrete category (there exists a forgetful
functor $F:\mathscr{C}\to$Sets) and $i:X\to F(X)$, where $X$ is
a set and $A\in\ob(\mathscr{C})$, is a function, then $A$ is free
on $X$ if for all $j:X\to F(B)$ $\exists!$ $\phi:A\to B$ such
that $\phi\circ i=j$. (Note $B\in\ob(\mathscr{C})$ and $\phi\in\mor(\mathscr{C})$). 

Compare:
\begin{itemize}
\item Vector Spaces
\item Commutative $k$-algebra
\item $R$-modules\end{itemize}
\begin{cor}
Every group $G$ is the homomorphic image of the a free group.\end{cor}
\begin{proof}
Let $X$ be a set of generators of $G$.The inclusion $f:X\to G$
gives a map $\overline{f}:\underset{x\mapsto x}{F(X)\to G}$. The
map $\overline{f}$ is surjective since $X$ is a set of generators.
So $G\cong F(X)/\ker(f)$. \end{proof}
\begin{defn}
Let $G$ be a group and let $Y$ be a subset of $G$. The \emph{normal
subgroup} $N=N(Y)$ of $G$ generated by $Y$ is the intersection
of all normal subgroups of $G$ containing $Y$.
\end{defn}
Check that it is well defined. (That is check $N(Y)$ is non-empty,
it relies on the fact $G$ is normal)
\begin{defn}
Let $X$ be a set and let $Y$ be a set of (reduced) words on $X$.
A group $G$ is said to be \emph{defined by generators $X$ and relations}
$w=e$ for $w\in Y$ if $G\cong F(X)/N(Y)$. (We say $(X|Y)$ is a
\emph{presentation} of $G$)\end{defn}
\begin{example*}
$\left\langle x|x^{6}\right\rangle \cong\zz/6\zz$

$\left\langle x,y|x^{4},y^{2},(xy)^{2}\right\rangle \cong D_{4}$
(or $D_{8}$ depending of your notation)\end{example*}
\begin{note*}
Presentations are not unique, e.g., $\left\langle x,y|x^{3},y^{2},xyx^{-1}y^{-1}\right\rangle \cong\zz/6\zz\cong\zz/2\zz\times\zz/3\zz$,
$\left\langle x,y|xy^{-5}\right\rangle \cong\zz$, $\left\langle x,y|x^{2},y^{2},(xy)^{4}\right\rangle \cong D_{4}$
(the Coveter presentation)

Given a presentation $G=\left\langle X|R\right\rangle \cong F(X)/N(R)$.
The \uline{word problem} asks if a given word $w\in F(X)$ equals
the identity of $G$. This is undecidable! {[}Novikav 1955{]}.\end{note*}
\begin{example*}
Burnside groups, $B(m,n)=\left\langle x_{1},\dots,x_{m}|w^{n}\mathrm{\, for\, any\, word\,}w\right\rangle $.
Question (Burnside 1902) Is $B(n,m)$ finite? In the case $B(1,n)\cong\zz/n\zz$
and $B(m,2)\cong(\zz/2\zz)^{n}$.
\end{example*}
Question: What are free objects in the category of abelian groups?
$\bigoplus_{x\in X}\zz$.

\newpage{}


\section{Tensor Product}

We'll work in the category of $R$-modules (no assumptions are made
on $R$, including whether it has $1$ or not). (Cross-reference this
whole chapter with Commutative Algebra Chapter 2)
\begin{elabeling}{00.00.0000}
\item [{\textbf{Recall}}] $M$ is a \emph{left} $R$-module if $R\times M\to M$
with $(r,m)\mapsto rm$ and $r(s(m))=(rs)m$. And $M$ is a \emph{right}
$R$-module if $R\times M\to M$ with $(r,m)\mapsto mr$ and $(mr)s=m(rs)$

\begin{example*}
$R=M_{n}(\cc)$ and $M=\cc^{n}$ is left ($M$ is columns vectors)
or right ($M$ is row vectors) $R$-module
\end{example*}

If $R$ is commutative a left $R$-module structure gives rise to
a right $R$-module structure, i.e., we define $mr=rm$. $M$ is an
$S-R$ bimodule if $M$ is a left $S$-module and a right $R$-module
and $(sm)r=s(mr)$, e.g., $\cc^{n}$ is a $M_{n}(\cc)-\cc$ bimodule.

\end{elabeling}
Suppose we have $f:A\oplus B\to C$ such that $f(a_{1}+a_{2},b)=f(a_{1},b)+f(a_{2},b)$,
$f(a,b_{1}+b_{2})=f(a,b_{1})+f(a,b_{2})$ and $f(ar,b)=f(a,rb)$.
We show $A\times B\to A\otimes_{R}B\to C$.
\begin{example*}
$f:\rr^{2}\oplus\rr^{2}\to\rr$, $f\left(\begin{pmatrix}a\\
b
\end{pmatrix},\begin{pmatrix}c\\
d
\end{pmatrix}\right)=4ac+bc+ad+4bd$. (Easy to check the above relations holds)\end{example*}
\begin{defn}
Let $A$ be a right $R$-module and $B$ a left $R$-module. Let $F$
be the free abelian group on the set $A\times B$. Let $K$ be the
subgroup generated by all elements:
\begin{enumerate}
\item $(a+a',b)-(a,b)-(a',b)$
\item $(a,b+b')-(a,b)-(a,b')$
\item $(ar,b)-(a,rb)$
\end{enumerate}
for all $a,a'\in A,b,b'\in B$ and $r\in R$. The quotient $F/K$
is called the \emph{tensor product} of $A$ and $B$ and is written
$A\otimes_{R}B$. Note: $(a,b)+K$ is written $a\otimes b$ and $(0,0)+K$
is written $0$. This is an abelian group.
\end{defn}
\textbf{Warning}: Not every element of $A\otimes_{R}B$ has the form
$a\otimes b$. A general element is (finite) $\sum n_{i}(a_{i}\otimes b_{i})$
with $n_{i}\in\zz$.

We have relations $(a_{1}+a_{2})\otimes b=a_{1}\otimes b+a_{2}\otimes b$,
$a\otimes(b_{1}+b_{2})=a\otimes b_{1}+a\otimes b_{2}$ and $ar\otimes b=a\otimes rb$.
If $A$ is a $S-R$ bimodule then $A\otimes_{R}B$ is a left $S$-module
since $F$ is an $S$-module by $s(a,b)=(sa,b)$ and $K$ is an $S$-submodule.
\begin{example*}
$\zz/2\zz\otimes\zz/2\zz\cong\zz/2\zz$. (c.f. Commutative Algebra)
\end{example*}
There is a function $\pi:A\times B\to A\otimes_{R}B$ defined by $(a,b)\mapsto a\otimes b$.
Note: $\pi$ is not a group homomorphism as $(a_{1}+a_{2},b_{1}+b_{2})\mapsto a_{1}\otimes b_{1}+a_{1}\otimes b_{2}+a_{2}\otimes b_{1}+a_{2}\otimes b_{2}$.
However $\pi(a_{1}+a_{2},b)=\pi(a_{1},b)+\pi(a_{2},b)$ and $\pi(a,b_{1}+b_{2})=\pi(a,b_{1})+\pi(a,b_{2})$
and $\pi(ar,b)=\pi(a,rb)$. (Call these relations ``middle linear'')
\begin{namedthm}[The universal property of Tensor Product ]
 Let $A_{R},_{R}B$ be $R$-module and $C$ an abelian group. If
$g:A\times B\to C$ is ``middle linear'' then $\exists!\overline{g}:A\otimes_{R}B\to C$
such that $\overline{g}\pi=g$.
\[
\xymatrix{A\times B\ar[d]_{\pi}\ar[dr]^{g}\\
A\otimes_{R}B\ar@{-->}[r]_{\exists!} & C
}
\]

\end{namedthm}
If $A$ is an $S-R$ bimodule and $C$ is an $S$-module then $\overline{g}$
is a $S$-module homomorphism.
\begin{proof}
Let $F$ be the free abelian group on $A\times B$. There is a unique
group homomorphism $g_{1}:F\to C$ determined by $(a,b)\mapsto g(a,b)$.
Since $g$ is ``middle linear'', $g_{1}((a+a',b)-(a,b)-(a',b))=g(a+a',b)-g(a,b)-g(a',b)=0$.
Similarly, the other generators of $K$live in $\ker g_{1}$, so we
get an induced map $\overline{g}:\underset{=F/K}{\underbrace{A\otimes_{R}B}}\to C$.
Note that $\overline{g}(a\otimes b)=g_{!}((a,b))=g(a,b)$ so $\overline{g}\pi=g$.

If $h:A\otimes B\to C$ is a group homomorphism with $h\pi=g$ then
$h(a\otimes b)=h\pi(a,b)=g(a,b)=\overline{g}(\pi(a,b))=\overline{g}(a\otimes b)$.
So $h$ and $\overline{g}$ agree on generators $a\otimes b$ of $A\otimes_{R}B$,
so $h=\overline{g}$.\end{proof}
\begin{example*}
$R=\zz,A=\zz/4\zz\oplus\zz$, $B=\qq$. Then $A\otimes_{R}B=\qq$. 

To prove this, define $f:\zz/4\zz\oplus\zz\times\qq\to\qq$ by $f((a,b),c)=bc$.
Then $f((a_{1},b_{1})+(a_{2},b_{2}),c)=b_{1}c+b_{2}c=f((a_{1},b_{1}),c)+f((a_{2},b_{2}),c)$,
$f((a,b),c_{1}+c_{2})=b(c_{1}+c_{2})=bc_{1}+bc_{2}$, $f((a,b)n,c)=f((na,nb),c)=nbc=f((a,b),nc)$.
So $f$ is ``middle linear'', so by the proposition there exists
a unique $\overline{f}:A\otimes B\to\qq$ with $\overline{f}((a,b)\otimes c)=bc$.
We have that $\overline{f}$ is surjective since $\overline{f}((0,1)\otimes c)=c$
for all $c\in\qq$. Now consider $d=\sum n_{i}(a_{i},b_{i})\otimes c_{i}$
in $\ker(\overline{f})$. So $\overline{f}(d)=\sum n_{i}b_{i}c_{i}=0$.
Now $(a,b)\otimes c=(a,b)4\otimes\frac{c}{4}=(0,4b)\otimes\frac{c}{4}=(0,1)4b\otimes\frac{c}{4}=(0,1)\otimes bc$.
Hence $d=\sum n_{i}(0,1)\otimes b_{i}c_{i}=(0,1)\otimes\sum n_{i}b_{i}c_{i}=(0,1)\otimes0=0$. 

Tensor products of vector spaces. If $V$ is a vector space over $k$
with basis $e_{1},\dots,e_{n}$ and $W$ is a vector space over $k$
with basis $f_{1},\dots,f_{m}$, then $V\otimes_{k}W$ is a vector
space with basis $\{e_{i}\otimes f_{j}\}$ (so dimension is $nm$)

To prove this, let $U$ be a vector space with basis $\{g_{ij}:1\leq i\leq n,1\leq j\leq m\}$.
Let $h:V\times W\to U$ be given by $\left(\sum a_{i}e_{i},\sum b_{j}f_{j}\right)\mapsto\sum a_{i}b_{j}g_{ij}$.
Check: $h$ is ``middle-linear''. So by the proposition there exists
a unique $\overline{h}:V\otimes_{k}W\to U$. The $k$-module homomorphism
$\overline{h}$ is surjective since $\overline{h}(e_{i}\otimes f_{j})=g_{ij}$.
Note that if $a=\sum a_{i}e_{i}$ and $b=\sum b_{j}f_{j},$ then $a\otimes b=(\sum a_{i}e_{i})\otimes(\sum b_{j}f_{j})=\sum a_{i}b_{j}(e_{i}\otimes f_{j})$.
So if $\overline{h}(\sum n_{ij}e_{i}\otimes f_{j})=0$, then $\sum n_{ij}g_{ij}=0$
so $n_{ij}=0$ for all $i,j$, hence $\sum n_{ij}(e_{i}\otimes f_{j})=0$
so $\overline{h}$ is injective. 
\end{example*}
Consider $\rr\otimes_{\qq}V$. Since $\rr$ is a $\rr-\qq$ bimodule,
this is a left $\rr$-module, so a vector space.
\begin{xca*}
If $\{e_{i}\}$ is a basis for $V$, then $\{1\otimes e_{i}\}$ is
a basis for $\rr\otimes_{\qq}V$ as a $\rr$-vector space.\end{xca*}
\begin{lem}
Let $R$ be a ring with $1$ and $A$ be a unitary left $R$-module.
Then $R\otimes_{R}A\cong A$ as a left $R$-module. \end{lem}
\begin{proof}
The map $f:R\times A\to A$ defined by $(r,a)\mapsto ra$ is ``middle-linear''
(check!), so $\exists!\overline{f}:R\otimes_{R}A\to A$ with $\overline{f}(r\otimes a)=ra$.
Since $r(r'\otimes a)=rr'\otimes a$. So $\overline{f}(r(r'\otimes a))=rr'a=r\overline{f}(r'\otimes a)$,
so $\overline{f}$ is an $R$-module homomorphism. Since $1\otimes a\mapsto a$,
$\overline{f}$ is surjective. Note that $r\otimes a=1\otimes ra$,
so if $\overline{f}(\sum n_{i}(r_{i}\otimes_{R}b_{i}))=0$ then since
we have 
\begin{eqnarray*}
\sum n_{i}(r_{i}\otimes b_{i}) & = & \sum n_{i}(1\otimes r_{i}b_{i})\\
 & = & \sum1\otimes n_{i}r_{i}b_{i}\\
 & = & 1\otimes\sum n_{i}r_{i}b_{i}
\end{eqnarray*}
we find $\overline{f}(\sum n_{i}(r_{i}\otimes b_{i}))=\sum n_{i}r_{i}b_{i}=0$,
so $\sum n_{i}(r_{i}\otimes b_{i})=1\otimes0=0$. So $\overline{f}$
is injective.
\end{proof}
In general, if $M$ is a left $R$-module, and $\phi:R\to S$ a ring
homomorphism, then $S\otimes_{R}M$ is a left $S$-module. This is
often called \emph{extension of scalars} or sometime \emph{base change.}
If $R$, $S$ are fields then $S\otimes_{R}M$ is a vector space with
the same dimension of $M$. 
\begin{xca*}
$K\subsetneq L$ fields, $L\otimes_{K}K[x_{1},\dots,x_{n}]\cong L[x_{1},\dots,x_{n}]$
(as vector spaces)
\end{xca*}

\subsection{Functoriality}

Suppose $\phi:M_{R}\to N_{R}$ and $\psi:_{R}M'\to_{R}N'$ are $R$-module
homomorphisms. We will now construct $\phi\otimes\psi:M\otimes_{R}M'\to N\otimes_{R}N'$
as follows: The map $f:M\times M'\to N\otimes_{R}N'$ given by $(m,m')\mapsto\phi(m)\otimes\psi(m')$
is ``middle-linear''. Check this yourself but we can see that 
\begin{eqnarray*}
(m_{1}+m_{1},m')\mapsto\phi(m_{1}+m_{2})\otimes\psi(m') & = & (\phi(m_{1})+\phi(m_{2}))\otimes\psi(m')\\
 & = & \phi(m_{1})\otimes\psi(m')+\phi(m_{2})\otimes\psi(m')\\
 & = & f(m_{1},m')+f(m_{2},m')
\end{eqnarray*}
This gives an induced map $\overline{f}=\phi\otimes\psi:M\otimes_{R}M'\to N\otimes_{R}N'$
defined by $\phi\otimes\psi(m\otimes m')=\phi(m)\otimes\psi(m')$ 

If $M,N$ are $S-R$ bimodules and $\phi$ is a bimodule homomorphism
then $\phi\otimes\psi$ is an $S$-module homomorphism. 

Then, given a right $R$-module $A$, we get a functor $A\otimes-:_{R}$Mod$\to$Groups,
it act on objects by $B\mapsto A\otimes_{R}B$ and on morphisms it
acts by $(f:B\to C)\mapsto(1\otimes f:A\otimes B\to A\otimes C)$.
Similarly, a left $R$-module $B$ gives a functor $-\otimes_{R}B:$Mod$_{R}\to$Groups. 

If $A$ is an $S-R$ bimodule, we replace Groups by $_{S}$Mod. 
\begin{thm}
\label{thm:Prove later}Let $R,S$ be rings, let $A$ be a right $R$-module,
$B$ an $R-S$ bimodule, and $C$ a right $S$-module. Then $\hom_{S}(A\otimes_{R}B,C)\cong\hom_{R}(A,\hom_{S}(B,C))$.\end{thm}
\begin{note*}
{}
\begin{itemize}
\item Write $F$ for the functor $-\otimes B$ and $G$ for the functor
$\hom(B,-)$. Then the theorem says $\hom_{S}(F(A),C)=\hom_{R}(A,G(C))$.
When we have such a situation for a pair of functors $F$ is called
\emph{left adjoint} to $G$ and $G$ is \emph{right adjoint} to $F$
\item If $B$ is an $R-S$ bimodule and $C$ is a right $S$ module, then
$\hom_{S}(B,C)$ is a right $R$-module, under the map $\psi r\in\hom_{S}(B,C)$
is given by $(\psi r)(b)=\psi(rb)$. Check: $(\psi r)s=\psi(rs)$
since $((\psi r)s)(b)=(\psi r)(sb)=\psi(rsb)=\psi(rs)(b)$. 
\item $\hom_{S}(B,C)$ is an abelian group $(\phi+\psi)(b)=\phi(b)+\psi(b)$.
Identity: $\phi(b)=0\forall b\in B$. 
\end{itemize}
\end{note*}
\begin{example*}
$R=S=C=K$ then $(A\otimes B)^{\mathrm{op}}\cong\hom(A,B^{\mathrm{op}})$\end{example*}
\begin{proof}[of Theorem \ref{thm:Prove later}]
 Given $\phi:A\otimes B\to C$, define $\Psi(\phi)=\psi:A\to\hom_{S}(B,C)$
by $\psi(a)(b)=\phi(a\otimes b)$. We check:
\begin{enumerate}
\item For each $a,\psi(a)\in\hom_{S}(B,C)$, 
\begin{eqnarray*}
\psi(a)(b+b') & = & \phi(a\otimes(b+b'))\\
 & = & \phi(a\otimes b+a\otimes b')\\
 & = & \phi(a\otimes b)+\phi(a\otimes b')\\
 & = & \psi(a)(b)+\psi(a)(b')
\end{eqnarray*}
\begin{eqnarray*}
\psi(a)(bs) & = & \phi(a\otimes bs)\\
 & = & \phi((a\otimes b)s)\\
 & = & \phi(a\otimes b)s\mathrm{\, since\,}\phi\mathrm{\, is\, an\,}S\mathrm{-module\, homomorphism}\\
 & = & \psi(a)(b)s
\end{eqnarray*}

\item $\psi$ is an $R$-module homomorphism: 
\begin{eqnarray*}
\psi(a+a')(b) & = & \phi((a+a')\otimes b)\\
 & = & \phi(a\otimes b+a'\otimes b)\\
 & = & \phi(a\otimes b)+\phi(a'\otimes b)\\
 & = & \psi(a)(b)+\psi(a')(b)\,\forall b
\end{eqnarray*}
So $\psi(a+a')=\psi(a)+\psi(a')\in\hom_{S}(B,C)$
\begin{eqnarray*}
\psi(ar)(b) & = & \phi(ar\otimes b)\\
 & = & \psi(a)(rb)\\
 & = & (\psi(a)r)(b)
\end{eqnarray*}
So $\psi(ar)=\psi(a)r$.
\item $\Psi$ is a group homomorphism
\begin{eqnarray*}
\Psi(\phi+\phi')(a)(b) & = & (\phi+\phi')(a\otimes b)\\
 & = & \phi(a\otimes b)+\phi'(a\otimes b)\\
 & = & \Psi(\phi)(a)(b)+\Psi(\phi')(a)(b)
\end{eqnarray*}
This is true for all $a,b$ so $\Psi(\phi+\phi')=\Psi(\phi)+\Psi(\phi')$.
Hence $\Psi$ is a group homomorphism.
\end{enumerate}
For the inverse, given an $R$-module homomorphism $\psi:A\to\hom_{S}(B,C)$
define the function $f:A\times B\to C$ by $f(a,b)=\psi(a)(b)$. This
is ``middle linear'' (Check!). So $f$ defines $\phi:A\otimes_{R}B\to C$
with $\phi(a\otimes b)=\psi(a)(b)$. This gives an inverse to $\Psi$. \end{proof}
\begin{example*}
Of Adjoints. Let $F:$Sets$\to$Groups defined by $X\mapsto F(X)$
(the free group) and $G:$Groups$\to$Sets the forgetful functor.
Then $\hom_{\mathrm{Groups}}(FX,H)\cong\hom_{\mathrm{Sets}}(X,GH)$.
\end{example*}
The point of all this is: if $F$ is a left adjoint functor, then
$F$ preserves coproduct.
\begin{example*}
$-\otimes B$ preserves direct sums of modules. $(A\oplus B)\otimes C\cong(A\otimes C)\oplus(B\otimes C)$. \end{example*}
\begin{prop}
Let $A$ be a right $R$-module, $B$ an $R-S$ bimodule and $C$
a left $S$-module. Then $(A\otimes_{R}B)\otimes_{S}C\cong A\otimes_{R}(B\otimes_{S}C)$.\end{prop}
\begin{proof}
{}
\begin{elabeling}{00.00.0000}
\item [{Sketch~1}] Fix $C$, define $A\otimes B\to A\otimes(B\otimes C)$
and define $a\otimes b\mapsto a\otimes(b\otimes c)$. This means that
the map $(A\otimes B)\times C\to A\otimes(B\otimes C)$ given by $(a\otimes b,c)=a\otimes(b\otimes c)$
is well define and ``middle linear''. Then do the same thing for
the other direction and we have an isomorphism.
\item [{Sketch~2}] We construct $A\otimes_{R}B\otimes C$ by $F(A\times B\times C)/R$
where $R$ is a set of relations. (For example $(a+a',b,c)=(a,b,c)+(a',b,c)$
or $(ar,b,c)=(a,rb,c)$ or $(a,bs,c)=(a,b,sc)$ etc.) Then use universal
properties of categories. 
\end{elabeling}
\end{proof}
\begin{defn}
Let $R$ be a commutative ring with identity. An \emph{$R$-algebra
}is a ring $A$ with identity and a ring homomorphism $f:R\to A$
mapping $1_{R}$ to $1_{A}$ such that $f(R)$ is in the centre of
$A$ ($f(r)a=af(r)\,\forall r\in R,a\in A$). This makes $A$ a left
and right $R$ -module.\end{defn}
\begin{example*}
$R=k$ a field, $A=k[x_{1},\dots,x_{n}]$

$K\subseteq L$ fields, $R=K,A=L$

$R=K$, $A=M_{n}(K)$\end{example*}
\begin{defn*}
An $R$-algebra \emph{morphism} is a ring homomorphism $\phi:A\to B$
with $\phi(1_{A})=1_{B}$ that is also an $R$-module homomorphism.
So $\phi(ra)=r\phi(a)$. \end{defn*}
\begin{prop}
Let $R$ be a commutative ring with $1$, and let $A,B$ be $R$-algebras.
Then $A\otimes_{R}B$ is an $R$-algebra with multiplication induced
from $(a\otimes b)(a'\otimes b')=aa'\otimes bb'$.\end{prop}
\begin{proof}
Once we have shown that the multiplication is well-defined, then $1\otimes1$
is the identity and we have $f:R\to A\otimes B$ given by $f(r)=r\otimes1=1\otimes r$.
This satisfies $(r\otimes1)(a\otimes b)=ra\otimes b=ar\otimes b=(a\otimes b)(r\otimes1)$. 

To show that the multiplication is well-defined and distributive we
can construct a homomorphism $A\otimes B\otimes A\otimes B\to A\otimes B$
by $a\otimes b\otimes a'\otimes b'\mapsto aa'\otimes bb'$. We construct
this map in stages: fix $a',b'$ and construct $\phi_{a',b'}:A\otimes B\to A\otimes B$.
Use $\phi_{a',b'}$to construct $\psi_{b}:A\otimes B\otimes A\to A\otimes B$
and thus the above map. This homomorphism is induced by a ``middle
linear'' map $f:A\otimes B\times A\otimes B\to A\otimes B$ defined
by $(a\otimes b,a'\otimes b')\mapsto(aa',bb')$ which is our multiplication.
The ``middle-linearity'' shows distributivity. 
\end{proof}

\subsection{Tensor Algebras}

Let $R$ be a commutative ring with $1$ and let $M$ be an $R$-module
\begin{defn}
For each $k\geq1$, set $T^{k}(M)=\underset{k}{\underbrace{M\otimes_{R}M\otimes_{R}\dots\otimes_{R}M}}$.
So $T^{0}(M)=R$. Define $T(M)=R\oplus M\oplus(M\otimes M)\oplus\dots=\bigoplus_{k=0}^{\infty}T^{k}(M)$.
By construction this is a left and right $R$-module.\end{defn}
\begin{thm}
$T(M)$ is an $R$-algebra containing $M$ defined by $(m_{1}\otimes\dots\otimes m_{i})(m_{1}'\otimes\dots\otimes m_{j}')=m_{1}\otimes\dots\otimes m_{i}\otimes m_{1}'\otimes\dots\otimes m_{j}'$
and extend via distributivity. For this multiplication, $T^{i}(M)T^{j}(M)\subseteq T^{i+j}(M)$.
If $A$ is any $R$-algebra and $\phi:M\to A$ is a $R$-module homomorphism,
then there exists a unique $R$-algebra homomorphism $\Phi:T(M)\to A$
such that $\Phi|_{M}=\phi$. \end{thm}
\begin{proof}
The map $T^{i}(M)\times T^{j}(M)\to T^{i+j}(M)$ defined by $(m_{1}\otimes\dots\otimes m_{i},m_{1}'\otimes\dots\otimes m_{j}')\mapsto(m_{1}\otimes\dots\otimes m_{i}\otimes m_{1}'\otimes\dots\otimes m_{j}')$
is ``middle-linear'' (check that this is well-defined.) So multiplication
is defined and distributive. Suppose $A$ is an $R$-algebra and $\phi:M\to A$
is an $R$-module homomorphism. Then $M\times M\to A$ defined by
$(m_{1},m_{2})\mapsto\phi(m_{1})\phi(m_{2})$ is middle linear. So
it defines an $R$-module homomorphism $M\otimes M\to A$. (Exercise:
check actually we get $T^{k}(M)\to A$). We thus get a $R$-module
homomorphism $\Phi:T(M)\to A$ with $\Phi|_{M}=\phi.$ This respect
multiplication, so is a ring homomorphism. Now $\Phi(1)=1$ by construction,
so we have a $R$-algebra homomorphism. Since $\Phi(m)=\phi(m)$ for
all $m\in M$, if $\Psi:T(M)\to A$ were another such $R$-algebra
homomorphism, $\Psi(m_{1}\otimes\dots\otimes m_{i})=\Phi\Psi(m_{1})\dots\Psi(m_{i})=\phi(m_{1})\dots\phi(m_{i})=\Phi(m_{1}\otimes\dots\otimes m_{i})$,
so $\Psi=\Phi$. \end{proof}
\begin{example*}
$R=K$ a field, $M=V$ a $d$ dimensional vector space with basis
$e_{1},\dots,e_{d}$. $T^{j}(M)$ is a vector space with basis $e_{i_{1}}\otimes\dots\otimes e_{i_{j}}$
and hence has dimension $d^{j}$. Multiplication is concatenation,
so $T(M)$ consists of ``non-commutative polynomials'' in the variables
$e_{1},\dots,e_{d}$. This is either called the ``non-commutative
polynomial algebra'' or ``free associative algebra''. 

$R=\zz$ and $M=\zz/6\zz$. Now $T^{j}(M)\cong\zz/6\zz$, so $T(M)\cong\zz\oplus\zz/6\zz\oplus\zz/6\zz\oplus\dots\cong\zz[x]/\left\langle 6x\right\rangle $.

We work out$T(\qq/\zz)$. Now $\qq/\zz\otimes_{\zz}\qq/\zz=0$, to
see this $\frac{a}{b}\otimes\frac{c}{d}=\frac{a}{b}\otimes\frac{b}{b}\frac{c}{d}=\left(\frac{a}{b}\right)b\otimes\frac{1}{b}\frac{c}{d}=a\otimes\frac{c}{bd}=0\otimes\frac{c}{bd}=0$.
So $T(\qq/\zz)=\zz\oplus\qq/\zz$. \end{example*}
\begin{defn}
A ring $S$ is $(\nn)$-\emph{graded }if $S\cong S_{0}\oplus S_{1}\oplus\dots=\oplus_{k\geq0}S_{k}$
(as groups) with $S_{j}S_{i}\subseteq S_{i+j}$ and $S_{i}S_{j}\subseteq S_{i+j}$
$\forall i,j\geq0$. The elements of $S_{i}$ are called \emph{homogeneous
of degree $i$}. A homomorphism $\phi:S\to T$ of graded rings is
\emph{graded} if $\phi(S_{k})\subseteq T_{k}$ for all $k$.\end{defn}
\begin{example*}
$S=k[x_{1},\dots,x_{n}]$ or $S=T^{k}(M)$\end{example*}
\begin{note*}
$S_{0}S_{0}\subseteq S_{0}$, so $S_{0}$ is a subring. Also $S_{0}S_{j}\subseteq S_{j}$,
so each $S_{j}$ is an $S_{0}$-module. If $S$ has an identity $1$,
it lives in $S_{0}$. (if not $1=e_{k}+e$, $e\in\oplus_{j=0}^{k+1}S_{j}$,
for $s\in S_{1},s=1\cdot s=(e_{k}+e)s=e_{k}s+es\rightarrow e_{k}s=0$).
If $S_{0}$ is in the centre of $S$, then $S$ is an $S_{0}$-algebra. 
\end{note*}

\subsection{Symmetric and Exterior Algebras}
\begin{defn}
The \emph{symmetric algebra} of an $R$-module $M$ is $S(M)=T(M)/C(M)$
where \\
$C(M)=\left\langle m_{1}\otimes m_{2}-m_{2}\otimes m_{1}:m_{1},m_{2}\in M\right\rangle $
(two-sided ideal generated by).
\end{defn}
Since $T(M)$ is generated as an $R$-algebra by $T$ and $M$, and
the images of $m_{1}\otimes m_{2}$ and $m_{2}\otimes m_{1}$ agrees
in $S(M)$, we have $S(M)$ is a commutative ring. (Exercise: think
about universal properties)
\begin{example*}
$V$ is a $d$-dimensional vector space over $k$, spanned by $e_{1},\dots,e_{d}$.
Then $S(V)\cong k[x_{1},\dots,x_{n}]$. \end{example*}
\begin{defn}
The \emph{exterior algebra} of an $R$-module $M$ is the $R$-algebra
$\wedge(M)=T(M)/A(M)$ where $A(M)=\left\langle m\otimes m:m\in M\right\rangle $.
The image of $m_{1}\otimes\dots\otimes m_{j}$ in $\wedge M$ is written
$m_{1}\wedge m_{2}\wedge\dots\wedge m_{j}$. Multiplication is called
\emph{exterior }or \emph{wedge product.}\end{defn}
\begin{example*}
$M=V$ a $d$-dimensional vector space over $k$ (of characteristic
not 2), spanned by $e_{1},\dots,e_{d}$. Then $\wedge M=T(M)/A(M)=$
\{non-commutative polynomials\}$/\left\langle l^{2}\right\rangle $
where $l=\sum a_{i}e_{i}$. We see that this forces $x_{i}x_{j}=-x_{j}x_{i}$
(consider $(x_{i}+x_{j})^{2}$). So in this case $\wedge V=k\left\langle x_{1},\dots,x_{j}\right\rangle /\left\langle x_{i}x_{j}+x_{j}x_{i}\right\rangle $.
In characteristic $2$, we have that $x_{i}x_{i}\notin\left\langle x_{i}x_{j}+x_{j}x_{i}\right\rangle $
\end{example*}
Write $\wedge^{k}M$ for the image of $T^{k}(M)$ in $\wedge(M)$ 
\begin{xca*}
This is a graded component, i.e., $\wedge(M)=\oplus\wedge^{k}(M)$. 
\end{xca*}
If $f\in\wedge^{k}M,g\in\wedge^{l}M$ then $fg=(-1)^{kl}gf$. This
is referred as ``graded commutative''


\subsection{Summary}

What we should remember/understand
\begin{itemize}
\item $\zz/m\zz\otimes_{\zz}\zz/n\zz$
\item $L\otimes K[x_{1},\dots,x_{n}]$
\item $V\otimes_{k}W$ where $V,W$ are vector-space over $k$
\item $\wedge^{k}V$ where $V$ is a vector-space over $k$.
\end{itemize}
\newpage{}


\section{Homological Algebra}
\begin{defn}
A sequence $\xyC{1.5pc}\xymatrix{\cdots\ar[r] & M_{1}\ar[r]^{\phi_{1}} & M_{2}\ar[r]^{\phi_{2}} & M_{3}\ar[r]^{\phi_{3}} & M_{4}\ar[r] & \cdots}
$ of groups/$R$-modules/$\dots$ is a \emph{complex} if $\phi_{i+1}\circ\phi=0$,
i.e., $\phi_{i}\subseteq\ker(\phi_{i+1})$.

It is \emph{exact} if $\ker(\phi_{i+1})=\im(\phi_{i})\,\forall i$.

A \emph{short exact sequence} is $\xymatrix{0\ar[r] & A\ar[r]^{\phi} & B\ar[r]^{\psi} & C\ar[r] & 0}
$. This means:
\begin{enumerate}
\item $\phi$ is injective
\item $\im\phi=\ker\psi$
\item $\psi$ is surjective
\end{enumerate}
A \emph{morphism of complexes} 
\[
\xyC{2pc}\xymatrix{\cdots\ar[r] & A_{i}\ar[r]^{\delta_{i}}\ar[d]_{f_{i}} & A_{i+1}\ar[r]^{\delta_{i+1}}\ar[d]_{f_{i+1}} & A_{i+2}\ar[r]^{\delta_{i+2}}\ar[d]_{f_{i+2}} & \cdots\\
\cdots\ar[r] & B_{i}\ar[r]^{\mu_{i}} & B_{i+1}\ar[r]^{\mu_{i+1}} & B_{i+2}\ar[r]^{\mu_{i+2}} & \cdots
}
\]
is a sequences of maps $f_{i}:A_{i}\to B_{i}$ such that the diagram
\[
\xyC{2pc}\xymatrix{A_{i}\ar[r]^{\delta_{i}}\ar[d]_{f_{i}} & A_{i+1}\ar[d]_{f_{i+1}}\\
B_{i}\ar[r]^{\mu_{i}} & B_{i+1}
}
\]
commutes. If all the $f_{i}$ are isomorphism then the complexes are
\emph{isomorphic}.\end{defn}
\begin{namedthm}[Short 5-lemma]
 Let
\[
\xyC{2pc}\xymatrix{0\ar[r] & A\ar[r]^{\psi}\ar[d]^{\alpha} & B\ar[r]^{\phi}\ar[d]^{\beta} & C\ar[r]\ar[d]^{\gamma} & 0\\
0\ar[r] & A'\ar[r]^{\psi'} & B'\ar[r]^{\phi'} & C'\ar[r] & 0
}
\]
be morphism of short exact sequences of groups.
\begin{itemize}
\item If $\alpha$ and $\gamma$ are injective so is $\beta$.
\item If $\alpha$ and $\gamma$ is surjective so is $\beta$
\item If $\alpha$ and $\gamma$ are isomorphism so is $\beta$
\end{itemize}
\end{namedthm}
\begin{proof}
Suppose that $\alpha$ and $\gamma$ are injective, and $\beta(b)=0$
for some $b\in B'$. Then $\phi'\beta(b)=0=\gamma\phi(b)$. Since
$\gamma$ is injective, $\phi(b)=0$, so $b\in\ker\phi,$ hence there
exists $a\in A$ such that $b=\psi(a)$. So $\beta(b)=\beta\psi(a)=\psi'\alpha(a)=0$.
Since $\psi'$ is injective, $\alpha(a)=0$, but $\alpha$ is also
injective, so $\alpha=0$. So $b=\psi(a)=0$ and thus $\beta$ is
injective. 

Suppose $\alpha$ and $\gamma$ are surjective and consider $b'\in B'$.
Since $\gamma$ is surjective, there exists $c\in C$ with $\gamma(c)=\phi'(b')$.
Since $\phi$ is surjective, there exists $b\in B$ with $\phi(b)=0$,
so $\gamma\phi(b)=\phi'\beta(b)=\phi'(b')$. Thus $\beta(b)=b'\in\ker\phi'=\im\psi'$.
So there exists $a'\in A'$ with $\psi'(a')=\beta(b)-b'$. Since $\alpha$
is surjective, there exists $a\in A$ such that $\alpha(a)=a'$, so
$\psi'\alpha(a)=\beta(b)-b'$. Thus $\beta\psi(a)=\beta(b)-b'$, so
$b'=\beta(b-\psi(a))\in\im\beta$. 
\end{proof}
Question: Given $A,C$ what can you say about $B$ with $\xyC{1.5pc}0\to A\to B\to C\to0$
exact? One obvious answer is $0\to A\overset{a\mapsto(a,0)}{\to}A\oplus C\underset{(a,c)\mapsto c}{\to}C\to0$
is always exact. 
\begin{defn}
Let $R$ be a ring and let $\xyC{1.5pc}\xymatrix{0\ar[r] & A\ar[r]^{\psi} & B\ar[r]^{\phi} & C\ar[r] & 0}
$ be a short exact sequence of $R$-modules. The sequence is said to
\emph{split} (or \emph{be split}) if there exists an $R$-submodule
$D\subseteq B$ such that $B=D+\psi(A)$ with $D\cap\psi(A)=\{0\}$
(i.e., $B\cong D\oplus\psi(A)$). ``There exists $R$-module complement
of $\psi(A)$ in $B$''

Equivalently $\xyC{2pc}\xymatrix{0\ar[r] & A\ar[r]\ar[d]^{\cong} & B\ar[r]\ar[d] & C\ar[r]\ar[d]^{\cong} & 0\\
0\ar[r] & A\ar[r] & A\oplus C\ar[r] & C\ar[r] & 0
}
$ is an isomorphism of complexes.\end{defn}
\begin{lem}
The short exact sequence $\xyC{1.5pc}\xymatrix{0\ar[r] & A\ar[r]^{\psi} & B\ar[r]^{\phi} & C\ar[r] & 0}
$ splits if and only if there exists $\mu:C\to B$ (called a \emph{section})
such that $\phi\circ\mu=\id_{C}$, if and only if there exists $\lambda:B\to A$
such that $\lambda\circ\psi=\id_{A}$\end{lem}
\begin{note*}
If there exists an $R$-module complement $D$ to $\psi(A)$ in $B$,
then $D\cong B/\psi(A)\cong C$
\begin{proof}[Proof of Note]
Consider $\phi:B\to C$. Note that $\phi|_{D}$ is injective, since
$D\cap\psi(A)=\{0)$ (as $\psi(A)=\ker\phi$). Since $\phi$ is surjective,
for any $c\in C$, there exists $b\in B$ with $\phi(b)=c$. Write
$b=d+\psi(a)$ for some $d\in D,a\in A$. But then $\phi(b)=\phi(d+\psi(a))=\phi(d)+\phi(\psi(a))=\phi(d)$,
so $\phi|_{D}$ is surjective.
\end{proof}
\end{note*}
\begin{proof}[Proof of Lemma]
If the sequence splits, there exists an $R$-module complement $D$
for $\phi(A)$. By above, $D\cong C$, so let $\mu:C\to D$ be the
isomorphism $(\mu=\phi^{-1})$. By construction $\phi(\mu(c))=c$
so $\phi\circ\mu=\id_{c}$.

Conversely, if there exists $\mu:C\to B$ such that $\phi\circ\mu=\id_{C}$,
let $D=\mu(C)$. We need to show that $D$ is an $R$-module complement
for $\psi(A)$. Let $b\in D\cap\psi(A)$. Then $b=\mu(c),$ so $\phi(B)=\phi(\mu(c))=c$,
but since $b=\psi(a)$ for some $a$, we have $\phi(b)=\phi(\psi(a))=0$,
so $c=0$ and thus $b=\mu(0)=0$. Given $b\in B$, let $d=\mu(\phi(b))$.
Then $\phi(b-d)=\phi(b-\mu\phi(b))=\phi(b)-\phi\mu\phi(b)=\phi(b)-\phi(b)=0$,
so $b-d\in\ker\phi=\im\psi$. So there exists $a\in A$ such that
$b=d+\phi(a)$, so $B=D+\psi(A)$\end{proof}
\begin{example*}
Given $A$ and letting $C=\zz$, what are the options for $B$ with
$\xyC{1.5pc}\xymatrix{0\ar[r] & A\ar[r]^{\psi} & B\ar[r]^{\phi} & C\ar[r] & 0}
$? In this case, $B\cong A\oplus\zz$, since the map $\mu:C\cong\zz\to B$,
given by $\mu:1\mapsto b$ where $b$ is a fixed choice of $b\in B$
with $\phi(b)=1$, is a splitting. As $\phi\circ\mu(n)=\phi(nb)=n\phi(b)=n\cdot1=n$. 
\end{example*}
Recall that $\hom_{R}(D,-)$ is a functor, $f:A\to B,$$f:\hom_{R}(D,A)\to\hom_{R}(D,B)$
defined by $\phi\mapsto f\circ\phi$.

Given $\xyC{1.5pc}\xymatrix{0\ar[r] & A\ar[r]^{\psi} & B\ar[r]^{\phi} & C\ar[r] & 0}
$, we can apply $\hom_{R}(D,-)$ to it: 
\[
\xyC{1.5pc}\xymatrix{0\ar[r] & \hom_{R}(D,A)\ar[r]^{\psi} & \hom_{R}(D,B)\ar[r]^{\phi} & \hom_{R}(D,C)\ar[r] & 0}
\]
. WARNING: no claims this is a complex yet.

Claim: $\xyC{1.5pc}0\to A\overset{\phi}{\to}B$ is exact, then $0\to\hom_{R}(D,A)\overset{\phi}{\to}\hom_{R}(D,B)$
is exact.
\begin{proof}
Consider $f\in\hom_{R}(D,A)$. If $f\neq0$, then there exists $d\in D$
with $f(d)=a\neq0$. Then $\phi(f)(d)=\phi(f(d))=\phi(a)\neq0$ since
$\phi$ is injective. So $\phi(f)|neq$0 , so $\hom_{R}(D,A)\overset{\phi}{\to}\hom_{R}(D,B)$
is injective.
\end{proof}
Claim: If $0\to A\overset{\psi}{\to}B\overset{\phi}{\to}C$ is exact
then $\hom_{R}(D,A)\overset{\psi}{\to}\hom_{R}(D,B)\overset{\phi}{\to}\hom_{R}(D,C)$
is exact
\begin{proof}
Let $f\in\hom_{R}(D,A)$. Then for all $d\in D$, $\phi\circ\psi(f)(d)=\phi(\psi(f(d))=0$,
so $\phi\circ\psi:\hom_{R}(D,A)\to\hom_{R}(D,C)=0$ (i.e., this is
a complex). Now consider $f\in\hom_{R}(D,B)$ with $\phi(f)=0$. Then
for any $d\in D$, $\phi(f(d))=0$, so $f(d)\in\ker\phi$ , and thus
there exists $a\in A$ with $\phi(a)=f(d)$. The choice of $a$ is
forced since $\psi$ is injective.

Define $g:D\to A$ by $g(d)=a$. We now check this is an $R$-module
homomorphism. Then $\psi g=f$, so $f\in\im\psi$. Suppose $g(d)=a,g(d')=a'$,
then $\psi(a)=f(d),\psi(a')=f(d')$. So $\psi(a+a')=\psi(a)+\psi(a')=f(d)+f(d')=f(d+d')$.
So we must have (since $\psi$ is injective) $g(d+d')=a+a'=g(d)+g(d')$.
(Check $g(rd)=rg(d)$)
\end{proof}
However if $0\to A\to B\to C\to0$ we do not necessarily have $\hom_{R}(D,B)\to\hom_{R}(D,C)\to0$
is exact.
\begin{example*}
$0\to\zz\overset{\times2}{\to}\zz\to\zz/2\zz\to0$, and $D=\zz/2\zz$. \end{example*}
\begin{defn}
We say that $\hom_{R}(D,-)$ is a \emph{left exact functor.}

If $F$ is a covariant functor, $F:R$-module $\to R$-module, then
$F$ is \emph{left exact} if $\xyC{1.5pc}\xymatrix{0\ar[r] & A\ar[r] & B\ar[r] & C\ar[r] & 0}
$ implies $\xyC{1.5pc}\xymatrix{0\ar[r] & F(A)\ar[r] & F(B)\ar[r] & F(C)}
.$

It is \emph{right exact }if $\xyC{1.5pc}\xymatrix{0\ar[r] & A\ar[r] & B\ar[r] & C\ar[r] & 0}
$ implies $\xyC{1.5pc}\xymatrix{F(A)\ar[r] & F(B)\ar[r] & F(C)\ar[r] & 0}
$

Hence it is \emph{exact} if $\xyC{1.5pc}\xymatrix{0\ar[r] & A\ar[r] & B\ar[r] & C\ar[r] & 0}
$ implies $\xyC{1.5pc}\xymatrix{0\ar[r] & F(A)\ar[r] & F(B)\ar[r] & F(C)\ar[r] & 0}
$
\end{defn}
Question: For which $D$ is $\hom_{R}(D,-)$ exact? Let $D=R$ (a
ring with identity) If $\xyC{1.5pc}\xymatrix{0\ar[r] & A\ar[r] & B\ar[r] & C\ar[r] & 0}
$ is exact then $\xyC{1.5pc}\xymatrix{0\ar[r] & \hom_{R}(R,A)\ar[r] & \hom_{R}(R,B)\ar[r] & \hom(R,C)\ar[r] & 0}
$ is exact because $\hom_{R}(R,A)\cong A$ by $\phi\mapsto\phi(1)$
and 
\[
\xyC{2pc}\xymatrix{f(1) & A\ar[r]^{\psi} & B & \psi f(1)\\
f\ar@{|->}[u] & \hom_{R}(R,A)\ar[r]\ar[u] & \hom_{R}(R,A)\ar[u] & \psi f\ar@{|->}[u]\\
 & f\ar@{|->}[r] & \psi f
}
\]
 commutes.
\begin{defn}
An $R$-module $P$ is projective if for any surjection $\phi:M\to N$
of $R$-modules and an $R$-module map $f:P\to N$ such that there
exists $g:P\to M$ such that $f=\phi\circ g$
\[
\xymatrix{ & P\ar[d]^{f}\ar@{-->}[ld]_{g}\\
M\ar@{->>}[r]^{\phi} & N\ar[r] & 0
}
\]
\end{defn}
\begin{example*}
$P=R^{n}$ for some $n$\end{example*}
\begin{prop}
Let $P$ be an $R$-module. The following are equivalent:
\begin{enumerate}
\item $P$ is projective
\item For all short exact sequence $\xyC{1.5pc}\xymatrix{0\ar[r] & A\ar[r]^{\psi} & B\ar[r]^{\phi} & C\ar[r] & 0}
$ we have \\
$\xyC{1.5pc}\xymatrix{0\ar[r] & \hom_{R}(P,A)\ar[r] & \hom_{R}(P,B)\ar[r] & \hom_{R}(P,C)\ar[r] & 0}
$ is exact. (That is $\hom_{R}(P,-)$ is an exact functor)
\item There exist an $R$-module $Q$ and set $S$ such that $P\oplus Q$$\cong\oplus_{S}R$.
``$P$ is a direct summand of a free module''
\end{enumerate}
\end{prop}
\begin{note*}
If $R=\zz,R=k[x_{1},\dots,x_{n}]$ then all projective modules are
free. (The second is Serre's conjecture and proven by Quillen-Suslin)\end{note*}
\begin{proof}
{}
\begin{elabeling}{00.00.0000}
\item [{$1\Rightarrow2$}] Let $\xyC{1.5pc}\xymatrix{0\ar[r] & A\ar[r]^{\psi} & B\ar[r]^{\phi} & C\ar[r] & 0}
$ be a short exact sequence. Then we know \\
$\xyC{1.5pc}\xymatrix{0\ar[r] & \hom_{R}(P,A)\ar[r] & \hom_{R}(P,B)\ar[r] & \hom_{R}(P,C)}
$ is exact. Now given $f\in\hom_{R}(P,C)$ we have 
\[
\xymatrix{ & P\ar[d]^{f}\ar@{-->}[ld]_{g}\\
B\ar@{->>}[r]^{\phi} & C\ar[r] & 0
}
\]
so there exists a unique $g:P\to B$ such that $f=\phi\circ g$, i.e.,
$f=\phi(g)$. Thus $\phi:\hom_{R}(P,B)\to\hom_{R}(P,C)$ is surjective.
\item [{$2\Rightarrow3$}] Suppose $\hom_{R}(P,-)$ is exact. Write $P=\oplus_{s\in S}R/K$
as a quotient of a free module. Then $0\to K\to\oplus R\overset{\pi}{\to}P\to0$
is exact. Since $\hom_{R}(P,\oplus R)\to\hom_{R}(P,P)$ is surjective,
there exists $\mu:P\to\oplus R$ such that $\pi(\mu)=\id:P\to P$,
i.e., for all $p\in P$ we have $\pi\circ\mu(p)=p$. Thus $\oplus R\cong P\oplus K$.
\item [{$3\Rightarrow1$}] Suppose $P\oplus Q=F$ (where $F$ is a free
$R$-module, i.e., $F\cong\oplus_{s\in S}R$, $S$ a set and let $i:S\to F$)
and 
\[
\xyC{1.5pc}\xymatrix{ & P\ar[d]^{f}\\
M\ar@{->>}[r]^{\phi} & N\ar[r] & 0
}
\]
Let $\pi:F\to P$ for the projection map. Then $f\circ\pi\in\hom(F,N)$.
For each $s\in S$, let $n_{s}$ be $f(\pi(i(s))$. Choose $m_{s}\in M$
with $\phi(m_{s})=n_{s}$. By the universal property there exists
a unique $g:F\to M$ such that $\phi\circ g=f\circ\pi$. So we have
the following commutative diagram
\[
\xyC{2pc}\xymatrix{ & F\ar[d]^{\pi}\ar[ddl]_{g}\\
 & P\ar[d]^{f}\\
M\ar@{->>}[r]^{\phi} & N\ar[r] & 0
}
\]
So define $h:P\to M$ by $h(p)=g(p,0)$. Check: This is an $R$-module
homomorphism. Then $\phi(h(p))=\phi(g(p,0))=f(\pi(p,0))=f(p)$. So
$\phi\circ h=f$ and so $P$ is projective.
\end{elabeling}
\end{proof}
Question: What about other functors? For example $\hom(-,D)$ or $A\otimes-$?
\begin{example*}
$0\to\zz\overset{\times2}{\to}\zz\to\zz/2\zz\to0$ and apply $\hom(-,\zz/2\zz)$.
Then applying $\hom(-,\zz/2\zz)$, we get \\
$\xyC{1.5pc}\xymatrix{0 & \zz/2\zz\ar[l] & \zz/2\zz\ar[l] & \zz/2\zz\ar[l] & 0\ar[l]}
$, but this is not exact. To see this note that we must have \\
$\xyC{1.5pc}\xymatrix{ & \zz/2\zz & \zz/2\zz\ar[l]_{0} & \zz/2\zz\ar[l]_{\id} & 0\ar[l]}
$ , showing the failure of surjectivity. \end{example*}
\begin{lem}
Let $\psi:A\to B,$ $\phi:B\to C$ be $R$-module homomorphism. If
$\xyC{1.5pc}\xymatrix{0\ar[r] & \hom(C,D)\ar[r] & \hom(B,C)\ar[r] & \hom(A,D)}
$ is exact for \uline{all }$R$-modules $D$, then $\xyC{1.5pc}\xymatrix{ & A\ar[r]^{\psi} & B\ar[r]^{\phi} & C\ar[r] & 0}
$ is exact.\end{lem}
\begin{proof}
We need to show:
\begin{enumerate}
\item $\phi$ is surjective, use $D=C/\im\phi$


Set $D=C/\phi(B)$, let $\phi_{1}:C\to D$ be the projection map.
Then $\pi_{1}\circ\phi:B\to C/\phi(B)$ is the zero map by construction.
So $\phi(\pi_{1})=0\in\hom(B,D)$. Since $\hom(C,D)\to\hom(B,D)$
is injective, $\pi_{1}=0$, so the projection $C\to C/\phi(B)$ is
the zero map. So $C/\phi(B)=0$ and thus $\phi(B)=C$ so it is surjective.

\item $\im\psi\subseteq\ker\phi$, use $D=C$, $\id:C\to C$


Exercise

\item $\ker(\phi)\subseteq\im\psi$, use $D=B/\im\psi$.


Exercise

\end{enumerate}
\end{proof}
\begin{prop}
Let $\xyC{1.5pc}\xymatrix{0\ar[r] & A\ar[r]^{\psi} & B\ar[r]^{\phi} & C\ar[r] & 0}
$ be an exact sequence of $R$-modules. Then \\
$\xyC{1.5pc}\xymatrix{D\otimes A\ar[r]^{1\otimes\psi} & D\otimes B\ar[r]^{1\otimes\phi} & D\otimes C\ar[r] & 0}
$ is exact.\end{prop}
\begin{proof}
Recall $\hom(F\otimes G,H)\cong\hom(F,\hom(G,H))$. Now by left exactness
of $\hom(-,E)$, for any $E$ we have \\
$\xyC{1.5pc}\xymatrix{0\ar[r] & \hom(C,E)\ar[r] & \hom(B,E)\ar[r] & \hom(A,E)}
$. Then for all $D$ \\
$\xyC{1.5pc}\xymatrix{0\ar[r] & \hom(D,\hom(C,E))\ar[r] & \hom(D,\hom(B,E))\ar[r] & \hom(D,\hom(A,E))}
$

So $\xyC{1.5pc}\xymatrix{0\ar[r] & \hom(D\otimes C,E)\ar[r] & \hom(D\otimes B,E)\ar[r] & \hom(D\otimes A,E))}
$ is exact. So by the lemma $D\otimes A\to D\otimes B\to D\otimes C\to0$
is exact (Check the maps are what you think they are)
\end{proof}
Recall: $M^{\bullet}$ : $\xyC{1.5pc}\xymatrix{\dots\ar[r] & M^{i-1}\ar[r]^{\partial_{i}} & M^{i}\ar[r]^{\partial_{i+1}} & M^{i+1}\ar[r]^{\partial_{1+2}} & M^{i+2}\ar[r] & \dots}
$ is a \emph{complex} (or \emph{cochain complex}) if $\partial_{j+1}\circ\partial_{j}=0$
for all $j$
\begin{defn}
Given a (cochain) complex $M$, the $n^{\mathrm{th}}$ \emph{cohomology}
group is $H^{n}(M)=\ker\partial_{n+1}/\im\partial_{n}$\end{defn}
\begin{notation*}
If $M_{\bullet}$ \foreignlanguage{english}{$\xyC{1.5pc}\xymatrix{\dots\ar[r] & M_{i+1}\ar[r]^{\partial_{i+1}} & M_{i}\ar[r]^{\partial_{i}} & M_{i-1}\ar[r]^{\partial_{i-1}} & M_{i-2}\ar[r] & \dots}
$ is a (chain) \emph{complex}, we write $H_{n}(M)=\ker\partial_{n}/\im\partial_{n+1}$
and call this the $n^{\mbox{th}}$ homology group.}\end{notation*}
\selectlanguage{english}%
\begin{defn}
Let $A$ be an $R$-module. A \emph{projective resolution }of $A$
is an exact sequence \foreignlanguage{british}{$\mathscr{P}$ }
\[
\xyC{1.5pc}\xymatrix{\dots\ar[r] & P_{n}\ar[r]^{\partial_{n}} & P_{n-1}\ar[r]^{\partial_{n-1}} & \dots\ar[r]^{\partial_{1}} & P_{0}\ar[r]^{\epsilon} & A\ar[r] & 0}
\]
such that each $P_{i}$ is a projective module.\end{defn}
\selectlanguage{british}%
\begin{example*}
$0\to\zz\overset{\times2}{\to}\zz\to\zz/2\zz\to0$ is a projective
resolution of the $\zz$-module $\zz/2\zz$\end{example*}
\begin{note*}
For $R$-module, we can actually ask that the $P_{i}$ be \uline{free}\emph{\uline{
}}$R$-modules. These always exists for $R$-modules
\end{note*}
Let $F:R$-modules$\to R$-modules be a covariant right exact functor
or a contravariant left exact functor. Then applying $F$ to $P_{n}\to P_{n-1}\to\dots\to P_{0}\to0$
(forget $A$) gives a complex $F(\mathscr{P})$. Then the $n^{\mbox{th}}$
\emph{derived functor }of $F$ is $H^{n}(F(\mathscr{P}))$ 
\begin{example*}
$F$ is $\hom(-,D)$. Then $F(\mathscr{P})$ is, $\xymatrix{0\ar[r] & \hom(P_{0},D)\ar[r]^{\partial_{1}} & \hom(P_{1},D)\ar[r]^{\partial_{2}} & \hom(P_{2},D)}
$\end{example*}
\begin{defn}
With the above setting $\ext^{n}(A,D)=\ker\partial_{n+1}/\im\partial_{n}$
for $n\geq1$. $\ext^{0}(A,D)=\ker\partial_{1}$\end{defn}
\begin{example*}
$0\to\zz\overset{\times2}{\to}\zz\to\zz/2\zz\to0$, what is $\ext^{n}(\zz/2\zz,\zz/2\zz)$?
$\xyC{1.5pc}\xymatrix{0 & \zz/2\zz\ar[l]_{\partial_{2}} & \zz/2\zz\ar[l]_{0=\partial_{1}} & \zz/2\zz\ar[l]_{\id} & 0\ar[l]}
$. So:
\begin{itemize}
\item $\ext^{0}(\zz/2\zz,\zz/2\zz)=\ker\partial_{1}=\zz/2\zz$
\selectlanguage{english}%
\item $\ext^{1}(\zz/2\zz,\zz/2\zz)=\ker\partial_{2}/\im\partial_{1}=\zz/2\zz$
\item $\ext^{n}(\zz/2\zz,\zz/2\zz)=0$ for $n\geq2$
\end{itemize}
\end{example*}
\begin{thm}
$\ext^{n}(A,D)$ does not depend on the choice of projective resolution\end{thm}
\begin{rem*}
$\ext_{R}^{1}(C,A)$ is in bijection with the equivalence classes
of $B$ such that $\xyC{1.5pc}\xymatrix{0\ar[r] & A\ar[r] & B\ar[r] & C\ar[r] & 0}
$ is exact. ``Extension of $C$ by $A$''\end{rem*}
\begin{example*}
$D\otimes-$. Let $\mathscr{P}$ be a projective resolution of $A$:
$\xyC{1.5pc}\xymatrix{0 & A\ar[l] & P_{0}\ar[l]_{\epsilon} & P_{1}\ar[l]_{\partial_{1}} & P_{2}\ar[l]_{\partial_{2}} & \dots\ar[l]_{\partial_{3}}}
$. Apply $D\otimes-$ to $\mathscr{P}$ $\xyC{1.5pc}\xymatrix{0 & D\otimes P_{0}\ar[l] & D\otimes P_{1}\ar[l]_{1\otimes\partial_{2}} & D\otimes P_{2}\ar[l]_{1\otimes\partial_{2}} & \dots\ar[l]_{1\otimes\partial_{3}}}
$d\end{example*}
\begin{defn}
The $n^{\mathrm{th}}$ derived functor of $D\otimes-$ is called $\tor_{n}^{R}(D,-)$.
So $\tor_{n}^{R}(D,A)=\ker\partial_{n}/\im\partial_{n+1}$ and $\tor_{0}^{R}(D,A)=D\otimes P_{0}/\im\partial$\end{defn}
\begin{example*}
$R=\zz$, $A=\zz/7\zz$ and $\xyC{1.5pc}\xymatrix{0 & A\ar[l] & \zz\ar[l] & \zz\ar[l]_{\times7} & 0\ar[l]}
$ and $D=\zz/7\zz$. So we get\\
 \foreignlanguage{english}{$\xyC{1.5pc}\xymatrix{0 & \zz/7\zz\otimes_{\zz}\zz\ar[l] & \zz/7\zz\otimes_{\zz}\zz\ar[l]_{\partial_{1}=0} & 0\ar[l]}
,$ but $\zz/7\zz\otimes_{\zz}\zz\cong\zz/7\zz$. So}
\selectlanguage{english}%
\begin{itemize}
\item $\tor_{0}^{\zz}(\zz/7\zz,\zz/7\zz)=(\zz/7\zz)/\im\partial_{1}=\zz/7\zz$
\item $\tor_{1}^{\zz}(\zz/7\zz,\zz/7\zz)=(\zz/7\zz)/0=\zz/7\zz$
\end{itemize}
\end{example*}
\selectlanguage{english}%
\begin{rem*}
\selectlanguage{english}%
If $A$ is a $\zz$-module (abelian group) then $A$ is torsion free
if and only if $\tor_{1}(A,B)=0$ for \uline{every} abelian group.\end{rem*}
\selectlanguage{british}%
\begin{defn}
A \emph{short exact sequence of complexes} $\xyC{1.5pc}\xymatrix{0\ar[r] & \mathscr{A}\ar[r]^{\psi} & \mathscr{B}\ar[r]^{\phi} & \mathscr{C}\ar[r] & 0}
$ is a set oh homomorphism of complexes such that $\xyC{1.5pc}\xymatrix{0\ar[r] & A_{n}\ar[r]^{\psi_{n}} & B_{n}\ar[r]^{\phi_{n}} & C_{n}\ar[r] & 0}
$ is exact for every $n$.
\[
\xymatrix{ & 0\ar[d] & 0\ar[d] & 0\ar[d]\\
\dots & A_{n+1}\ar[l]\ar[d]_{\psi_{n+1}} & A_{n}\ar[l]\ar[d]_{\psi_{n}} & A_{n-1}\ar[l]\ar[d]_{\psi_{n-1}} & \ar[l]\dots\\
\dots & B_{n+1}\ar[l]\ar[d]_{\phi_{n+1}} & B_{n}\ar[l]\ar[d]_{\phi_{n}} & B_{n-1}\ar[l]\ar[d]_{\phi_{n-1}} & \dots\ar[l]\\
\dots & C_{n+1}\ar[l]\ar[d] & C_{n}\ar[l]\ar[d] & C_{n-1}\ar[l]\ar[d] & \dots\ar[l]\\
 & 0 & 0 & 0
}
\]
This diagrams commutes, the rows are complexes and the columns are
exacts.\end{defn}
\begin{thm}[Long exact sequence of cohomology]
 Let $\xyC{1.5pc}\xymatrix{0\ar[r] & \mathscr{A}\ar[r]^{\psi} & \mathscr{B}\ar[r]^{\phi} & \mathscr{C}\ar[r] & 0}
$ be a short exact sequence of complexes. Then there is a long exact
sequence 
\[
\xyC{1.3pc}\xymatrix{0\ar[r] & H^{0}(\mathscr{A})\ar[r] & H^{0}(\mathscr{B})\ar[r] & H^{0}(\mathscr{C})\ar[r]^{\partial_{0}} & H^{1}(\mathscr{A})\ar[r] & H^{1}(\mathscr{B)}\ar[r] & H^{1}(\mathscr{C})\ar[r]^{\partial_{1}} & H^{2}(\mathscr{A})\ar[r] & \dots}
\]

\end{thm}
What are the maps? Given
\[
\xymatrix{A_{n-1}\ar[r]^{\partial_{n}}\ar[d]_{\psi_{n-1}} & A_{n}\ar[r]^{\partial_{n+1}}\ar[d]_{\psi_{n}} & A_{n+1}\ar[d]_{\psi_{n+1}}\\
B_{n-1}\ar[r]_{\mu_{n}} & B_{n}\ar[r]_{\mu_{n+1}} & B_{n+1}
}
\]
We want $H^{n}(\mathscr{A})\to H^{n}(\mathscr{B})$. Let $a\in\ker\partial_{n+1}$.
Then $\psi_{n+1}\circ\partial_{n+1}(a)=0$, so $\mu_{n+1}\circ\psi_{n}(a)=0$,
hence $\psi_{n}(a)\in\ker\mu_{n+1}$. We want $\ker\partial_{n+1}/\im\partial_{n}\to\ker\mu_{n+1}/\im\mu_{n}$.
It suffices to check $\psi(\im\partial_{n})\subseteq\im(\mu_{n})$.
If $a\in A_{n-1}$ then $\psi_{n}\circ\partial_{n}(a)=\mu_{n}\circ\psi_{n-1}(a)\im(\mu_{n})$.
So we get a map $H^{n}(\mathscr{A})\to H^{n}(\mathscr{B})$ and similarly
$H^{n}(\mathscr{B})\to H^{n}(\mathscr{C})$

For the other map, we use the Snake Lemma
\begin{namedthm}[Snake Lemma]
 Let 
\[
\xymatrix{ & A\ar[r]^{\psi}\ar[d]_{f} & B\ar[r]^{\phi_{n+1}}\ar[d]_{g} & C\ar[d]_{h}\ar[r] & 0\\
0\ar[r] & A'\ar[r]_{\psi'} & B'\ar[r]_{\phi'} & C'
}
\]
 be a commutative diagram with exact rows. Then there is an exact
sequence
\[
\xyC{1.5pc}\xymatrix{\ker f\ar[r] & \ker g\ar[r] & \ker h\ar[r]^{\partial} & \cok f\cong A'/\im f\ar[r] & \cok g\ar[r] & \cok h}
\]
\end{namedthm}
\begin{proof}
Define $\delta$: $\ker h\to\cok f$. Let $c\in\ker h$. Then there
is $b\in B$ with $\phi(b)=c$ since $\phi$ is surjective. By commutativity
$0=h(c)=h\circ\phi(b)=\phi'\circ g(b)$. So $g(b)\in\ker\psi'$. By
exactness there exists $a'\in A'$ such that $\psi'(a')=g(b)$. Set
$\partial(c)=a'+\im f\in\cok f$.

We need to show that $\partial$ is well defined. Given another choice
$\widetilde{b}$ with $\phi(\widetilde{b})=c$, the difference $b-\widetilde{b}\in\ker\phi=\im\psi$.
So there exists $a\in A$ such that $\psi(a)=b-\widetilde{b}$. But
then $g\psi(a)'=g(b)-g(\widetilde{b})=\psi'f(a)$. So $g(\widetilde{b})=\psi'(a'-f(a))$.
We then would set $\partial(c)=a'-f(a)+\im f=a'+\im f$. 
\end{proof}
\newpage{}


\section{Representation Theory}

Check in this chapter whether each proposition or lemma rely on the
fact that rings need to have $1$.


\subsection{Ring Theory for Representation Theory}

Recall: Let $G$ be a group. The group algebra $R[G]=\left\{ \sum a_{g}g:a_{g}\in R,g\in G\right\} =\oplus_{g\in G}R$.

Representation theory is the study of modules over $R[G]$
\begin{defn}
An $R$-module $M\neq\{0\}$ is \emph{simple} if $M$ has no proper
submodules, i.e., $N\subseteq M$ then $N=M$ or $N=\{0\}$\end{defn}
\begin{prop}
Let $M$ and $M'$ be simple left $R$-modules. Then every $R$-module
homomorphism $f:M\to M'$ is either $0$ or an isomorphism.\end{prop}
\begin{proof}
Suppose $f$ is not zero. Then $\ker(f)$ is a submodule of $M$ that
is not equal to $M$, so it must be the $0$ module, i.e., $f$ is
injective. Similarly $\im f$ is a submodule of $M'$ that is not
the zero module, so $\im f=M'$ and thus $f$ is surjective. So $f$
is an isomorphism.
\end{proof}
A ring $R$ is simple if $_{R}R$ is a simple $R$-module, e.g, $R=k$
a field
\begin{defn}
A left $R$-module is \emph{semi-simple} if it is the direct sum of
simple modules.
\end{defn}

\begin{defn}
A left ideal $I\subseteq R$ is \emph{minimal} if there exists no
left ideal $J$ of $R$ such that $0\subsetneq J\subsetneq I$. \end{defn}
\begin{example*}
$R=M_{n}(k)$. $I=\begin{pmatrix}\begin{matrix}*\\
\vdots\\
*
\end{matrix} & 0\end{pmatrix}$ is a minimal left ideal. \end{example*}
\begin{defn}
A ring $R$ is (left) \emph{semisimple} if it is isomorphic as an
$R$-module to a direct sum of minimal left ideals. (i.e. $_{R}R$
is a semisimple $R$-module)\end{defn}
\begin{example*}
$M_{n}(K)\cong\oplus_{j=0}^{n}I_{j}$ where $I_{j}=\begin{pmatrix}0 & \begin{matrix}*\\
\vdots\\
*\\
j^{\mathrm{th}}\mbox{-column}
\end{matrix} & 0\end{pmatrix}$\end{example*}
\begin{prop}
A left $R$-module $M$ is semisimple if and only if every submodule
of $M$ is a direct summand.\end{prop}
\begin{proof}
$\Rightarrow$) Suppose $M$ is semisimple, so $M\cong\oplus_{j\in J}S_{j}$.
For any subset $I\subseteq J$, define $S_{I}=\oplus_{j\in I}S_{j}$.
Let $B$ be a submodule of $M$. Then by Zorn's lemma, there is $K\subseteq J$
maximal with respect to the property that $S_{K}\cap B=\{0\}$. (Suppose
$K_{1}\subsetneq K_{2}\subsetneq\dots$ with $S_{k_{i}}\cap B=\{0\}$.
Set $K'=\cup K_{i}$ and consider $S_{K'}$. If $b\in S_{K'}\cap B$
then $b\in\oplus_{j\in K'}S_{j}$, so $b=s_{j_{1}}+\dots+s_{j_{r}},$$s_{j_{i}}\in S_{j_{i}}$.
There is $K_{s}$ with $j_{1},\dots,j_{r}\in K_{s}$ , so $B\in S_{K_{s}}\cap B$
which is a contradiction) 

We claim $M=B\oplus S_{K}$, we just need to show that $m\in M\Rightarrow m=b+s_{k}$
for $b\in b,s_{k}\in S_{K}$. If $j\in K$, then $S_{j}\subseteq B+S_{K}$. 

If $j\notin K$, then by maximality, $(S_{K}+S_{j})\cap B\ni b\neq0$.
So there exists $s_{K}\in S_{K},s_{j}\in S_{j}$ such that $s_{K}+s_{j}=b$,
so $s_{j}=b-s_{K}\in S_{j}\cap(B+S_{K})\neq0$ since $s_{j}\neq0$
as $b\notin S_{K}$. Thus all $S_{j}$ are contained in $B+S_{K}$,
so $M\subseteq B+S_{K}$. 

$\Leftarrow$) Suppose every submodule of $M$ is a direct summand.
We first show that every non-zero submodule $B$ of $M$ contains
a simple summand.

Fix $b\neq0$ with $b\in B$. By Zorn's lemma there exists a submodule
$C$ of $B$ maximal with respect to $b\notin C$. If $C=\{0\}$,
then $B=Rb$ and $B$ must be simple (otherwise any proper non-zero
submodule would not contain $b$). 

Otherwise write $M=C\oplus C'$, then $B=C\oplus(\underset{D}{\underbrace{C'\cap B}})$.
(Since $C\cap(C'\cap B)=\{0\}$ and $b\in B\Rightarrow b=c+c'$ for
$c\in C,c'\in C'$, since $b,c\in B$, $c'\in B$.) We claim that
the non-zero submodule $D$ is simple. If not by the above argument
we can write, $D=D'\oplus D''$ where $D'$, $D''$ are non-zero submodules
of $D$. We claim that we do not have $b\in(C\oplus D')\cap(C\oplus D'')$.
If we did, we could write $b=c+d'=c'+d''$ for $c,c'\in C$ and $d'\in D',d''\in D''$.
But then $c-c'=d''-d'\in C\cap D=\{0\}$, so $d'=d''\in D'\cap D''=\{0\}$,
hence $c=c'=b$ contradicting $b\notin C$. But this means one of
$C\oplus D'$ and $C\oplus D''$ does not contain $b$, contradicting
the choice of $C$. Thus $B=C\oplus D$ contains the simple summand
$D$.

We now show that $M$ is semisimple. By Zorn's lemma there is a family
$\left\{ S_{j}:j\in I\right\} $ of simple submodules of $M$ maximal
with respect to the property that the submodule $U$ that they generated
is their direct sum. By hypothesis, $M=U\oplus V$. If $V=\{0\}$,
$M$ is a direct sum of simple modules, so we are done. Otherwise
$V$ has a non-zero simple summand $S$, $V\cong S\oplus V'$. Then
$U\cap S=\{0\}$, so $\sum S_{j}+S=\oplus S_{j}\oplus S$ contradicting
the maximality of $U$. So $V=\{0\}$ and $M$ is the direct sum of
simple submodules.\end{proof}
\begin{namedthm}[Maschke's Theorem]
 If $G$ is a finite group and $k$ a field with $\mbox{char}(K)\nmid|G|$,
then $k[G]$ is semisimple. (i.e. $k[G]$ is a direct sum of simple
$k[G]$-modules)\end{namedthm}
\begin{proof}
It suffices to show that every submodule (ideal) $I$ of $kG$ is
a direct summand. We have 
\[
\xymatrix{0\ar[r] & I\ar@<1ex>[r]^{i} & kG\ar[r]\ar@<1ex>[l]^{\lambda} & \cok\ar[r] & 0}
\]
so it suffices to construct $\lambda:kG\to I$ such that $\lambda\circ i=\id_{I}$.
Since both $kG$ and $I$ are vector space over $k$, there exists
$V\subseteq kG$ such that $kG\cong I\oplus V$ as vector space. Let
$\pi:kG\to I$ be the projection map (it is a linear map). 

Define $\lambda:kG\to kG$ by $\lambda(u)=\frac{1}{|G|}\sum_{g\in G}g\pi(g^{-1}u)$.
Note that $\lambda(u)\in I$, since $\pi(g^{-1}u)\in I$ and $g\pi(g^{-1}u)\in I$
as $I$ is a left ideal. Note also that if $b\in I$ then $\lambda(b)=b$.
Indeed $\lambda(b)=\frac{1}{|G|}\sum_{g\in G}g\pi(g^{-1}b)=\frac{1}{|G|}\sum_{g\in G}gg^{-1}b=\frac{|G|}{|G|}b=b$. 

Finally we check that $\lambda$ is a $kG$-module homomorphism. It
is straightforward to check that $\lambda$ is a $k$-linear map,
since $\pi$ is. Also for $h\in G$, 
\begin{alignat*}{2}
\lambda(hu) & =\frac{1}{|G|}\sum_{g\in G}g\pi(g^{-1}hu)\\
 & =\frac{h}{|G|}\sum_{g\in G}h^{-}g\pi(g^{-1}hu)\\
 & =\frac{h}{|G|}\sum_{g'\in G}g'\pi(g'^{-1}u) &  & \mbox{where }g^{'}=h^{-1}g\\
 & =h\lambda(u)
\end{alignat*}
so $\lambda$ is a $kG$-module homomorphism with $\lambda\circ i=\id_{I}$.
So $I$ is a direct summand. \end{proof}
\begin{example*}
$\cc[\zz/2\zz]=\{a(0)+b(1):a,b\in\cc\}=\underset{=\{a(0)+a(1):a\in\cc\}}{\underbrace{\cc\left((0)+(1)\right)}}\oplus\underset{=\left\{ a(0)-a(1):a\in\cc\right\} }{\underbrace{\cc\left((0)-(1)\right)}}$\end{example*}
\begin{defn}
Let $G$ be a group. A \emph{representation }of $G$ is a group homomorphism,
$\phi:G\to\GL(V)$ where $V$ is a vector space. It is \emph{finite
dimensional} if $V$ is a finite dimensional vector space. $V$ is
a \emph{simple} $kG$-module if $V$ has no $G$-invariant subspace.
\end{defn}
Point: If $V$ is a vector space over $k$, then $V$ is a $kG$-module,
via $g\cdot v=\phi(g)\cdot v$. 
\begin{example*}
Let $G=S_{3}$ and $\phi:S_{3}\to\GL_{3}(\cc)$ send a permutation
to its permutation matrix. $\phi((1,2))=\begin{pmatrix}0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 1
\end{pmatrix}$ and $\phi((1,2,3))=\begin{pmatrix}0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0
\end{pmatrix}$. This makes $\cc^{3}$ into a $\cc[S_{3}]$-module. Is it simple?
The answer is no because we notice that $\begin{pmatrix}1\\
1\\
1
\end{pmatrix}$ is a common subspace to both matrix. So we have $\cc^{3}\cong_{\cc[S_{3}]}\mbox{span}\begin{pmatrix}1\\
1\\
1
\end{pmatrix}\oplus V$, where $V$ is a 2-dimensional submodule. In fact $V=\mbox{span}\left(\begin{pmatrix}0\\
1\\
-1
\end{pmatrix},\begin{pmatrix}1\\
0\\
-1
\end{pmatrix}\right)$. In the basis $\begin{pmatrix}1\\
1\\
1
\end{pmatrix},\begin{pmatrix}0\\
1\\
-1
\end{pmatrix},\begin{pmatrix}1\\
0\\
-1
\end{pmatrix}$ for $V$ we have $(1,2)\to\begin{pmatrix}1 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0
\end{pmatrix}$, $(1,2,3)\to\begin{pmatrix}1 & 0 & 0\\
0 & 0 & 1\\
0 & -1 & -1
\end{pmatrix}$.
\end{example*}

\begin{example*}
$\cc[\zz/2\zz]$, $\ff_{2}[\zz/2\zz]$ and let $V=\cc((0)+(1))$ and
$W=\ff_{2}((0)+(1))$. Maschke's theorem tells us that $\cc[\zz/2\zz]$
is a direct summand (see previous example) but nothing about $\ff_{2}[\zz/2\zz]$.
In fact we can not write $\ff_{2}[\zz/2\zz]\cong W\oplus?$. \end{example*}
\begin{prop}
{} 
\begin{enumerate}
\item Every submodule and every quotient of a semisimple module is semisimple
\item If $R$ is semisimple, then every left $R$-module $M$ is semisimple.
\end{enumerate}
\end{prop}
\begin{proof}
{}
\begin{enumerate}
\item Let $B$ be a submodule of $M$. Every submodule $C$ of $B$ is a
submodule of $M$, so $M\cong C\oplus D$ for some $D$. Let $\pi:M\to C$
be the projection map and let $\lambda:B\to C$ be given by $\lambda=\pi|_{B}$.
Then 
\[
\xymatrix{0\ar[r] & C\ar@<1ex>[r]^{i} & B\ar[r]\ar@<1ex>[l]^{\lambda} & \cok\ar[r] & 0}
\]
so $B\cong C\oplus\cok$. So every submodule of $B$ is a direct summand
so $B$ is semisimple.


Let $M/H$ be a quotient of $M$. Since $M$ is semisimple we have
$M\cong H\oplus H'$ for some submodule $H'$. By the first part $H'$
is semisimple so $M/H\cong H'$ is semisimple. 

\item Suppose $R$ is semisimple. Then any free $R$-module is semisimple.
($R\cong\oplus M_{i}$ so $\oplus R\cong\oplus\oplus M_{i}$ ) But
every $R$-module is a quotient of a free module, so every $R$-module
is semisimple.
\end{enumerate}
\end{proof}
\begin{cor}
Let $G$ be a finite group and $k$ a field with $\mbox{char}k\nmid|G|$.
Then every $kG$-module is a direct sum of simple $kG$-modules, so
every representation is a direct sum of irreducible representation.\end{cor}
\begin{prop}
Let $R\cong_{R}\oplus_{i\in I}M_{i}$ be a semisimple ring, where
the $M_{i}$ are simple modules and let $B$ be a simple $R$-module.
Then $B\cong M_{i}$ for some $i$. \end{prop}
\begin{proof}
We have $0\neq B\cong\Hom_{R}(R,B)\cong\oplus_{i\in I}\Hom_{R}(M_{i},B)$.
However by Schur's Lemma $\Hom_{R}(M_{i},B)=0$ unless $M_{i}\cong B$. \end{proof}
\begin{cor}
Let $G$ be a finite group and $k$ a field with $\mbox{char}k\nmid|G|$.
Then there are only a finite number of simple $kG$-modules up to
isomorphism, and thus only a finite number of irreducible representation
of $G$.\end{cor}
\begin{example*}
Let $G=S_{3}$. 
\begin{itemize}
\item $\phi_{1}:G\to\cc^{*}$, $\phi_{1}(g)=1$ for all $g$. This corresponds
to the $\cc[S_{3}]$ submodule $\cc(\sum_{g\in S_{3}}g)$. 
\item $\phi_{2}:G\to\cc^{*}$, $\phi_{2}(g)=\sgn(g)=\begin{cases}
1 & g\mbox{ is even}\\
-1 & g\mbox{ is odd }
\end{cases}$. This corresponds to the $\cc[S_{3}]$ submodule $\cc\left(\sum_{g\in S_{3}}\sgn(g)g\right)$.
\item $\phi_{3}:G\to\GL_{2}(\cc)$, $\phi_{3}((1,2))=\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}$ and $\phi_{3}((1,2,3))=\begin{pmatrix}0 & -1\\
1 & -1
\end{pmatrix}$. 


Exercise:
\begin{enumerate}[label=\alph{enumi}]
\item  Check that this is an irreducible representation
\item Find a two dimensional submodule of $\cc[S_{3}]$ that this is isomorphic
to.
\end{enumerate}
\end{itemize}
So $\cc[S_{3}]\cong\underset{\cong\phi_{1}}{\cc}\oplus\underset{\cong\phi_{2}}{\cc}\oplus\underset{\cong\phi_{3}}{\cc^{2}}\oplus\underset{\cong\phi_{3}}{\cc^{2}}$
\end{example*}
Question: What are the possibilities for semisimple rings?

e.g.: $k[G]$, $G$ finite, good characteristic. $M_{n}(k)$. $M_{n}(D)$
where $D$ is a division ring. From these we can create more for example
$M_{n_{1}}(D_{1})\times\dots\times M_{n_{r}}(D_{r})$. 
\begin{thm}[Wedderburn-Artin]
 A ring $R$ (with 1) is semisimple if and only if $R$ is isomorphic
to a direct sum/product of matrix rings over division rings. $R\cong M_{n_{1}}(D_{1})\times\dots\times M_{n_{r}}(D_{r})$.
The $n_{i}$ , $D_{i}$ are unique up to permutation.\end{thm}
\begin{proof}
We've just discuss ``if''

Suppose $R$ is semisimple, so $R\cong_{R}\oplus_{i\in I}M_{i}$.
We first note that $|I|<\infty$, since $1\in R$, $1=m_{i_{1}}+\dots+m_{i_{s}}$
for some $m_{i_{j}}\in M_{i_{j}}$. So $R=R1\subseteq M_{i_{1}}\oplus\dots\oplus M_{i_{s}}\subseteq R$,
so we have equality. After reordering, we may assume that $M_{i}\ncong M_{j}$
for $i\neq j$, $1\leq i,j\leq r$ and for all $j<r$ there exists
$i\leq r$ with $M_{j}\cong M_{i}$. Write $B_{i}=\oplus_{M_{j}\cong M_{i}}M_{j}$,
so $R\cong B_{1}\oplus\dots\oplus B_{r}$.

We have $R^{\mathrm{op}}\cong\Hom_{R}(R,R)$ (as a ring) with the
map $\xyC{1pc}\xymatrix{f(1) & f\ar@{|->}[l]}
,$ then $f(r)=f(r\cdot1)=rf(1)$ and \foreignlanguage{english}{}\\
\foreignlanguage{english}{$\xyC{1pc}\xymatrix{f\circ g(1)=f(g(1))=g(1)f(1) & f\circ g\ar@{|->}[l]}
$ So $R^{\mathrm{op}}\cong\Hom_{R}(R,R)\cong\Hom_{R}(\oplus_{i=1}^{r}B_{i},\oplus_{i=1}^{r}B_{i})\cong\oplus_{i,j=1}^{r}\Hom_{R}(B_{i},B_{j})$.
Now $\Hom_{R}(B_{i},B_{j})=\Hom_{R}(\oplus_{l=1}^{n_{i}}M_{l},\oplus_{k=1}^{n_{j}}M_{k})=\oplus_{l=1}^{n_{i}}\oplus_{k=1}^{n_{j}}\Hom_{R}(M_{i},M_{j})=0$
if $i\neq j$ by Schur's Lemma. Since every non-zero function in $\Hom_{R}(M_{i},M_{i})$
is an isomorphism by Schur's lemma $\Hom_{R}(M_{i},M_{i})$ is a division
ring with multiplication being function composition. Call this $D_{i}^{\mathrm{op}}$.
Then $\Hom_{R}(B_{i},B_{i})\cong\oplus_{k,l}^{n_{i}}D_{i}^{\mathrm{op}}\underset{\mathrm{check}}{\cong}M_{n_{i}}(D_{i}^{\mathrm{op}})$.
So $R^{\mathrm{op}}\cong M_{n_{1}}(D_{1}^{\mathrm{op}})\times\dots\times M_{n_{r}}(D_{r}^{\mathrm{op}})$,
hence $R\cong M_{n_{1}}(D_{1}^{\mathrm{op}})^{\mathrm{op}}\times\dots\times M_{n_{r}}(D_{r}^{\mathrm{op}})^{\mathrm{op}}=M_{n_{1}}(D_{1})\times\dots\times M_{n_{r}}(D_{r})$. }

\selectlanguage{english}%
Proof omits uniqueness.\end{proof}
\begin{xca*}
$M_{n}(D^{\mathrm{op}})^{\mathrm{op}}\cong M_{n}(D)$\end{xca*}
\begin{cor}[Molien]
 If $G$ is a finite group, and $k$ is algebraically closed, with
$\mbox{char}k\nmid|G|$, then $k[G]\cong M_{n_{1}}(k)\times\dots\times M_{n_{r}}(k)$
and thus $\sum n_{i}^{2}=|G|$.\end{cor}
\begin{example*}
$\cc[\zz/3\zz]\cong M_{n_{1}}(\cc)\times\dots\times M_{n_{r}}(\cc)$.
Now $3=n_{1}^{2}+\dots+n_{r}^{2}$ implies $r=3$ and $n_{1}=n_{2}=n_{3}=1$.
So $\cc[\zz/3\zz]\cong\cc\times\cc\times\cc$. 

Let us look at the irreducible representation. We always have the
``trivial representation'', $\phi_{1}:\zz/3\zz\to\cc^{*}$ defined
by $\phi_{1}(g)=1$ for all $g$.

We then have $\phi_{2}((0))=1,\phi_{2}((1))=\omega$ and $\phi_{2}((2))=\omega^{2}$
where $\omega=e^{\frac{2\pi i}{3}}$, similarly we also get $\phi_{3}((0))=1,$
$\phi_{3}((1))=\omega^{2}$ and $\phi_{3}((2))=\omega$

So then $\cc[\zz/3\zz]\cong\underset{\phi_{1}}{\underbrace{\cc((0)+(1)+(2))}}\times\underset{\phi_{2}}{\underbrace{\cc((0)+\omega^{2}(1)+\omega(2))}}\times\underset{\phi_{3}}{\underbrace{\cc((0)+\omega(1)+\omega^{2}(2))}}$
. Check that this is a ring isomorphism e.g. $((0)+(1)+(2))((0)+\omega(1)+\omega^{2}(2))=0$. \end{example*}
\begin{proof}[Proof of Corollary]
 By Maschke's theorem $k[G]$ is semisimple, so by Wedderburn-Artin
theorem $k[G]\cong M_{n_{1}}(D_{1})\times\dots\times M_{n_{r}}(D_{r})$
where $D_{i}\cong\Hom_{KG}(M_{i},M_{i})^{\mathrm{op}}$ for simple
$kG$-module $M_{i}$. First note that $k\subseteq\Hom(M_{i},M_{i})^{\mathrm{op}}$,
for $a\in k$, $u\in M_{i}$ we set $a(u)=au$. Then $a(gu)=agu=g(au)$
so this is $kG$-homomorphism. Consider any $f\in D_{i}^{\mathrm{op}}$,
since $f$ is a $kG$-homomorphism it is a linear transformation,
so $f(au)=af(u)$, i.e., $(af)(u)=(fa)(u)$, so $f$ commutes with
any $a\in k$. Let $k(f)$ be the smallest sub division ring of $D_{i}^{\mathrm{op}}$
that contains $k$ and $f$. The division ring $k(f)$ is a finite
dimensional vector space over $k$. 

Thus $1,f,f^{2},f^{3},\dots$ are linearly dependent over $k$, so
there exists $g\in k[X]$ with $g(f)=0$. Take $g$ with minimal degree.
But then $\{a_{0}+a_{1}f+\dots+a_{r}f^{\deg(g)-1}:a_{i}\in k\}$ is
closed under addition, multiplication. Also $g$ is an irreducible
polynomial, since otherwise $g=g_{1}g_{2}$ would imply $\underset{\neq0}{\underbrace{g_{1}(f)}}\underset{\neq0}{\underbrace{g_{2}(f)}}=0$
in the division ring $D_{i}^{\mathrm{op}}$. We show that this is
closed under division. Given $h=\sum a_{i}f^{i}$, the elements $1,h,h^{2},h^{3},\dots$
are linearly dependant over $k$. So there exists $b_{i}\in k$ with
$\sum_{i=j\geq0}^{s}b_{i}h^{i}=0$, where we may assume that $b_{j}=1$,
then $\frac{1}{h}=-\sum_{i=j+1}^{s}b_{i}h^{i-j-1}$ and this can be
written as $\sum_{i=0}^{r}c_{i}f^{i}$. Then the multiplication in
$k(f)$ is commutative (since $k$ commutes with $f$), so $k(f)$
is a field containing $k$. Since $f$ is algebraic over the algebraically
closed field $k$, $f\in k$. 
\end{proof}
Question: We now have (for good $k$) $kG\cong M_{n_{1}}(k)\times\dots\times M_{n_{r}}(k)$.
What is $r$?

Answer: It is the number of conjugacy class of $G$

Recall: A \emph{conjugacy class} of a group $G$ is a set $C_{h}=\{ghg^{-1}:g\in G\}$
of all conjugates of an element of $h$. The \emph{class sum} corresponding
to $C_{h}$ is $z_{h}=\sum_{g'\in C_{h}}g'$. The \emph{centre }of
a ring $R$ is $Z(R)=\{a\in R:ab=ba\,\forall b\in R\}$. e.g. The
centre of $M_{n}(k)$ is $\{\lambda I:\lambda\in k\}$
\begin{lem}
Let $G$ be a finite group. Then the class sum $z_{h}$ form a $k$-basis
for $Z(k[G])$\end{lem}
\begin{proof}
First consider $z_{h}=\sum_{g'=ghg^{-1}}g'\in k[G]$. For any $\widetilde{g}\in G$
we have 
\[
\widetilde{g}z_{h}=\sum_{g'=ghg^{-1}}\widetilde{g}g'=\sum_{g'=ghg^{-1}}(\widetilde{g}g)g(g^{-1}\widetilde{g}^{-1})\widetilde{g}=z_{h}\widetilde{g}
\]
 since if $g_{1}\neq g_{2}\in C_{h}$ then $\widetilde{g}g_{1}\widetilde{g}\neq\widetilde{g}g_{2}\widetilde{g}$.
Hence $z_{h}\in Z(K[G])$

Now suppose $z\in\sum a_{g}g\in Z(k[G])$. Then for all $\widetilde{g}\in G$,
$\widetilde{g}z\widetilde{g}^{-1}=\sum a_{g}\widetilde{g}g\widetilde{g}^{-1}=\sum a_{g}g$,
so $a_{\widetilde{g}g\widetilde{g}^{-1}}=a_{g}$ and thus the coefficients
of $z$ are constant on conjugacy classes. So $z$ is a linear combination
of class sums.\end{proof}
\begin{cor}
Let $G$ be a finite group and $k$ a field with $k=\overline{k}$
and $\mbox{char}K\nmid|G|$. Then $kG\cong M_{n_{1}}(k)\times\dots\times M_{n_{r}}(k)$
where $r=$ number of conjugacy class of $G$.\end{cor}
\begin{proof}
The centre of $M_{n_{1}}(k)\times\dots\times M_{n_{r}}(k)$ has dimension
$r$ over $k$, so $r=$ number of conjugacy classes.\end{proof}
\begin{defn}
Let $\phi:G\to\GL(V)$ be a representation of $G$. The \emph{character}
of $\phi$ is $\chi_{\phi}:G\to k$, $\chi_{\phi}(g)=\Tr\phi(g)$.
(Note $\Tr(A)=\sum a_{ii}$)
\end{defn}
Warning: This is \uline{not} a group homomorphism unless $\dim V=1$. 
\begin{note*}
$\chi(ghg^{-1})=\Tr\phi(ghg^{-1})=\Tr(\phi(g)\phi(h)\phi(g)^{-1})=\Tr(\phi(h)\phi(g)\phi(g)^{-1})=\Tr\phi(h)=\chi_{\phi}(h)$,
so characters are constant on conjugacy classes.\end{note*}
\begin{defn}
The \emph{character table} of a finite group $G$ is the $r\times r$
table (where $r$ is the number of conjugacy classes) with columns
indexed by conjugacy classes and rows indexed by irreducible representation
recording the character.\end{defn}
\begin{example*}
{}
\begin{elabeling}{00.00.0000}
\item [{$G=S_{3}$}]~
\item [{~}] %
\begin{tabular}{c|c|c|c}
 & $(1)$ & $(1,2),(1,3),(2,3)$ & $(1,2,3),(1,3,2)$\tabularnewline
\hline 
$\phi_{1}$ & $1$ & $1$ & $1$\tabularnewline
\hline 
$\phi_{2}$ & $1$ & $-1$ & $1$\tabularnewline
\hline 
$\phi_{3}$ & $2$ & $0$ & $-1$\tabularnewline
\end{tabular}
\item [{$G=\zz/3\zz$}]~
\item [{~}] %
\begin{tabular}{c|c|c|c}
 & $(0)$ & $(1)$ & $(2)$\tabularnewline
\hline 
$1$ & $1$ & $1$ & $1$\tabularnewline
\hline 
$\omega$ & $1$ & $\omega$ & $\omega^{2}$\tabularnewline
\hline 
$\omega^{2}$ & $1$ & $\omega^{2}$ & $\omega$\tabularnewline
\end{tabular}
\end{elabeling}
\newpage{}
\end{example*}

\section{Galois Theory}
\begin{defn}
A \emph{field extension $L$ }of a field $K$ is a field $L$ containing
$K$. We'll write $L/K$ or $L:K$. Given a subset $X$ of $L$ the
intersection of all subfields of $L$ containing $K$ and $X$ is
denoted $K(X)$.\end{defn}
\begin{example*}
$K=\qq,L=\rr$ and $X=\{\sqrt{2}\}$ then $K(\sqrt{2})=\{a+b\sqrt{2}:a\in\qq\}$.

$X=\pi$, $K(\pi)=$ set of all rational functions of $\pi$\end{example*}
\begin{defn}
An extension field $L/K$ is \emph{simple} if $L=K(\alpha)$ for some
$\alpha\in L$\end{defn}
\begin{example*}
$L=\qq(i,\sqrt{5})=\qq(i+\sqrt{5})$. The inclusion one way is clear.
For the other way notice that $(i+\sqrt{5})^{2}=4+2\sqrt{5}i\in L\Rightarrow\sqrt{5}i\in L$.
Also $-\sqrt{5}+5i\in L\Rightarrow6i\in L$ so $i\in L$.\end{example*}
\begin{defn}
An element $\alpha\in L$ is \emph{algebraic} over $K$ if there exists
a \uline{monic} polynomial $g\in K[x]$ with $g(\alpha)=0$. The
$g$ of lowest degree is called the \emph{minimal polynomial}. If
$\alpha$ is not algebraic, it is said to be \emph{transcendental. }\end{defn}
\begin{example*}
$\overline{\qq}=$ the algebraic closure of $\qq=$ the set of all
algebraic number over $\qq$. This is countable. (So transcendental
elements of $\cc$ exists)\end{example*}
\begin{defn}
An extension $L/K$ is algebraic if every element of $L$ is algebraic.
\end{defn}
In general if $\alpha$ is algebraic over $\qq$ with a minimal polynomial
$f$ of degree $d$ and $\beta$ is algebraic over $\qq$ with a minimal
polynomial $g$ of degree $e$, what can you say about $\alpha+\beta$?
\begin{defn}
The \emph{degree }of $L/K$ written $[L:K]$ is the dimension of $L$
as a vector space over $K$.\end{defn}
\begin{note*}
If $L=K(\alpha)$ for $\alpha$ algebraic with minimal polynomial
$g$ then $[L:K]=\deg g$ since $\{1,\alpha,\alpha^{2},\dots,\alpha^{\deg g-1}\}$
is a basis. If $\alpha$ is transcendental then $L\cong K(t)$ and
$[L:K]=\infty$ (define $\phi:K(t)\to K(\alpha),t\mapsto\alpha$)\end{note*}
\begin{namedthm}[The Tower Law]
Let $K,L,M$ be fields with $K\subseteq L\subseteq M$. Then $[M:K]=[M:L][L:K]$\end{namedthm}
\begin{proof}
Let $\{x_{\alpha}:\alpha\in I\}$ be a basis for $L/K$ and let $\{y_{\beta}:\beta:J\}$
be a basis for $M/L$. Define $z_{\alpha\beta}=x_{\alpha}y_{\beta}\in M$.
We claim that $\{z_{\alpha\beta}\}$ is a basis for $M/K$.

We show that they are linearly independent. If $\sum_{\alpha,\beta}a_{\alpha\beta}z_{\alpha\beta}=0$
with finitely many $a_{\alpha\beta}\in K$ non-zero. Then $\sum_{\beta}(\sum_{\alpha}a_{\alpha\beta}x_{\alpha})y_{\beta}=0$,
since the $y_{\beta}$ are linearly independent over $L$ we have
$\sum_{\alpha}a_{\alpha\beta}x_{\alpha}=0$ for all \textbf{$\beta$}.
Since the $x_{\alpha}$ are linearly independent over $K$ we have
$a_{\alpha\beta}=0$ for all $\alpha,\beta$.

We show spanning. If $z\in M$, then $z=\sum\lambda_{\beta}y_{\beta}$
for $\lambda_{\beta}\in L$. For each $\lambda_{\beta}=\sum a_{\alpha\beta}x_{\alpha}$.
So $x=\sum_{\beta}(\sum_{\alpha}a_{\alpha\beta}x_{\alpha})y_{\beta}=\sum_{\alpha,\beta}a_{\alpha\beta}x_{\alpha}y_{\beta}=\sum a_{\alpha\beta}x_{\alpha\beta}$.

So $\{z_{\alpha\beta}\}$ is a basis for $M$ over $K$, so $[M:K]=[M:L][L:K]$\end{proof}
\begin{example*}
$[\qq(i,\sqrt{5}):\qq]=[\qq(i,\sqrt{5}):\qq(i)][\qq(i):\qq]=2\times2=4$.
The minimal polynomial of $i+\sqrt{5}$ over $\qq$ is $x^{4}-8x^{2}+36$.
(Note that this is \uline{not} $(x^{2}+1)(x^{2}-5)$)\end{example*}
\begin{defn}
An \emph{automorphism }of $L$ is a field isomorphism $\phi:L\to L$
(so $\phi(0)=0$ and $\phi(1)=1$). We say $\phi$ \emph{fixes} $K$
if $\phi(a)=a$ for all $a\in K$.\end{defn}
\begin{example*}
$\phi:\cc\to\cc$. $\phi(a+bi)=a-bi$ complex conjugation. 

$\phi:\qq(\sqrt{5},i)\to\qq(\sqrt{5},i)$ defined by $\phi(a+b\sqrt{5}+ci+d\sqrt{5}i)=a-b\sqrt{5}+ci-d\sqrt{5}i$.
Note $\phi$ fixes $\qq(i)$ but not $\qq(\sqrt{5})$. \end{example*}
\begin{defn}
The Galois group $\mbox{Gal}(L/K)$ of $L/K$ is the group of all
automorphisms of $L$ fixing $K$.\end{defn}
\begin{example*}
Using the $\phi$ defined in the second part of the previous example,
we have $\phi\in\mbox{Gal}(\qq(\sqrt{5},i)/\qq(i))$ but not in $\mbox{Gal}(\qq(\sqrt{5},i)/\qq(\sqrt{5}))$.

$\mbox{Gal}(\cc/\rr)\cong\zz/2\zz$ (generated by complex conjugation)
(Because $\phi(a+bi)=a+b\phi(i)$ and $\phi(i)^{2}=\phi(-1)=-1$)
\end{example*}
Note that $\mbox{Gal}(L/K)$ is a group under function composition.
$\phi:L\to L,\psi:L\to L$, $\phi(a)=\psi(a)=a$ for $a\in K$. $\phi\circ\psi:L\to L$
is an isomorphism and $\phi\psi(a)=\phi(\psi(a))=\phi(a)=a$ for $a\in K$
\begin{example*}
$\mbox{Gal}(\qq(\sqrt{5},i)/\qq)=\zz/2\zz\times\zz/2\zz$. 

$\mbox{Gal}(\qq(\sqrt[3]{2})/\qq)=1$, $[\qq(\sqrt[3]{2}):\qq]=3$\end{example*}
\begin{defn}
For a subgroup $H$ of $\mbox{Gal}(L/K)$ we denote by $L^{H}$ the
set $L^{H}=\{\alpha\in L:\phi(\alpha)=\alpha$ for all $\alpha\in H\}$.
This is a subfield of $L$ called the \emph{fixed field of $H$}\end{defn}
\begin{example*}
$H=\mbox{Gal}(\cc/\rr)$, $\cc^{H}=\rr$.

$\mbox{Gal}(\qq(\sqrt{5},i)/\qq)=\left\langle \phi_{1},\phi_{2}\right\rangle $
where $\phi_{1}(i)=-i,\phi_{1}(\sqrt{5})=\sqrt{5}$ and $\phi_{2}(i)=i,\phi_{2}(\sqrt{5})=-\sqrt{5}$.
Let $H_{i}=\left\langle \phi_{i}\right\rangle $. Then $\qq(\sqrt{5},i)^{H_{1}}=\qq(i),\qq(\sqrt{5},i)^{H_{2}}=\qq(\sqrt{5}),\qq(\sqrt{5},i)^{\mbox{Gal}}=\qq$.
Define $H_{3}=\left\langle \phi_{3}\right\rangle $, where $\phi_{3}(i)=-i$
and $\phi_{3}(\sqrt{5})=-\sqrt{5}$. 

\[
\xymatrix{ & \mbox{Gal}\ar@{-}[dr]\ar@{-}[dl]\ar@{-}[d]\\
H_{1}\ar@{-}[dr] & H_{3}\ar@{-}[d] & H_{2}\ar@{-}[dl]\\
 & 1
}
\qquad\xyR{3.5pc}\xyC{1pc}\xymatrix{ & \qq(\sqrt{5},i)=\qq(\sqrt{5},i)^{1}\ar@{-}[dr]\ar@{-}[d]\ar@{-}[dl]\\
\qq(\sqrt{5})=\qq(\sqrt{5},i)^{H_{2}}\ar@{-}[dr] & \qq(i\sqrt{5})=\qq(\sqrt{5},i)^{H_{3}}\ar@{-}[d] & \qq(i)=\qq(\sqrt{5},i)^{H_{1}}\ar@{-}[dl]\\
 & \qq=\qq(\sqrt{5},i)^{\mbox{Gal}}
}
\]
\end{example*}
\begin{note*}
For any subgroup $H$ of $\mbox{Gal}(L/K)$ we have $K\subseteq L^{H}\subseteq L$
and $H\leq\mbox{Gal}(L/L^{H})\leq\mbox{Gal}(L/K)$\end{note*}
\begin{defn}
A polynomial $f\in K[x]$ splits over $K$ if $f=a\prod_{i=1}^{d}(x-b_{i})$,
$a,b_{1},\dots,b_{d}\in K$\end{defn}
\begin{example*}
e$f=x^{3}-2$ splits over $\cc$ Note $f=(x-\sqrt[3]{2})(x-\sqrt[3]{2}\omega)(x-\sqrt[3]{2}\omega^{2})$
where $\omega=e^{\frac{2\pi i}{3}}$. So we see that $f$ does not
split over $\qq(\sqrt[3]{2})$\end{example*}
\begin{defn}
A field $L$ is a splitting field for a polynomial $f\in K[x]$ if
$K\subseteq L$ and 
\begin{enumerate}
\item $f$ splits over $L$
\item If $K\subseteq M\subseteq L$ and splits over $M$ then $M=L$
\end{enumerate}
(Equivalently $L=K(\sigma_{1},\dots,\sigma_{d})$ where $\sigma_{1},\dots,\sigma_{d}$
are the roots of $f$ in $L$)
\end{defn}
These always exist, and are unique up to isomorphism. The proof uses
induction on $\deg f$, where we use the intermediate field $M=K[x]/(f)$.
\begin{example*}
$\qq(\sqrt[3]{2},\sqrt[3]{2}\omega,\sqrt[3]{2}\omega^{2})=\qq(\sqrt[3]{2},\omega)$
is a splitting field for $f=x^{3}-2$ (where $\omega=e^{\frac{2\pi i}{3}}$)\end{example*}
\begin{defn}
An extension $L/K$ is \emph{normal} if every irreducible polynomial
$f$ over $K$ which has at least one root in $L$ splits over $L$. \end{defn}
\begin{example*}
$\cc/\rr$ is normal

$\qq(\sqrt[3]{2})/\qq$ is not normal.\end{example*}
\begin{defn}
An irreducible polynomial $f\in K[x]$ is \emph{separable} over $K$
if it has no multiple zeros in a splitting field, (i.e, the $b_{i}$
are distinct). Otherwise it is \emph{inseparable}\end{defn}
\begin{example*}
$x^{4}+x^{3}+x^{2}+x+1$ is separable, its roots are $\omega^{j}$,
$j=1,\dots,4$ where $\omega^{5}=1$

$K=\ff_{2}(x)$, $f(t)=t^{2}+x$ is inseparable. $K\subseteq L$ where
$y\in L$ satisfies $f(y)=0$. $f(y)=y^{2}+x=0$, $x=y^{2}$, so $f=t^{2}+y^{2}=(t+y)^{2}$\end{example*}
\begin{prop}
If $K$ is a field of characteristic $0$, then every irreducible
polynomial is separable over $K$.

If $K$ has characteristic $p>0$, then $f$ is separable unless $f=g(x^{p})$.
\end{prop}
Recall: A polynomial $f\in K[x]$ has a double root if and only if
$f$ and $f'$ (the formal derivative) have a common factor. If $f$
had a double root and $f'\neq0$, $f$ and $f'$ would have a common
factor in $K[x]$ (by the Euclidean algorithm). But since $\deg(f')<\deg(f)$,
this factor is not $f$, contradicting $f$ being irreducible, unless
$f'=0$. We only have $f'=0$ if $\mbox{char}K=p$ and $f=g(x^{p})$. 
\begin{defn}
An algebraic extension $L/K$ is \emph{separable} if for $\alpha\in L$,
its minimal polynomial is separable over $K$.\end{defn}
\begin{thm}[Fundamental Theorem of Galois Theory]
Let $L/K$ be a finite separable normal field extension with $[L:K]=n$
and $\mbox{Gal}(L/K)=G$ then
\begin{enumerate}
\item $|G|=n$
\item For $K\subseteq M\subseteq L$ we have $M=L^{\mathrm{Gal}(L/M)}$
and $H=\mathrm{Gal}(L/L^{H})$, so $H\mapsto L^{H}$ is an order reversing
bijection between the poset of subgroups of $G$ and subfields $K\subseteq M\subseteq L$
\item If $K\subseteq M\subseteq L$ then $[L:M]=|\mbox{Gal}(L/M)|$ and
$[M:K]=|G|/|\mbox{Gal}(L/M)|$
\item $M/K$ is normal if and only if $\mbox{Gal}(L/M)$ is a normal subgroup
of $\mbox{Gal}(L/K)$. In that case $\mbox{Gal}(M/K)\cong G/\mbox{Gal}(L/M)$
\end{enumerate}
\end{thm}
\begin{cor}
{}
\begin{enumerate}
\item If $L/K$ is finite, normal, separable then there are only finitely
many fields $M$ with $K\subseteq M\subseteq L$.
\item If $\mbox{Gal}(L/K)$ is abelian, then for any $K\subseteq M\subseteq L$
we have $M/K$ normal.
\end{enumerate}
\end{cor}
We see from this that $\mbox{Gal}(\qq(\sqrt[3]{2},\omega)/\qq)=S_{3}$
since $\qq(\sqrt[3]{2})/\qq$ is not normal (exercise: show this directly
by writing down 6 automorphism)

Galois' Application: $\mbox{Gal}(L/K)$ simple implies no intermediate
normal $M/K$, This in terns implies no highest degree polynomial
``solved by radicals''
\begin{proof}[Proof of the Fundamental Theorem of Galois Theory (in ideas)]

\begin{lem}
If $\phi_{i}:M\to L$, $i=1,\dots,r$ are distinct inclusion of fields,
then the $\phi_{i}$ are linearly independent over $L$, i.e., $\sum a_{i}\phi_{i}(m)=0\,\forall m$
then $a_{i}=0\forall i$.
\end{lem}
We apply this to $M=L$. For $H$ a subgroup of $\mbox{Gal}(L/K)$
we use the lemma to show that $[L:L^{H}]=|H|$. ($\{\phi\in H\}$
are linearly independent $\phi:L\to L$). So $[L^{H}:K]=[L:K]/|H|$.
Next we use the following propositions
\begin{prop}
If $L/K$ is normal and $K\subseteq M\subseteq L$ has $M/K$ normal
then for all $\phi\in\mbox{Gal}(L/K),\phi(M)=M$ 
\end{prop}
We use twice. Once for 4. and first to show that the $[L:K]$ maps
$L\to N$ (where $N$ is a bigger field) we construct by hand have
image in $L$. Use this to show $|\mbox{Gal}(L/K)|=[L:K]$, thus $L^{\mathrm{Gal}(L/K)}=K$,
because $[L:L^{\mathrm{Gal}(L/K)}]=|G|=[L:K]$. \end{proof}
\begin{example*}
{}
\begin{itemize}
\item $K=\qq$, $L=$ splitting field of $x^{3}-2$, that is $L=\qq(\sqrt[3]{2},\sqrt[3]{2}\omega,\sqrt[3]{2}\omega^{2})=\qq(\sqrt[3]{2},\omega)$
where $\omega=e^{\frac{2\pi i}{3}}$. Now $K\leq L$ if finite, separable
(characteristic 0) and normal. What is $\mbox{Gal}(L/K)=G$? Any elements
of $G$ permutes \foreignlanguage{english}{$\sqrt[3]{2},\sqrt[3]{2}\omega,\sqrt[3]{2}\omega^{2}$.
So $G\subseteq S_{3}$, but since $G$ is of order $6=[L:K]$, we
must have $G=S_{3}$.
\[
\xyC{2.5pc}\xyR{1.5pc}\xymatrix{ & L\ar@{-}[ddl]_{3}\ar@{-}[d]_{2}\ar@{-}[dr]_{2}\ar@{-}[drr]^{2}\\
 & \qq(\sqrt[2]{3}\omega^{2})\ar@{-}[dd]_{3} & \qq(\sqrt[3]{2}\omega)\ar@{-}[ddl]_{3} & \qq(\sqrt[3]{2})\ar@{-}[ddll]^{3}\\
\qq(\omega)\ar@{-}[dr]_{2}\\
 & \qq
}
\qquad\xymatrix{ & \{\id\}\ar@{-}[ddl]_{3}\ar@{-}[d]_{2}\ar@{-}[dr]_{2}\ar@{-}[drr]^{2}\\
 & \left\langle (12)\right\rangle \ar@{-}[dd]_{3} & \left\langle (13)\right\rangle \ar@{-}[ddl]_{3} & \left\langle (23)\right\rangle \ar@{-}[ddll]^{3}\\
\left\langle (123)\right\rangle \ar@{-}[dr]_{2}\\
 & S_{3}
}
\]
}
\item Let $K=\qq$ and $L$ be the splitting field of $x^{3}-3x-1=(x-\alpha)(x-\beta)(x-\gamma)$.
Then $\qq\leq L=\qq(\alpha,\beta,\gamma)$. What is $\mbox{Gal}(L/K)=?$
So $G\subseteq S_{3}$. Using the fact about discriminant (see below)
we have that no transposition is in $G$. (Since $(\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)=\pm9$.
) Hence we have $|G|=3$ so $G=\zz/3\zz$. 
\end{itemize}
\end{example*}
\begin{fact*}
If $L$ is the splitting field of a cubic then $\mbox{Gal}(L/K)=\begin{cases}
\zz/3\zz & \Delta(f)\mbox{ is a square in }\qq\\
S_{3} & \mbox{otherwise}
\end{cases}$\end{fact*}
\begin{defn}
The \emph{discriminant} of a polynomial $f$ with roots $\alpha_{1},\alpha_{2},\dots,\alpha_{n}$
is $\Delta(f)=\prod_{i<j}(\alpha_{i}-\alpha_{j})^{2}$\end{defn}
\begin{fact*}
You can express $\Delta(f)$ as a polynomial on the coefficients of
$f$\end{fact*}
\begin{example*}
If $f=x^{3}-ax^{2}+bx-c$, $\Delta(f)=a^{2}b^{2}+18abc-27c^{2}-4a^{3}c-4b^{3}$\end{example*}

\end{document}
