An emailed question asked
I’m currently revising Commutative Algebra and I am going through
the proof of proposition 5.4 in the lecture notes, which proves
that if $0 \rightarrow M_1 \rightarrow M_2 \rightarrow M_3
\rightarrow 0$ is a short exact sequence, then $M_2$ is Artinian if
and only if $M_1$ and $M_3$ are both Artinian.
We have a short exact sequence so we know that $\alpha$ is
injective and $\beta$ is surjective and
$Im(\alpha)=Ker(\beta)$. During the proof it refers to
$\alpha^{1}$ and $\beta^{1}$, but surely it only makes sense to
talk about inverses if the maps are bijective? For example if $M_1$
was a submodule of $M_2$ and $\alpha$ was the inclusion map, then
it wouldn’t make sense to talk about $\alpha^{1}$ because elements
in $M_2\setminus M_1$ don’t have a preimage in $M_1$ (obviously in
this case a submodule would clearly be Artinian but still, the
inverse doesn’t exist).
Also in the proof of Theorem 4.12, I presume the maps $\alpha$ and
$\beta$ are the localization maps i.e $\alpha(t)=t/1 \in
A_p$. During the proof it refers to $\alpha^{1}$ and $\beta^{1}$,
but again I’m not sure these have inverses because what is the
image under $\alpha{^1}$ of $r/s$?
Perhaps you could tell me where I am going wrong.
to which I replied
If $f$ is any function from a set $A$ to a set $B$ and $C\subseteq
B$, then we can define $f^{1}(C)$ to be the subset of $A$
consisting of those $a \in A$ such that $f(a) \in C$ (and not just
$\in B$). This is standard mathematical notation for what is known
as the "inverse image of $C$ under $f$", and using it does not
imply that $f$ is invertible.
A followup question follwed:
So in the case of Proposition 5.4, say $N_k$ is a submodule of
$M_3$, then $\beta^{1}(N_k)$ is the elements of $M_2$ which map
into $N_k$. So assuming you’ve proved that these elements are a
submodule, then the chain $\beta^{1}(N_1)$, $\beta^{1}(N_2)\dots$
terminates in $M_2$, so $\beta^{1}(N_k)= \beta^{1}(N_{k+1})\dots$
for some $k$. And because $\beta$ is surjective it follows that
$\beta(\beta^{1}(N_k))=N_k=
\beta(\beta^{1}(N_{k+1}))=N_{k+1}\dots$, so the chain in $M_3$
terminates. Is that correct?
to which my reply was:
Yes, you are completely correct. And it's not hard to see that the
inverse image of a submodule (of the codomain) is a submodule of
the domain.
MA3G6 Commutative Algebra
MA3G6 Commutative Algebra Question about inverse maps
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