Mathematics » MA3G6 Commutative Algebra
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The latest posts to Mathematics » MA3G6 Commutative Algebraen-GB(C) 2021 University of WarwickWed, 19 Feb 2014 20:34:52 GMThttp://blogs.law.harvard.edu/tech/rssSiteBuilder2, University of Warwick, http://go.warwick.ac.uk/sitebuilder2014 EXAM FEEDBACK
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f546298e7901462f9bd17e06d4
<p>I have just put a copy of Thursday's exam, with model solutions and feedback from me after marking your scripts, on the module resources web page.</p>
<p> </p>Sat, 24 May 2014 19:01:19 GMTJohn Cremona094d43f546298e7901462f9bd17e06d4typo in proof of Hilbert Basis Theorem
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f546150ae101461dd955b61a8b
<p>I was asked:</p>
<p> In the proof of the Hilbert Basis Theorem when defining $g_1$ shouldn't it say that $g_1=\sum(r_i X^{\deg(g)-n} f_{i,n})$ ? Otherwise we won't necessarily get the same leading coefficient.</p>
<p>and this is correct. In the printed notes (page 27 line 11) it incorrectly has $f_{1,n}$ instead of $f_{i,n}$.</p>
<p> </p>Wed, 21 May 2014 08:15:21 GMTJohn Cremona094d43f546150ae101461dd955b61a8bDedekind Domains are not in the exam
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f546150c1a01461a155e1b2c17
<p>Several people have asked me whether Dedekind Domains are examinable, and I want to make sure that everyone has the same answer (since I was stopped in the corridor and asked this yesterday and think I was rather evasive).</p>
<p>I told Michael when he gave the last lecture, which was on Dedekind Domains, that if asked he could say that the material of the last lecture was not examinable. I think I started DDs in the lecture before that, which it probably what is worrying you. Hence I would now like everyone to know that</p>
<p><strong>There is nothing about Dedekind Domains in this year's exam.</strong></p>
<p>I hope this is clearer than saying whether or not the are "examinable" since a cruel lecturer (not me!) might be very pedantic and say that something was examinable even though it was not in fact on this year's exam. Let's make this quite unambiguous:</p>
<p> </p>
<p><strong>There is nothing about Dedekind Domains in this year's Commutative Algebra exam!</strong></p>
<p>Good luck on Thursday.</p>Tue, 20 May 2014 14:42:26 GMTJohn Cremona094d43f546150c1a01461a155e1b2c17Re: 2012 Q4 part e
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545f642cf014605e76d2103d8
<p>I'm glad it is now sorted. Thanks for posting this on the forum, I'm sure there will be others who will find this useful.</p>
<p> </p>
<p> </p>Fri, 16 May 2014 16:39:51 GMTJohn Cremona094d43f545f642cf014605e76d2103d8Re: 2012 Q4 part e
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545f642d30146058f192c04d8
<p>Oh! I had just completely neglected that for R to even be a valuation ring it must be an integral domain, hence the confusion when I got to part (e). Sorry that it wasn't obvious which bit I got wrong but thanks for pointing it out!</p>Fri, 16 May 2014 15:03:22 GMTA guest user094d43f545f642d30146058f192c04d8Re: 2012 Q4 part e
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545f642cf01460588443667b4
<p>Yes we did (use that $R$ is an integral domain in part (d)). It refers throughout to the field of fractions, which is only defined when $R$ is an integral domain. The question not how to characterise valuation rings amongst all rings, just how to characterise them amongst integral domains.</p>
<p> </p>
<p>In (e) you do have to say why $P/P$ and $R_P$ are both integral domains (as the solution does).</p>
<p> </p>
<p> </p>Fri, 16 May 2014 14:55:54 GMTJohn Cremona094d43f545f642cf01460588443667b4Re: 2012 Q4 part e
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545f642cf01460584276a6776
<p>I am still a bit confused, because in part (d) did we not have to use that R was an integral domain? Whereas in part (e) we don't necessarily have that R is an integral domain.</p>Fri, 16 May 2014 14:51:25 GMTA guest user094d43f545f642cf01460584276a6776Re: 2012 Q4 part e
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<p>Isn't that exactly what part (d) tells you? The original definition of valuation ring is that for any two elements $x,y$ either $x$ divides $y$ or vice versa, so that either $(x)\supseteq(y)$ or vice versa, or in other words that any for two nonzero <em>principal</em> ideals one contains the other. Part (d) of the question was to prove that for <em>any</em> two ideals $I$ amd $J$, one must contain the other, and also prove the converse, that this implies that $R$ is a valuation ring. (The converse was the easier part, since if something is true for any two nonzero ideals it is certainly true for two principal ideals).</p>
<p>In words one could say that the result of part (d) is that an integral domain is a valuation ring if and only if the lattice of nonzero ideals is "linearly ordered" by inclusion. Passing from $R$ to $R/P$ or to $R_P$ we cuts down from the lattice of ideals of the new ring in a way which preserves this ordering by inclusion, hence the result.</p>
<p>I hope I haven't made it less clear by saying too much!</p>
<p> </p>Fri, 16 May 2014 14:40:14 GMTJohn Cremona094d43f545f642d301460579e9557a0b2012 Q4 part e
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<p>Hi could you possibly explain the solution to question 4e of the 2012 paper, proving R/P is a valuation ring?</p>
<p>I understand the correspondence that I/P is contained in J/P if and only if I is contained in J. So if we show that for all I and J containing P that I contains J or vice versa then we must have by (d) that R/P is a valuation ring. But the solution then just seems to say that this must be true for all such I and J in a valuation ring R with prime ideal P and I don't understand why.</p>
<p> </p>
<p>Cheers</p>Fri, 16 May 2014 14:11:16 GMTA guest user094d43f545f642d30146055f64a377d6Re: Revision class in Monday the 19th from 13:00-14:00 in MS.05
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545f642d30145fa0695990542
<p><strong>Revision classes</strong></p>
<p>Thursday 15 May 14.00-16.00 in MS.03</p>
<p>Monday 19 May 13.00-14.00 in MS.05</p>Wed, 14 May 2014 09:18:26 GMTJohn Cremona094d43f545f642d30145fa0695990542Question about inverse maps
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545f642d30145f9eb7b094ced
<p>An emailed question asked</p>
<p> </p>
<p> I’m currently revising Commutative Algebra and I am going through<br> the proof of proposition 5.4 in the lecture notes, which proves<br> that if $0 \rightarrow M_1 \rightarrow M_2 \rightarrow M_3<br> \rightarrow 0$ is a short exact sequence, then $M_2$ is Artinian if<br> and only if $M_1$ and $M_3$ are both Artinian.<br> <br> We have a short exact sequence so we know that $\alpha$ is<br> injective and $\beta$ is surjective and<br> $Im(\alpha)=Ker(\beta)$. During the proof it refers to<br> $\alpha^{-1}$ and $\beta^{-1}$, but surely it only makes sense to<br> talk about inverses if the maps are bijective? For example if $M_1$<br> was a submodule of $M_2$ and $\alpha$ was the inclusion map, then<br> it wouldn’t make sense to talk about $\alpha^{-1}$ because elements<br> in $M_2\setminus M_1$ don’t have a preimage in $M_1$ (obviously in<br> this case a submodule would clearly be Artinian but still, the<br> inverse doesn’t exist).<br> <br> Also in the proof of Theorem 4.12, I presume the maps $\alpha$ and<br> $\beta$ are the localization maps i.e $\alpha(t)=t/1 \in<br> A_p$. During the proof it refers to $\alpha^{-1}$ and $\beta^{-1}$,<br> but again I’m not sure these have inverses because what is the<br> image under $\alpha{^-1}$ of $r/s$?<br> <br> Perhaps you could tell me where I am going wrong.<br> <br>to which I replied</p>
<p> If $f$ is any function from a set $A$ to a set $B$ and $C\subseteq<br> B$, then we can define $f^{-1}(C)$ to be the subset of $A$<br> consisting of those $a \in A$ such that $f(a) \in C$ (and not just<br> $\in B$). This is standard mathematical notation for what is known<br> as the "inverse image of $C$ under $f$", and using it does not<br> imply that $f$ is invertible.</p>
<p>A follow-up question follwed:<br><br> So in the case of Proposition 5.4, say $N_k$ is a submodule of<br> $M_3$, then $\beta^{-1}(N_k)$ is the elements of $M_2$ which map<br> into $N_k$. So assuming you’ve proved that these elements are a<br> submodule, then the chain $\beta^{-1}(N_1)$, $\beta^{-1}(N_2)\dots$<br> terminates in $M_2$, so $\beta^{-1}(N_k)= \beta^{-1}(N_{k+1})\dots$<br> for some $k$. And because $\beta$ is surjective it follows that<br> $\beta(\beta^{-1}(N_k))=N_k=<br> \beta(\beta^{-1}(N_{k+1}))=N_{k+1}\dots$, so the chain in $M_3$<br> terminates. Is that correct?<br><br>to which my reply was:<br><br> Yes, you are completely correct. And it's not hard to see that the<br> inverse image of a submodule (of the codomain) is a submodule of<br> the domain.</p>Wed, 14 May 2014 08:48:50 GMTJohn Cremona094d43f545f642d30145f9eb7b094cedRe: test post, please ignore
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545f642cf0145f9e231a471bb
<p>Answer: all three work!</p>
<p> </p>
<p> </p>Wed, 14 May 2014 08:38:41 GMTJohn Cremona094d43f545f642cf0145f9e231a471bbtest post, please ignore
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545f642cf0145f9e1c21c71b7
<p>Just testing whether sitebuilder can understand simple latex dollar signs for formulas: $a^n+b^n=c^n$ with $a$, $b$, $c\in\mathbb{N}$.</p>
<p> </p>
<p>Or latex brackets for maths: \(a^n+b^n=c^n\) with \(a\), \(b\), \(c\in\mathbb{N}\).</p>
<p> </p>
<p>Or where we need the tedious long-tyle markup: <img src="https://mathtex.warwick.ac.uk/cgi-bin/mathtex.cgi?a%5En%2Bb%5En%3Dc%5En" alt="a^n+b^n=c^n"> with <img src="https://mathtex.warwick.ac.uk/cgi-bin/mathtex.cgi?a" alt="a">, <img src="https://mathtex.warwick.ac.uk/cgi-bin/mathtex.cgi?b" alt="b">, <img src="https://mathtex.warwick.ac.uk/cgi-bin/mathtex.cgi?c%5Cin%5Cmathbb%7BN%7D" alt="c\in\mathbb{N}">.</p>Wed, 14 May 2014 08:38:13 GMTJohn Cremona094d43f545f642cf0145f9e1c21c71b7Re: Revision class in Monday the 19th from 13:00-14:00 in MS.05
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545f642d30145f9dda5034c81
<p>Sorry if I caused confusion yesterday by the mass email (not via this forum) to make sure that everyone knew about the class on the 19th. There really are two! Michael and Chris will sort out between them what will be covered when.</p>
<p> </p>
<p>I cannot make another mass email since the university system which does those is down for maintenance today and tomorrow...</p>
<p>JEC</p>Wed, 14 May 2014 08:33:43 GMTJohn Cremona094d43f545f642d30145f9dda5034c81Re: Revision class in Monday the 19th from 13:00-14:00 in MS.05
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545f642cf0145f6cdd9682884
<p>Yeah Mike is doing one on the 15th and Ill do one on the 19th. </p>Tue, 13 May 2014 18:17:36 GMTA guest user094d43f545f642cf0145f6cdd9682884Re: Revision class in Monday the 19th from 13:00-14:00 in MS.05
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545f642d30145f67b3b1e2b36
<p>We got an email the other day saying that there would be a revision class on the 15th by Michael Selig. Is this in addition to that one, or instead of?</p>Tue, 13 May 2014 16:47:22 GMTAmit Chachlani094d43f545f642d30145f67b3b1e2b36Revision class in Monday the 19th from 13:00-14:00 in MS.05
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545db5d100145f59d9feb0351
<p>I have booked a room to do the revion class in, as stated in the title. Feel free to come and ask as many questions as you like, on anything from the lecture notes, problem sheets, past exams, etc... </p>Tue, 13 May 2014 12:45:19 GMTA guest user094d43f545db5d100145f59d9feb0351Re: Rings of small dimension chapter
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545db56940145f4ead6935159
<p>Correct. We did not do tensor products. Anything mentioning tensor products or the symbol <img src="https://mathtex.warwick.ac.uk/cgi-bin/mathtex.cgi?%5Cotimes" alt="\otimes"> is not examinable.</p>
<p> </p>
<p>[That's the trouble with trying to be explicit instead of leaving you clever people to use your common sense -- you take me too literally!]</p>Tue, 13 May 2014 09:30:02 GMTJohn Cremona094d43f545db56940145f4ead6935159Re: Rings of small dimension chapter
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545db5d100145f4986a555aba
<p>There are also a couple of propositions at the end of section 3.2 that are in smaller print; are these also non-examinable?</p>
<p>Thanks</p>Tue, 13 May 2014 08:00:00 GMTA guest user094d43f545db5d100145f4986a555abaRe: Classifying modules up to isomorphism
https://warwick.ac.uk/fac/sci/maths/undergrad/ughandbook/lecturers/ma3g6/ma3g6forum/?post=094d43f545db5d100145e5b50e7a3a9d
<p>No, that topic did not come up a lot in past papers. It only came up in 2011 which I have explicitly said was based on a different syllabus. There was nothing about that in our course, or in the 2012 or 2013 exams, and there will not be this year either.</p>
<p> </p>
<p>Having said that you may certainly be asked to prove that some isomorphism holds between modules -- but no "classification".</p>Sat, 10 May 2014 10:36:59 GMTJohn Cremona094d43f545db5d100145e5b50e7a3a9d