Timeline for Expressing $-\operatorname{adj}(A)$ as a polynomial in $A$?
Current License: CC BY-SA 4.0
36 events
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| Dec 11, 2020 at 15:39 | history | edited | YCor |
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| Dec 11, 2020 at 15:22 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
Fix TeX. This is not a vertically centered \sum-like large operator, which is what \mathop constructs.; deleted 22 characters in body
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| Dec 11, 2020 at 15:12 | history | edited | Denis Serre | CC BY-SA 4.0 |
edited title
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| Aug 13, 2015 at 18:47 | history | edited | darij grinberg | CC BY-SA 3.0 |
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| May 15, 2015 at 21:01 | history | rollback | user9072 |
Rollback to Revision 5
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| S May 15, 2015 at 19:12 | history | suggested | luisfelipe18 | CC BY-SA 3.0 |
change $p_1 I + p_2 A$ for $p_0 I + p_1 A$
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| May 15, 2015 at 18:43 | review | Suggested edits | |||
| S May 15, 2015 at 19:12 | |||||
| Nov 7, 2011 at 14:44 | answer | added | Marc van Leeuwen | timeline score: 6 | |
| Jul 21, 2010 at 5:46 | vote | accept | Laurent Lessard | ||
| Jul 20, 2010 at 10:17 | history | edited | darij grinberg | CC BY-SA 2.5 |
minor typo fix
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| Jul 20, 2010 at 9:35 | answer | added | Pierre-Yves Gaillard | timeline score: 5 | |
| Jul 20, 2010 at 4:14 | comment | added | Harry Gindi | @Tom Church: The tag [ring-theory] was merged into [ra.rings-and-algebras], and the tag [commutative-rings] was merged into [ac.commutative-algebra]. See tea.mathoverflow.net/discussion/34/2/tag-mergerename-requests/… | |
| Jul 20, 2010 at 4:12 | history | edited | Harry Gindi |
edited tags
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| Jul 19, 2010 at 6:17 | comment | added | Tom Church | I have restored the author's original tags to the question. I see no reason to have removed these (eminently applicable) tags. If the remover feels that they clearly deserve to be deleted, then he should have no trouble waiting for someone else to do so. | |
| Jul 19, 2010 at 6:14 | history | edited | Tom Church |
restored original tags
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| Jul 19, 2010 at 5:44 | history | edited | Harry Gindi |
edited tags
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| Jul 18, 2010 at 19:39 | answer | added | Pete L. Clark | timeline score: 9 | |
| Jul 18, 2010 at 18:13 | answer | added | Qiaochu Yuan | timeline score: 12 | |
| Jul 18, 2010 at 16:24 | comment | added | Qiaochu Yuan | What I meant is that there are these results of the form "if a first-order sentence is true in X, it must also be true in Y, Z, W...." and my intuition about proving a polynomial identity (with coefficients in Z) by proving it for C^n is that it is a statement of this sort. Maybe this is not a good way to think about things; in any case I appreciate the clarification. | |
| Jul 18, 2010 at 9:04 | comment | added | Victor Protsak | @Qiaochu: How can your intuition be "model theoretic" (I assume it means "true over an arbitrary commutative ring") and be "about diagonalizable matrices and their eigenvalues" (which makes sense only when the ground ring is a field) $\textit{at the same time}?$ For matrix identities (and polynomial identities more generally), the right kind of intuition does $\textit{not}$ come from eigenvalues or complex numbers. See Bill's remark at the end of his answer and my comment. | |
| Jul 18, 2010 at 8:16 | comment | added | Qiaochu Yuan | @Victor: the definition I know of the adjugate matrix implies that it is polynomial in the entries of the matrix, so one doesn't have to work over an arbitrary ring in the first place; each entry encodes a polynomial identity over Z which can be proven over C. @Pierre: I agree that it is valuable to do things from first principles. But much of my intuition about matrices is about diagonalizable matrices and their eigenvalues, and I like to take advantage of that intuition whenever I can. I guess you could say I'm more comfortable with maximal tori than their Lie groups. | |
| Jul 18, 2010 at 6:42 | answer | added | Bill Dubuque | timeline score: 14 | |
| Jul 18, 2010 at 4:35 | comment | added | Pierre-Yves Gaillard | Dear Qiaochu Yuan: I agree that the algebraic closures of Q are countable. But I think it's not necessary to use them in this context. I should have referred to the opposition algebraic/transcendent (or perhaps algebraic/analytic) instead of countable/uncountable. What surprised me in your comment is your invocation of R (hidden behind C). One completes Q to get R and do analysis on it. But this has nothing to do (in my opinion) with, say, Cayley-Hamilton. Apart from that, I liked your comment very much! | |
| Jul 17, 2010 at 23:34 | comment | added | Victor Protsak | Quiaochu (1st comment) and Martin (5th comment), you have to be a bit careful here: OP's statement is $\textit{a priori}$ stronger than CH theorem, so some care is required to conclude that you can "cancel" $A$ from both sides over an arbitrary ring. Just to give you an example of what could go wrong: in Boolean rings, $x^2=x$, but this doesn't imply $x=1$. | |
| Jul 17, 2010 at 19:03 | comment | added | Qiaochu Yuan | @Pierre: the algebraic closure of Q is countable! My intuition for this style of proof is that it is a model-theoretic phenomenon, but I'm sure those more qualified could make that much more precise. | |
| Jul 17, 2010 at 18:47 | answer | added | Victor Protsak | timeline score: 25 | |
| Jul 17, 2010 at 11:33 | comment | added | Pierre-Yves Gaillard | Dear Victor: I've just added a comment about your proof of CH on the link you gave. | |
| Jul 17, 2010 at 10:06 | comment | added | Pierre-Yves Gaillard | Dear Victor: I've started looking at the link you gave. It seems very interesting. I'll read it quietly, but I wanted to thank you right away! | |
| Jul 17, 2010 at 5:46 | comment | added | Victor Protsak | Pierre-Yves, I agree! See mathoverflow.net/questions/29271/…, where this was discussed $\textit{ad nausem}$ and Martin finally made a concession that it's not the most natural approach, but it's awesome (to him). | |
| Jul 16, 2010 at 11:40 | comment | added | Pierre-Yves Gaillard | In my humble opinion Qiaochu Yuan's answer is more convincing than Martin's. All you need to know (I think) is that $\mathbb Z[X_1,...,X_k]$ is a domain. I don't think things like the existence of algebraic closure, the density of diagonal matrices, Zariski's topology, ..., are necessary (or even helpful) in this context. I don't even think Qiaochu Yuan had to invoke complex numbers. I believe the statement is truly elementary, and can be (easily) proved by considering only countable sets. | |
| Jul 16, 2010 at 9:14 | comment | added | Martin Brandenburg | It should be noted that Michele's proof uses that $R$ is (w.l.o.g.) a field (othweise $p_0$ is just a non-unit). | |
| Jul 16, 2010 at 8:56 | comment | added | Laurent Lessard | Thank you all! So it looks like proving this result for invertible matrices in C is sufficient to extend the result to any commutative ring R. | |
| Jul 16, 2010 at 8:50 | comment | added | user47274 | When $A$ is not invertible $p_0=0$ and, again, it is the Cayley-Hamilton Theorem | |
| Jul 16, 2010 at 8:49 | comment | added | Martin Brandenburg | It's easy to reduce to the case that $R$ is an algebraically closed field. Now prove the claim for diagonal matrices and then use their density with respect to the Zariski topology. | |
| Jul 16, 2010 at 8:47 | comment | added | Qiaochu Yuan | This is a polynomial identity in n^2 variables A_{ij} with integer coefficients, so it holds over any commutative ring R if and only if it holds over a dense subset of C, so the invertible case is already enough. | |
| Jul 16, 2010 at 8:44 | history | asked | Laurent Lessard | CC BY-SA 2.5 |