Timeline for Algebraic geometry used "externally" (in problems without obvious algebraic structure).
Current License: CC BY-SA 2.5
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 28 at 0:07 | comment | added | LSpice | @VictorProtsak about the new questions 1 2. | |
| Jul 27 at 22:42 | comment | added | Qiaochu Yuan | I also asked about Victor's claim that the Zariski density argument is circular here: mathoverflow.net/questions/475892/… | |
| Jul 27 at 19:39 | comment | added | Qiaochu Yuan | I was very confused about whether Victor Protsak's argument actually works and it looks like I'm not the only one, so I asked a question about it on math.SE here: math.stackexchange.com/questions/4951342/… | |
| Jul 27, 2021 at 22:38 | comment | added | wlad | Final remark: The zariski-density argument (as I understand it) is still pedagogically useful. Also, I don't know if scheme technology (as opposed to varieties) can make the argument valid for general commutative rings. | |
| Jul 27, 2021 at 22:21 | comment | added | wlad | Observe that the same argument works for proving that $\mathcal X_{AB} = \mathcal X_{BA}$. In either case, the theorem for every commutative ring is reduced to the case of $\mathbb C$. | |
| Jul 27, 2021 at 22:19 | comment | added | wlad | ...continued 2. Prove that the theorem is true for free commutative rings by using the fact that it's true for $\mathbb C$. This is easy to do since the elements of a free commutative ring can be interpreted as polynomials over $\mathbb C$. 3. Prove that the theorem is true for all commutative rings using the fact that it's true for the free commutative rings. | |
| Jul 27, 2021 at 22:19 | comment | added | wlad | The algebraic geometric proof is pretty terrible, in that it only proves the theorem for the algebraically closed fields. A better proof is in three short stages: 1. Prove that the theorem is true for matrices over $\mathbb C$. This can be done by observing that the diagonalisable matrices are dense in the set of all square matrices. The proof of this for $\mathbb C$ is very easy. continued... | |
| Jul 17, 2010 at 11:30 | comment | added | Pierre-Yves Gaillard | Here is my phrasing of (the lemma underlying) Victor's proof of CH. Let $R$ be a non-commutative ring, let $X$ be an indeterminate, let $r$ be in $R$, let $R[X]\to R, f\mapsto f(r)$ be the evaluation, let $f,g\in R[X]$ be such that the coefficients of $g$ commute with $r$. Then $f(r)g(r)=(f g)(r)$. | |
| Jun 28, 2010 at 10:28 | comment | added | Martin Brandenburg | I don't understand your comments about generic points. Nothing like that is needed in the proof. @Victor: I get the message, I also don't think that the proof in my answer is the most natural one. But it's just awesome ;-). | |
| Jun 28, 2010 at 6:30 | comment | added | Victor Protsak | Cont'd: You may object that Bourbaki develops the whole classification of modules over a PID, inv factors, and Jordan decomp first, and then has CH over a field as a cor, but I don't consider this approach to CH either direct or elegant. I am sorry, but it is perverse to work so hard and apply so many tools in order to show a basic fact that holds in greater generality, unless, of course, the aim is to showcase the tools. BTW, in half a dozen times or so I've taught linear algebra, I never got to do the Jordan normal form (let alone prove it), but I've always stated, and sometimes proved, CH. | |
| Jun 28, 2010 at 6:15 | comment | added | Victor Protsak | Boyarsky: In your sketch, some work (Vandermonde det) is still needed to show that eigenvectors with different eigenvalues are lin indep and, of course, "eigenvalues are roots of the char polynomial" is a weak form of CH. So invoking only algebraic closure, eigenvalues, properties of the determinant, det tricks, and Zariski topology, one can show CH over a field, with extra work (half-page of small print in Bourbaki) remaining to establish it generically. And it's somehow supposed to be more natural than a direct proof "when you first step into linear algebra"??? | |
| Jun 28, 2010 at 5:49 | comment | added | Victor Protsak | Ryan: I believe it to be the original proof of the Cayley-Hamilton theorem due to Frobenius. It appears in Bocher's "Introduction to higher algebra" (1907), with specialization done by explicit manipulation of polynomials. BTW, Bocher was a Harvard professor :) The whole approach of Frobenius-Weierstrass-Kronecker to classification problems in linear algebra is refreshing when compared with abstract nonsense treatment that, sadly, became the norm later (besides Bocher, see Gantmacher's "Theory of matrices"). It also easily extends to the noncommutative setting (Capelli, etc). | |
| Jun 26, 2010 at 3:46 | comment | added | Ryan Reich | I don't know who first wrote the proof Victor gave, but I first heard it from Dennis Gaitsgory a few years ago and thought it quite novel, sort of the ideal argument. It's on the Wikipedia page in obfuscatory detail: en.wikipedia.org/wiki/… | |
| Jun 25, 2010 at 22:05 | comment | added | Boyarsky | Victor, how about the following: if the char. poly. splits with distinct roots, then we get an eigenline for each eigenvalue and hence have diagonalized the matrix with distinct eigenvalues. The converse is clear, and we haven't used Cayley-Hamilton. I think that addresses your #1. For #2, very nice! (Alas, it seems too slick to be used when teaching Cayley-Hamilton to inexperienced undergraduates.) | |
| Jun 25, 2010 at 22:00 | comment | added | Victor Protsak | 1 If you want to say that CH holds at the generic point then I agree, and generic semisimplicity is then a corollary which isn't needed in the proof of CH itself. I don't see a direct proof of Zariski density of diag matrices w/distinct eigenvalues (geom pts), because some form of CH is needed to relate them to the condition that the char polynomial has distinct roots. 2 Let $S$ be the commutant of $A$ in $M_n$, then $S[\lambda]\subset M_n[\lambda]$ contains $X=A-\lambda I_n$ and $adj X$ and spec'n is a unique ring hom $\phi:S[\lambda]\to M_n$ that is identity on $S$ and $\phi(\lambda)=A.$ | |
| Jun 25, 2010 at 16:09 | comment | added | Boyarsky | Victor, your suggested proof of Cayley-Hamilton looks like the standard incorrect proof: how can you specialize $\lambda$ to be a matrix (such as $A$) when the computation rests on it being an element of the commutative coefficient ring over which the matrices live? Please clarify what I am misunderstanding. | |
| Jun 25, 2010 at 10:20 | comment | added | Boyarsky | Victor, it is not backwards: semisimplicity is implied by the separability of the characteristic polynomial, which is to say that its geometric zero set is disjoint from that of its derivative, or equivalently the resultant of the pair is non-vanishing. The latter is a Zariski-open locus (even hypersurface complement) within the irreducible space of all $n \times n$ matrices (over a field) and so is Zariski-dense if it has a geometric point. A suitable diagonal matrix over an infinite extension provides such a point. So semisimplicity holds at the generic point, which is what we need. QED | |
| Jun 25, 2010 at 0:59 | comment | added | Victor Protsak | Awk, this is so backwards. Zariski density of semisimple matrices in all matrices $\textit{logically depends}$ on the Cayley-Hamilton theorem, which has a 1-line proof: $$ $$ Let $X=A-\lambda I_n,$ then $p_A(\lambda)I_n=(\det X)I_n=X(adj X)$ in the $n\times n$ matrix polynomials in $\lambda,$ now specialize $\lambda\to A,$ get $p_A(A)=0\ \square$ $$ $$ There is a similarly short proof for $p_{AB}(\lambda)=\lambda^{m-n}p_BA(\lambda).$ | |
| Jun 24, 2010 at 9:13 | history | answered | Martin Brandenburg | CC BY-SA 2.5 |