Skip to main content
7 events
when toggle format what by license comment
Jun 16 at 21:27 comment added Qiaochu Yuan @Victor: my apologies, I don't know if you're even checking MO anymore, but I don't see how this proof works. $\lambda$ is a scalar and $A - \lambda$ is $A$ minus a scalar matrix, it doesn't make any sense to substitute $A$ for $\lambda$ on the RHS. Doesn't this run into exactly the same problem as the naive proof where we substitute $\lambda = A$ directly into $\det(A - \lambda)$?
Aug 3, 2010 at 14:46 comment added Donu Arapura Victor, while I don't disagree with you that Cayley-Hamilton can and should be approached directly when teaching linear algebra, I still maintain that it is an instructive use of the Zariski topology. I don't know where other people learned this, I found it for myself while teaching a beginning algebraic class back in the mid 90's. I might have given as a homework problem.
Jul 31, 2010 at 23:08 comment added Victor Protsak As I have pointed out several times on MO already, to completely prove Cayley-Hamilton directly it suffices to write one line, $$p_A(\lambda)=(A-\lambda)\text{adj}(A-\lambda), \lambda\to A \implies p_A(A)=0,$$ whereas to prove that a general matrix is diagonalizable requires a lot more. I am just wondering about the origin of this misapprehension, apparently common among a certain generation of algebraic geometers (or AG students) at MO, that linear algebra results are to be proved in a roundabout way with AG tools: is there a canonical textbook that advocates this philosophy?
Apr 16, 2010 at 13:08 comment added Donu Arapura Yes for etale etc. [cf. Milne p 114]. Morally yes for analytic topology of a proper variety over C by GAGA.
Apr 16, 2010 at 12:58 comment added Qfwfq Is sheaf cohomology of coherent sheaves always the same in the Zariski topology and in finer topologies?
Apr 16, 2010 at 12:43 history edited Donu Arapura CC BY-SA 2.5
added 9 characters in body
Apr 16, 2010 at 12:37 history answered Donu Arapura CC BY-SA 2.5