Timeline for Does base extension reflect the property of being isomorphic?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 20, 2022 at 17:49 | comment | added | mhahthhh | if the finite dimensional representation is initially defined over K, then its image is always a finite dimensional K-algebra. | |
| Dec 17, 2013 at 19:21 | comment | added | Dag Oskar Madsen | @NoahSnyder I was thinking $A$ finitely generated, $M$ and $N$ finite dimensional, and wondered if the same proof would work. | |
| Dec 17, 2013 at 19:17 | comment | added | Noah Snyder | @DagOskarMadsen: Doesn't Julian's answer give a counterexample in the f.g. case? | |
| Dec 17, 2013 at 14:48 | comment | added | Edgardo | @DagOskarMadsen, at least the way I am thinking about it, it is important that everything is finite-dimensional over $K$. I don't know what happens in the general case. Also, it is a bit annoying that one has to treat the cases of K finite and infinite separately. Surely there is a uniform way of proceeding... | |
| Dec 17, 2013 at 12:29 | comment | added | Dag Oskar Madsen | Would this also work for $A$ finitely generated (as an algebra)? | |
| Dec 17, 2013 at 3:28 | vote | accept | Noah Snyder | ||
| Dec 17, 2013 at 2:57 | comment | added | Edgardo | Sorry, here it is: Modules that become isomorphic to $M$ over the algebraic closure of our finite field $K$ are classified by $H^1(G, \mathrm{Aut}(M'))$, where $M'$ is $M$ base-changed to the algebraic closure, and $G$ is the absolute Galois group of $K$. Now $\mathrm{Aut}(M')$ is the set of $\bar{K}$-points of a connected algebraic $K$-group, namely, the automorphism group of $M$ (considered as a $K$-variety). There is a theorem of Lang and Steinberg that says $H^1$ always vanishes in this setting. | |
| Dec 17, 2013 at 2:51 | comment | added | Noah Snyder | I follow the infinite case, very nice! Any chance you could flesh out the last paragraph a little more? Which Galois cohomology group? (I think $H^1$?) Why does vanishing of that Galois cohomology group yield the desired conclusion? Sorry for being slow. | |
| Dec 17, 2013 at 0:44 | comment | added | Edgardo | I don't think so. It comes down to this: Take a polynomial $f \in K[x_1, \dots, x_n]$. If there exists $(a_1, \dots, a_n) \in L^n$ such that $f(a_1, \dots, a_n) \neq 0$, then also there exists $(b_1, \dots, b_n) \in K^n$ with $f(b_1, \dots, b_n) \neq 0$. The existence of $a_i$ means that $f$ is not identically vanishing. | |
| Dec 17, 2013 at 0:42 | comment | added | Ben Wieland | In the infinite field case, you're using the separable hypothesis, right? | |
| Dec 17, 2013 at 0:13 | comment | added | Noah Snyder | Sorry, I was confused. | |
| Dec 16, 2013 at 23:54 | comment | added | Edgardo | I don't think it uses commutativity anywhere (?) We are just identifying $\mathrm{Hom}_A(M,N)$ with the linear subspace of $\mathrm{Hom}_K(M,N)$ which commutes with $A$. | |
| Dec 16, 2013 at 23:06 | history | answered | Edgardo | CC BY-SA 3.0 |