Can different modules have the same symmetric algebra? (answered: no) - MathOverflow

Can different modules have the same symmetric algebra? (answered: no)

2009-11-23T08:21:33Z

Algebraic geometry allows one to think of an $A$-module $M$ geometrically as a module of functions on the $A$-scheme $\mathrm{Spec}(\mathrm{Sym}(M))$.

I'm wondering if anything is lost in just replacing $M$ by this geometric object. Since nothing is lost in taking $\mathrm{Spec}$, this amounts to asking:

(1) Can two non-isomorphic $A$-modules $M$ , $N$ have isomorphic symmetric $A$-algebras $\mathrm{Sym}(M)$ ,
$\mathrm{Sym}(N)$?

(Clearly they are not isomorphic as graded $A$-algebras.)

If the answer is "No", great! If "Yes", I would like to see a specific example.

It may be interesting to have a second interpretation, even if it doesn't help solve the problem. Since we have the adjunction (of set-valued functors)

$hom_{A-alg}(\mathrm{Sym}(M),B) \simeq hom_{A\mathrm{-mod}}(M,B),$

by Yoneda's lemma, an equivalent question would be:

(2) If the (set-valued) functors $hom_{A-\mathrm{mod}}(M,-)$ and $hom_{A-\mathrm{mod}}(N,-)$ agree on $A$-algbras, do they agree on $A$-modules?

Edit: I emphasized "set-valued" above, thanks to a comment from Buzzard. Also, partially in response to Mark Hovey's comment, I removed "Is it safe to think of modules geometrically" from the quesiton statement, since I don't want to assert that this is "the correct" geometric interpretation of a module.

Answer by a-fortiori for Can different modules have the same symmetric algebra? (answered: no)

2009-11-23T08:59:54Z

EDIT: the following argument misinterprets question (2) as using internal homs of A-mod

For any $A$-module $Q$, equip $A\oplus Q$ with the structure of an $A$-algebra such that $Q$ is an ideal of square zero. Then, $\mathrm{Hom}(M,Q)=\ker(\mathrm{Hom}(M,A\oplus Q)\to\mathrm{Hom}(M,A))$. This is functorial in $Q$, so you can apply Yoneda.

Answer by auniket for Can different modules have the same symmetric algebra? (answered: no)

2009-11-23T11:50:06Z

EDIT: The argument does not work.

Answer to 1: No.

Let $\phi:$ Sym($M$) $\to$ Sym($N$) be an $A$-algebra isomorphism. We will see that it induces an $A$-module isomorphism $\tilde \phi: M \to N$. Pick a set of $A$-module generators $m_1, \ldots, m_k$ of $M$. These also generate Sym($M$) as an $A$-algebra. Let $\phi(m_i) = \sum_j n_{ij}$ with $n_{ij} \in N^j$. In particular each $n_{i1} \in N$. I claim that $\tilde \phi: m_i \to n_{i1}$ gives a well defined $A$-module map from $M$ to $N$. To see it, all you have to show (I think) is that if $\sum a_im_i = 0$ for $a_1, \ldots, a_k \in A$, then $\sum a_in_{i1} = 0$. But it is true, because $\phi$ is an $A$-algebra homomorphism and hence $0 = \phi(\sum a_i m_i) = \sum_j (\sum_i a_i n_{ij})$ and thus the inner sum is $0$ for each $j$, because it is the $j$-th graded component. The claim is proved.

In the same way you can show $\phi^{-1}$ also induces a map $N \to M$ and it should be inverse to $\tilde \phi$.

This argument seems to show that there is a map from $Hom_{A-alg}(Sym(M),Sym(N))$ to $Hom_{A-mod}(M,N)$. But I could not have seen it before writing it out.

Answer by Kevin Buzzard for Can different modules have the same symmetric algebra? (answered: no)

2009-11-23T16:15:14Z

I now believe a-fortiori's argument: translations are a problem, but, as a-fortiori observed, they are the only problem. Let me spell it out.

Say $f:Sym(M)\to Sym(N)$ is an isomorphism. For $m\in M$ write $f(m)=f_0(m)+f_1(m)+f_{\geq2}(m)$ with obvious notation: $f_0(m)$ is in $A$, $f_1(m)$ is in $N$ and $f_{\geq2}(m)$ is all of the rest. Now here's another $A$-algebra map $g:Sym(M)\to Sym(N)$. To define $g$ all I have to do is to say where $m\in M$ goes so let's say $g(m)=f_1(m)+f_{\geq2}(m)$.

Claim: $g$ is an $A$-algebra isomorphism.

The proof is that $g$ is just $f$ composed with the isomorphism $Sym(M)\to Sym(M)$ sending $m$ to $m-f_0(m)$ (one needs to check that this is an isomorphism but it is because there's an obvious inverse).

Claim: the isomorphism of rings inverse to $g$ also has "no constant terms", i.e. it's of the form $h:Sym(N)\to Sym(M)$ where $h(n)=h_1(n)+h_{\geq2}(n)$ with no constant term.

The proof is that $g$ sends terms of degree $d$ to terms of degree $d$ or higher, so applying $g$ to $h(n)=h_0(n)+h_1(n)+h_{\geq2}(n)$ we see $n=h_0(n)+f_1(h_1(n))+$(higher order terms).

Claim: $f_1$ and $h_1$ are mutual inverses. This is easy now.

So in fact all the ideas are in a-fortiori's comments.