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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 $\mathrm{Sym}(M)$ ,

(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.

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The reformulation (2) is a bit confusing. If we fix an isomorphism between Sym(M) and Sym(N) that happens not to be grading-preserving, then we get an induced isomomorphism between the set-valued functors from A-modules to sets that you mention. But it will not be in general induced by an isomorphism of A-modules M=N (even if they're isomorphic!)---indeed both of the functors you mention above are naturally group-valued, but the given isomorphism between them might not be an isomorphism of group-valued functors. Hence "do they agree" should be interpreted very carefully. – Kevin Buzzard Nov 23 '09 at 14:28
Algebraic geometry inspires me to think of M as the global sections of the quasi-coherent sheaf M-tilde on Spec A. You really can't go wrong with this interpretation. – Mark Hovey Nov 23 '09 at 15:24
up vote 6 down vote accepted

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.

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Actually translation could still be a problem. The map phi: m -> m - f_0(m) does not seem to be an A-algebra morphism: for m_1, m_2 in Sym(M), phi(m_1m_2) = f_0(m_1)f_1(m_2) + f_1(m_1)f_0(m_2) + higher order terms, whereas phi(m_1)phi(m_2) = f_1(m_1)f_2(m_2) + higher order terms. – auniket Nov 24 '09 at 0:35
@auniket: No, it's a module homomorphism from M, which defines an algebra homomorphism from Sym(M) by its universal property (just extend it linearly and multiplicatively). Also note that your argument doesn't really apply, since m_1m_2 won't be in M; it will be a degree 2 element in Sym(M). – Andrew Critch Nov 24 '09 at 8:58

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.

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This argument looks very cute but I do not believe it. If Sym(M) and Sym(N) are isomorphic as A-algebras then (after fixing an isomorphism) we get a natural isomorphism between the functors Hom_{A-mod}(M,-) and Hom_{A-mod}(N,-) from the category of A-algebras to the category of sets. But in general this natural isomorphism will not be an isomorphism of groups! [check it: set M=N=A so Sym(M)=Sym(N)=A[T], and now choose the isomorphism sending T to T+a with a non-zero in A]. Hence "ker" in the answer above isn't as well-behaved as we would like and I don't think this proof works. – Kevin Buzzard Nov 23 '09 at 13:35
It would work if the isomorphism Sym(M)~Sym(N) was compatible with the augmentations Sym(M)->A, but in this case we could recover M as the conormal sheaf anyway. – user2035 Nov 23 '09 at 14:45
Translations are the only problem: the composition Sym(M)->Sym(N)->A defines a group element in Spec(Sym(M))(A), hence by translation an automorphism of Sym(M), and composing with this automorphism makes Sym(M)~Sym(N) compatible with the augmentations. – user2035 Nov 23 '09 at 15:06
There is too much missing in the "Translations are the only problem..." comment for me to be able to follow it. I'm not saying it's wrong but can we back up a bit? What is a group element in Spec(Sym(M))(A) and why is our given map a group element? – Kevin Buzzard Nov 23 '09 at 15:32
It's OK, I figured out what you meant myself now. Very nice! – Kevin Buzzard Nov 23 '09 at 16:15

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.

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The map from the A-alg homs to the A-module homs is just "M lives in Sym(M) and there's a natural A-module projection from Sym(N) onto N". I think you're better off not picking generators---just imagine that M is generated by M. What I am confused about is that I don't think your argument is sufficiently functorial. For example given maps Sym(M)->Sym(N)->Sym(P), does your argument produce a commuting triangle of maps M->N->P? I don't see immediately that this is the case. – Kevin Buzzard Nov 23 '09 at 13:24
[and hence I don't see why your map M->N is definitely an isomorphism; that's the point of my comment.] – Kevin Buzzard Nov 23 '09 at 13:25
Here's an example. Consider the map f: Sym(k) = k[x] --> k[x] that takes x to x^2. The linear part of this map is 0. Consider also the map g: k[x] --> k[x] that takes x to x+1. The linear part of this map is the identity. So the composition of the linear parts is 0. However, the linear part of the composition gf, which takes x to 1 + 2x + x^2, is multiplication by 2. So this answer is not functorial, as buzzard points out, and is wrong. – Mark Hovey Nov 23 '09 at 15:22
I see that I was wrong. Sorry! – auniket Nov 24 '09 at 0:22
Your argument is close to working though. If you "remove the noise in degree 0" it's fine. – Kevin Buzzard Nov 24 '09 at 0:36

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