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.

notbe 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