TITLE: Can different modules have the same symmetric algebra? (answered: no)
QUESTION [11 upvotes]: 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.

REPLY [6 votes]: 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.