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.



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


