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7
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edited Dec 2 2009 at 1:06
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Can different modules have the same symmetric algebra? / Is it safe to think of modules geometrically? (answered: no/ yes)
Algebraic geometry inspires allows one to think of an $A$-module $M$ geometrically as a module of functions --- precisely speaking, 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.
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edited Nov 27 2009 at 9:26
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Can different modules have the same symmetric algebra? / Is it safe to think of modules geometrically? (answered: no / yes)
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5
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edited Nov 23 2009 at 20:44
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Algebraic geometry inspires one to think of an $A$-module $M$ geometrically as a module of functions --- precisely speaking, 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)
$Hom_{A-alg}(\mathrm{Sym}(M),B) hom_{A-alg}(\mathrm{Sym}(M),B) \simeq Hom_{A\mathrm{-mod}}(M,B),$hom_{A\mathrm{-mod}}(M,B),$
by Yoneda's lemma, an equivalent question would be:
(2) If the (set-valued) functors $Hom_{A-\mathrm{mod}}(M,-)$ hom_{A-\mathrm{mod}}(M,-)$ and $Hom_{A-\mathrm{mod}}(N,-)$ 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.
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4
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edited Nov 23 2009 at 9:15
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Can different modues modules have the same symmetric algebra? / Is it safe to think of modules geometrically?
Algebraic geometry inspires one to think of an $A$-module $M$ geometrically as a module of functions --- precisely speaking, 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
$Hom_{A-alg}(\mathrm{Sym}(M),B) \simeq Hom_{A\mathrm{-mod}}(M,B),$
by Yoneda's lemma, an equivalent question would be:
(2) If the functors $Hom_{A-\mathrm{mod}}(M,-)$ and $Hom_{A-\mathrm{mod}}(N,-)$ agree on $A$-algbras, do they agree on $A$-modules?
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3
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edited Nov 23 2009 at 8:33
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Algebraic geometry inspires one to think of an $A$-module $M$ geometrically as a module of functions --- precisely speaking, 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:
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
$2 \mathrm{Hom}{A-alg}(\mathrm{Sym}(M),B) Hom_{A-alg}(\mathrm{Sym}(M),B) \simeq \mathrm{Hom}{A\mathrm{-mod}}(M,B),$Hom_{A\mathrm{-mod}}(M,B),$
by Yoneda's lemma, an equivalent question would be:
If the functors $\mathrm{Hom}{A-\mathrm{mod}}(M,-)$ Hom_{A-\mathrm{mod}}(M,-)$ and $\mathrm{Hom}{A-\mathrm{mod}}(N,-)$ Hom_{A-\mathrm{mod}}(N,-)$ agree on $A$-algbras, do they agree on $A$-modules?
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2
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edited Nov 23 2009 at 8:28
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Algebraic geometry inspires one to think of an $A$-module $M$ geometrically as a module of functions --- precisely speaking, 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:
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
$\mathrm{Hom}2 \mathrm{Hom}{A\mathrm{-alg}}(\mathrm{Sym}(M),B) A-alg}(\mathrm{Sym}(M),B) \simeq \mathrm{Hom}{A\mathrm{-mod}}(M,B),$
by Yoneda's lemma, an equivalent question would be:
If the functors $\mathrm{Hom}{A\mathrm{-mod}}(M,-)$ A-\mathrm{mod}}(M,-)$ and $\mathrm{Hom}{A\mathrm{-mod}}(N,-)$ A-\mathrm{mod}}(N,-)$ agree on $A$-algbras, do they agree on $A$-modules?
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asked Nov 23 2009 at 8:21
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Can different modues have the same symmetric algebra? / Is it safe to think of modules geometrically?
Algebraic geometry inspires one to think of an $A$-module $M$ geometrically as a module of functions --- precisely speaking, 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:
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
$\mathrm{Hom}{A\mathrm{-alg}}(\mathrm{Sym}(M),B) \simeq \mathrm{Hom}{A\mathrm{-mod}}(M,B),$
by Yoneda's lemma, an equivalent question would be:
If the functors $\mathrm{Hom}{A\mathrm{-mod}}(M,-)$ and $\mathrm{Hom}{A\mathrm{-mod}}(N,-)$ agree on $A$-algbras, do they agree on $A$-modules?
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