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## Commutative algebras and Gamma-modules

A commutative algebra (with unity) over a field gives rise to the covariant functor F: Set_f->Vect from finite sets to vector spaces: F(E) := A^{otimes E}. Is it true that, over complex numbers, a finite dimensional algebra can be reconstructed from the corresponding functor?

(A Gamma-module is a functor from finite pointed sets to vector spaces; so F is not a Gamma-module. I use this term in the title just because I do not know the correct term for F: Set_f->Vect.)

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How is F defined on arrows? Like, what is the morphism in Vect corresponding to the only function {1,2}->{1}? – Mattia Talpo Oct 7 2010 at 18:53
I think it would help if (a) you were more explicit about how you define $F$ (b) you explained what you mean by reconstructing the algebra from the functor. I think I can guess the answer to both questions but it is hard to be sure I'm right. – Simon Wadsley Oct 7 2010 at 18:57

Now I can prove this, see http://www.pdmi.ras.ru/~ssp/te.pdf

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Let me clarify my question. For a commutative algebra $A$ we define a functor $F:\mathrm{Set}_\mathrm{f}\to\mathrm{Vect}$ by

$F(I)=A^{\otimes I}$ for a finite set $I$ and

$F(t):F(I)\to F(J)$, $\bigotimes_{i\in I}a_i\mapsto\bigotimes_{j\in J}\prod_{i\in t^{-1}(j)}a_i$ for a map $t:I\to J$ (exactly as Andreas Blass proposed).

Suppose now that two finite-dimensional algebras $A$ and $B$ over the complex numbers produce isomorphic functors $F$ and $G$. Is it true that then $A$ and $B$ are isomorphic?

The question is not trivial. Let $e:F\to G$ be an isomorphism of functors. Then $e_{\{1\}}:A\to B$ and $e_{\{1,2\}}:A\otimes A\to B\otimes B$ are isomorphisms of vector spaces. If we had $e_{\{1,2\}}=e_{\{1\}}\otimes e_{\{1\}}$, this would imply that $e_{\{1\}}$ is an isomorphism of algebras. The problem is that we have only linear naturality relations between $e_I$.

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I think you should have edited your question rather than posting this as an answer. Surely Andreas Blass has correctly answered this. Because the multiplication in $A$ is given by $F$ applied to the surjection from $[2]$ to $[1]$ your map $e_{[1]}$ is an iso of rings. This works for any field and f.d. doesn't make any difference either. – Simon Wadsley Oct 8 2010 at 13:13
Use naturality of $e$ with respect to the map {1,2} -> {1} to see that the map A -> B actually preserves products. – Martin Brandenburg Oct 8 2010 at 13:37
I agree with Semen, naturality only seems to give you $m(e_{[2]}(a\otimes a'))=e_{[1]}(aa')$, where $m\colon B\bigotimes B\to B$ is the multiplication. Further naturality gives $e_{[2]}(a\otimes1)=e_{[1]}(a)\otimes1$ but beyond that I don't see how to get something of interest. – Torsten Ekedahl Oct 8 2010 at 18:01

I'd guess that the intended functor uses the multiplication operation of $A$ to provide $F(s):A\otimes A\to A$ where $s$ is the surjection $\{1,2\}\to\{1\}$ mentioned by Mattia Talpo, that it uses the unity element of $A$ to provide $F(i):k\to A$ where $i$ is the injection $\emptyset\to\{1\}$ and $k$ is the field of scalars, and that $F$ is to be defined on arbitrary maps between finite sets by a natural generalization and combination of these two examples. The trouble with this guess is that it makes the question trivial, since the algebra structure is contained in $F(s)$ and $F(i)$, so no work is needed to reconstruct the algebra from $F$. Unfortunately, I have no alternative guess as to what $F$ the proposer might have intended.

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 My thoughts exactly :) – Mattia Talpo Oct 7 2010 at 23:59 @Andreas: I also thought that $F$ has this action on morphisms, but since $ab \otimes c$ and $a \otimes bc$ are not equal in $A^{\otimes 2}$, I doubt that this is functorial. – Martin Brandenburg Oct 8 2010 at 7:19