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Let $A,N$ be $A$-modules. I'm trying to prove $A^{\oplus n} \otimes_A N \cong N^{\oplus n}$. First we define $f: F(A,N) \rightarrow N^{\oplus n}$ where $F(A,N)$ is a free module with a basis $e_{a,\nu}, a \in A, \nu \in N$. We see that $f: e_{(a_1,...,a_n),\nu} \mapsto (a_1\nu,...,a_n\nu), a_i \in A, \nu \in N$ is a correctly defined homomorphism, because $e_{\alpha_1(\alpha_2+\alpha_3),\nu}-e_{\alpha_1\alpha_2,\nu}-e_{\alpha_1\alpha_3,\nu} \mapsto (0,...,0), \\ e_{\alpha,\nu_1(\nu_2+\nu_3)}-e_{\alpha,\nu_1\nu_2}-e_{\alpha,\nu_1\nu_3} \mapsto (0,...,0), \\ e_{a\alpha,\nu}- a e_{\alpha,\nu}\mapsto(0,...,0),\\ e_{\alpha,a\nu} - ae_{\alpha,\nu}\mapsto(0,...,0).$

Next we need to show that $f$ is an isomorphism. But how to do that?

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Do you know the universal property of the tensor product of $A$-modules? It yields, in particular, that the map defined by $(a_1,...,a_n)\otimes x\mapsto(a_1x,...,a_nx)$ is well-defined. An inverse is then given by $(x_1,...,x_n)\mapsto(1,0,...,0)\otimes x_1+...+(0,...,0,1)\otimes x_n$. The map does the same that donkey explained in his answer, I just thought you might be interested in how you can define explicitly. – InvisiblePanda Apr 20 '12 at 20:09
@Rand al'Thor: +1, but I am no donkey. – Jacob Bell Apr 20 '12 at 20:40
@Randal'Thor, thanks for your remark. But I don't understand how it follows from the universal property. Could you elaborate on that, please? – Sergey Filkin Apr 22 '12 at 8:05
I think he's alluding to the fact that maps $M \otimes N \to P$ correspond to bilinear maps $M \times N \to P$, and the morphism he defined is clearly bilinear. (it's a good point: when considering morphisms out of a tensor product, instead of taking a concrete model (for which you then have to bother with generators and relations), you only use its universal property (which automatically gives you well-defined morphisms). – Jacob Bell Apr 22 '12 at 10:33
@donkey kong: Sorry, I'm not a native English speaker, and after reading it again, I think it was a bit wrong there ;) don't know how to really fix it though, maybe replace "that" by "thing" (and also add a kong behind your name) ;) – InvisiblePanda Apr 26 '12 at 5:57
up vote 3 down vote accepted

In general $(M \oplus N) \otimes P \simeq (M \otimes P) \oplus (N \otimes P)$, one direction is as you say $(m,n) \otimes p \mapsto (m \otimes p, n \otimes p)$ with inverse $(m \otimes p_1, n \otimes p_2) \mapsto (m,0)\otimes p_1 + (0,n)\otimes p_2$.

In your case you also use $A \otimes_A M \simeq M$, that's given by $a \otimes m \mapsto am$ and $m \mapsto 1 \otimes m$.

makes sense?

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sure. Thank you. – Sergey Filkin Apr 20 '12 at 20:05

The easiest way is to show that $A^{\oplus n} \otimes_A N$ has the universal property of $N^{\oplus n}$. Indeed, by the universal property of $- \otimes_A N$, we have $\newcommand{\Hom}{\textrm{Hom}}$ $$\begin{split} \Hom (A^{\oplus n} \otimes_A N, M) & \cong \Hom_A (A^{\oplus n}, \Hom (N, M)) \\ & \cong \Hom_A (A, \Hom (N, M))^{\times n} \\ & \cong \Hom (N, M)^{\times n} \\ & \cong \Hom (N^{\oplus n}, M) \end{split}$$ Hence, by the Yoneda lemma, $A^{\oplus n} \otimes_A N \cong N^{\oplus n}$. The best part about this proof is that $n$ can be an infinite cardinal, and $A$ does not need not be commutative.

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thanks for your argument, but it's too advanced for me at the moment. But I will recall your answer later. – Sergey Filkin Apr 22 '12 at 8:08

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