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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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    @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) ;)2012-04-26

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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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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.2012-04-22