1. ## Question on proof

Here is a proof I am reading for class:

Let $A$ be a local ring, $M$ and $N$ finitely generated $A$-modules. If $M$ $\otimes_A N=0$, then $M=0$ or $N=0$.

Proof. Let $\mathfrak{m}$ be a the maximal ideal of $A$ and $k=\frac{A}{\mathfrak{m}}$ be the residue field. We may assume that $M \not = 0$ and prove that $N=0.$ By Nakayama's lemma, we see that $\mathfrak{m}M \subsetneq M$. Therefore $\frac{A}{\mathfrak{m}} \otimes M \cong \frac{M}{\mathfrak{m}M} \cong k^n \not = 0$ is naturally a $k$-vector space with rank $n>0$. Hence $M$ $\otimes_A N =0$ $\Rightarrow \frac{A}{\mathfrak{m}} \otimes (M \otimes_A N)=0 \text{ }$ $\Rightarrow {\color{red}(k \otimes_A N)^n \cong k^n \otimes_A N }\cong (\frac{A}{\mathfrak{m}} \otimes M) \otimes_A N = 0 \text{ }$ $\Rightarrow \frac{N}{\mathfrak{m}N}\cong \frac{A}{\mathfrak{m}} \otimes N= k \otimes_A N =0 \text{ }$ $\Rightarrow N=\mathfrak{m}N \text{ }$ $\Rightarrow N=0$ by Nakayama's Lemma. $\Box$

I don't understand the part in red. Can somebody explain why $(k \otimes_A N)^n \cong k^n \otimes_A N$?

2. Originally Posted by xianghu21
Here is a proof I am reading for class:

Let $A$ be a local ring, $M$ and $N$ finitely generated $A$-modules. If $M$ $\otimes_A N=0$, then $M=0$ or $N=0$.

Proof. Let $\mathfrak{m}$ be a the maximal ideal of $A$ and $k=\frac{A}{\mathfrak{m}}$ be the residue field. We may assume that $M \not = 0$ and prove that $N=0.$ By Nakayama's lemma, we see that $\mathfrak{m}M \subsetneq M$. Therefore $\frac{A}{\mathfrak{m}} \otimes M \cong \frac{M}{\mathfrak{m}M} \cong k^n \not = 0$ is naturally a $k$-vector space with rank $n>0$. Hence $M$ $\otimes_A N =0$ $\Rightarrow \frac{A}{\mathfrak{m}} \otimes (M \otimes_A N)=0 \text{ }$ $\Rightarrow {\color{red}(k \otimes_A N)^n \cong k^n \otimes_A N }\cong (\frac{A}{\mathfrak{m}} \otimes M) \otimes_A N = 0 \text{ }$ $\Rightarrow \frac{N}{\mathfrak{m}N}\cong \frac{A}{\mathfrak{m}} \otimes N= k \otimes_A N =0 \text{ }$ $\Rightarrow N=\mathfrak{m}N \text{ }$ $\Rightarrow N=0$ by Nakayama's Lemma. $\Box$

I don't understand the part in red. Can somebody explain why $(k \otimes_A N)^n \cong k^n \otimes_A N$?
Because $(k \otimes_A N)^{\oplus n} \cong k^{\oplus n} \otimes_A N$...direct sums commute with the tensor product.