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Thread: Question on proof

  1. #1
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    Question on proof

    Here is a proof I am reading for class:

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

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

    I don't understand the part in red. Can somebody explain why $\displaystyle (k \otimes_A N)^n \cong k^n \otimes_A N $?
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  2. #2
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    Quote Originally Posted by xianghu21 View Post
    Here is a proof I am reading for class:

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

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

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