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Math Help - Question on proof

  1. #1
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    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 ?
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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 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.
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