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On page 118 of Matsumura it is said that it is not true in general that $A[[X]]\otimes_A k(\mathfrak p)$ is isomorphic to $k(\mathfrak p)[[X]]$ where $\mathfrak p$ is a prime ideal of $A$ and $k(\mathfrak p)=A_{\mathfrak p}/\mathfrak pA_{\mathfrak p}$. Could you give me an example?

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Which Matsumura, Commutative Ring Theory or Commutative Algebra? –  Zev Chonoles Jan 26 '13 at 5:44
    
commutative ring theory –  Blu Jan 26 '13 at 9:56
    
See my answer at math.stackexchange.com/questions/274803 –  Martin Brandenburg Jan 26 '13 at 12:09
    
@MartinBrandenburg I see now why Matsumura's book has to mean that the isomorphism must be the canonical one. –  user26857 Jan 26 '13 at 12:36

1 Answer 1

Take $A=\mathbb Z$ and $\mathfrak p=(0)$. Then one has to prove that $\mathbb Z[[X]]\otimes_{\mathbb Z}\mathbb Q$ is not isomorphic to $\mathbb Q[[X]]$. In order to do this note first that $\mathbb Z[[X]]\otimes_{\mathbb Z}\mathbb Q$ is isomorphic to the fraction ring $S^{-1}\mathbb Z[[X]]$, where $S=\mathbb Z-\{0\}$. The ring $\mathbb Q[[X]]$ is local, while $S^{-1}\mathbb Z[[X]]$ is not.

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One can also observe that $\sum_{n\ge 0} 2^{-n}X^n\in \mathbb Q[[X]]$ is not in $\mathbb Z[[X]]\otimes_{\mathbb Z} \mathbb Q$ because the latter only contains formal powers series with bounded denominators. –  user18119 Jan 26 '13 at 11:55
    
@QiL An isomorphism could be different from the canonical one. –  user26857 Jan 26 '13 at 11:57
    
But of course Matsumura means that the canonical map $A[[X]] \otimes_A K \to K[[X]]$ is not an isomorphism. –  Martin Brandenburg Jan 26 '13 at 12:08
    
@MartinBrandenburg How do you know this? I've checked the book and he says exactly what the OP has written above. –  user26857 Jan 26 '13 at 12:11
    
Anyway I upvoted the answer because it proves there is no isomorphism at all. –  user18119 Jan 26 '13 at 18:33

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