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We say a homomorphism of Noetherian rings $\varphi:A\rightarrow B$ is regular if $\varphi$ is flat and for every prime ideal $p$ of $A$, the fiber ring $B\otimes_Ak(p)$ is geometrically regular over $k(p)$, that is, for every finite field extension $L$ of $k(p)$, the ring $B\otimes_AL$ is regular.

My question is: Why $B\otimes_AL$ is a finite $B$-module, or why $B\otimes_Ak(p)$ is a finite $B$-module?

Any comments or answers will be appreciated!

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$k(p)$ being...the field of fractions of $A/(p)$? –  Kevin Carlson Jul 30 '13 at 10:37
    
yes, $k(p)$ is the field of fractions of $A/p$, or $S^{-1}A/pS^{-1}A$ where $S=A-p$. –  nick Jul 30 '13 at 13:44
    
I do not understand. If A=B, then $B\otimes_A L=L$, which is not necessarily finite module over $B$. –  Marci Jul 30 '13 at 14:02
    
thanks, if $A=B$, $L$ may not be a finite module over $A$. In the proof of Theorem 32.1 in Matsumura's book "commutative ring theory", there is a statement "since $B_L$ is a finite $B$-module", maybe it is wrong. –  nick Jul 30 '13 at 16:08
    
To be honest, I do not understand the proof of Theorem 32.1 –  Marci Jul 31 '13 at 1:20

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