$A\rightarrow B$ a ring homomorphism of Noetherian rings, where $A$ is local. $M$, $N$ finitely generated and nonzero $A$- and $B$- modules, respectively. Then I seem to get $\mbox{dim}_ {B}(M\otimes_{A} N) = \mbox{dim}_ {B}N$. Could that be true? It seems a little strange that the dimension of $(M\otimes_{A} N)$ (as a $B$-module) is independent of $M$.
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1$\begingroup$ But isn't the (Krull) dimension of any vector space 0? $\endgroup$– ashpoolJun 25, 2010 at 23:18
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1$\begingroup$ Let $k$ be the residue field of $A$ such that dim A>0. Take N=B=A, M=k. Then LSH=0, RHS = dim A. $\endgroup$– Hailong DaoJun 25, 2010 at 23:59
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$\begingroup$ I think my mistake stems from not recognizing the fact that if (A,m) is a local ring and p is a non-maximal prime, A localized at p is not finitely generated as an A-module. $\endgroup$– ashpoolJun 26, 2010 at 14:02
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1$\begingroup$ The Krull dimension of a vector space is definitely not zero. The Krull dimension of $k[X]$ is $1$ ($k$ a field) for instance. You must be thinking finite-dimensional vector spaces. A module finite $k$-algebra is artinian, hence zero-dimensional. This is because it is a finitely-generated module over an artin ring, so it satisfies DCC on $k$-subspaces, and any ideal is clearly a $k$-subspace. $\endgroup$– Keenan KidwellJun 26, 2010 at 18:30
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$\begingroup$ The Krull dimension of k[x] as a module over itself(i.e. as a ring) is 1. But as a vector space(i.e. as a k-module) it is zero, since the dimension of a field (as a ring) is zero. $\endgroup$– ashpoolJun 29, 2010 at 0:34
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See Bruns and Herzog A.5(b) and A.11(b).
