Assume $A$ is a noetherian local ring with $\mathfrak{m}_y$ being the unique maximal ideal and $\dim A=0$. We have the exact sequence $$0\to \mathfrak{m}_y\to A\to k(y)\to 0,$$ where $k(y)$ is the residue field $A_{\mathfrak{m}_y}/\mathfrak{m}_y$.

Let $B$ be a finitely generated flat $A$-algebra by $\phi:A\to B$, then $$0\to \mathfrak{m}_y\otimes_A B\to B\to k(y)\otimes_A B\to 0$$ is an exact sequence of $A$-algebra. From the fact that $\dim A=0$, $\mathfrak{m}_y$ is nilpotent.

Let $\phi^{-1}(\mathfrak{p}_x)=\mathfrak{m}_y$ for a prime ideal $\mathfrak{p}_x$ in $B$. My question is: Why would this implies $$\dim (k(y)\otimes_A B)_{\mathfrak{p}_x}=\dim B_{\mathfrak{p}_x}?$$ An approach that I can think of is to consider the dimension of the objects in the second exact sequence above. But is it true that $\dim (\mathfrak{m}_{y}\otimes_A B)_{\mathfrak{p}_x}=0$ and does it make sense?

This question will be helpful in understanding Hartshorne III.9.5, the dimension formula of fiber of flat morphism.

  • $\begingroup$ updated, it just mean localization at $\mathfrak{p}_x$ $\endgroup$
    – Ivan So
    Jan 6 at 20:39
  • $\begingroup$ I think you need to use modding out by a nilpotent ideal does not change the dimension. $\endgroup$
    – Youngsu
    Jan 7 at 2:40
  • $\begingroup$ @Youngsu Explicitly? $\endgroup$
    – Ivan So
    Jan 7 at 8:29

As mentioned in the comments you need to use the fact that modding out by a niplotent ideal does not change the dimension. The reason is that a nilpotent ideal is contained in every prime ideal, or in other words the topological space defined by taking $\operatorname{Spec}$ are isomorphic (but not as schemes!).

Now $k(y)\otimes_A B\cong B/\mathfrak{m}_yB$ and $\mathfrak{m}_yB$ is nilpotent in $B$.

Thus your result follows.


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