# Finding the dimension of certain tensor product with flat $A$-algebra

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.

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