# Extension of rings decreasing Krull dimension

Let $A \subset B$ a ring extension. It is well known that if the extension is integral, then $\dim B=\dim A$. I can think of some examples where the Krull dimension increases (and by that I mean $\dim B>\dim A$), like $A \subset A[X]$ for any ring $A$. And this situation looks a bit natural or intuitive to me, even though Krull dimension has nothing to do with others notions of dimension I'm used to work with, like the dimension of a vector space. In the few exemples of extension decreasing the Krull dimension I've seen, $B$ is a field.

So here is my question:

can you give some examples of extensions decreasing Krull dimension where $B$ is not a field?

• Well, how about $\mathbb{Z}[x]\subset\mathbb{Q}[x]$? – Amitai Yuval Jul 17 '15 at 13:46

Here you are: let $n\ge 1$. Then for any $k\enspace(0\le k <n)$: $$A=\mathbf Z[X_1,\dots, X_n]\subset B=\mathbf Q(X_1,\dots,X_k)[X_{k+1},\dots, X_n]$$ satisfies $\dim A=n+1,\enspace\dim B=n-k$.
Given a ring $A$, the localization $A_{\mathfrak{p}}$, for $\mathfrak{p}$ a prime ideal, has $\dim A_{\mathfrak{p}}=\operatorname{ht}\mathfrak{p}$. So for example, $\mathbb{Z}[x]\to \mathbb{Q}[x]$ goes from $2$ to $1$.