Lemma 10.116.7. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $k \subset K$ be a field extension. Set $S_ K = K \otimes _ k S$. Let $\mathfrak q \subset S$ be a prime and let $\mathfrak q_ K \subset S_ K$ be a prime lying over $\mathfrak q$. Then

$\dim (S_ K \otimes _ S \kappa (\mathfrak q))_{\mathfrak q_ K} = \dim (S_ K)_{\mathfrak q_ K} - \dim S_\mathfrak q = \text{trdeg}_ k \kappa (\mathfrak q) - \text{trdeg}_ K \kappa (\mathfrak q_ K)$

Moreover, given $\mathfrak q$ we can always choose $\mathfrak q_ K$ such that the number above is zero.

Proof. Observe that $S_\mathfrak q \to (S_ K)_{\mathfrak q_ K}$ is a flat local homomorphism of local Noetherian rings with special fibre $(S_ K \otimes _ S \kappa (\mathfrak q))_{\mathfrak q_ K}$. Hence the first equality by Lemma 10.112.7. The second equality follows from the fact that we have $\dim _ x X = \dim _{x_ K} X_ K$ with notation as in Lemma 10.116.6 and we have $\dim _ x X = \dim S_\mathfrak q + \text{trdeg}_ k \kappa (\mathfrak q)$ by Lemma 10.116.3 and similarly for $\dim _{x_ K} X_ K$. If we choose $\mathfrak q_ K$ minimal over $\mathfrak q S_ K$, then the dimension of the fibre ring will be zero. $\square$

There are also:

• 2 comment(s) on Section 10.116: Dimension of finite type algebras over fields, reprise

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).