Lemma 10.112.6. Let $R \to S$ be a homomorphism of Noetherian rings. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$. Then

$\dim (S_{\mathfrak q}) \leq \dim (R_{\mathfrak p}) + \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}).$

Proof. We use the characterization of dimension of Proposition 10.60.8. Let $x_1, \ldots , x_ d$ be elements of $\mathfrak p$ generating an ideal of definition of $R_{\mathfrak p}$ with $d = \dim (R_{\mathfrak p})$. Let $y_1, \ldots , y_ e$ be elements of $\mathfrak q$ generating an ideal of definition of $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ with $e = \dim (S_{\mathfrak q}/\mathfrak pS_{\mathfrak q})$. It is clear that $S_{\mathfrak q}/(x_1, \ldots , x_ d, y_1, \ldots , y_ e)$ has a nilpotent maximal ideal. Hence $x_1, \ldots , x_ d, y_1, \ldots , y_ e$ generate an ideal of definition of $S_{\mathfrak q}$. $\square$

Comment #4320 by Zhiyu Zhang on

A little typo: in the last sentence, "generate an ideal of definition if" shall be " generate an ideal of definition of".

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