
Lemma 10.59.12. Suppose that $R$ is a Noetherian local ring and $x\in \mathfrak m$ an element of its maximal ideal. Then $\dim R \leq \dim R/xR + 1$. If $x$ is not contained in any of the minimal primes of $R$ then equality holds. (For example if $x$ is a nonzerodivisor.)

Proof. If $x_1, \ldots , x_{\dim R/xR} \in R$ map to elements of $R/xR$ which generate an ideal of definition for $R/xR$, then $x, x_1, \ldots , x_{\dim R/xR}$ generate an ideal of definition for $R$. Hence the inequality by Proposition 10.59.8. On the other hand, if $x$ is not contained in any minimal prime of $R$, then the chains of primes in $R/xR$ all give rise to chains in $R$ which are at least one step away from being maximal. $\square$

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