Lemma 10.60.13. 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 10.60.9. 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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