Lemma 10.114.4. Let k be a field. Let S be a finite type k-algebra which is an integral domain. Then \dim (S) = \dim (S_{\mathfrak m}) for any maximal ideal \mathfrak m of S. In words: every maximal chain of primes has length equal to the dimension of S.
Proof. Write S = k[x_1, \ldots , x_ n]/\mathfrak p. By Proposition 10.114.2 and Lemma 10.114.3 all the maximal chains of primes in S (which necessarily end with a maximal ideal) have length n - \text{height}(\mathfrak p). Thus this number is the dimension of S and of S_{\mathfrak m} for any maximal ideal \mathfrak m of S. \square
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