Lemma 10.114.7. Let $k$ be a field. Let $S$ be a finite type $k$ algebra. Assume that $S$ is Cohen-Macaulay. Then $\mathop{\mathrm{Spec}}(S) = \coprod T_ d$ is a finite disjoint union of open and closed subsets $T_ d$ with $T_ d$ equidimensional (see Topology, Definition 5.10.5) of dimension $d$. Equivalently, $S$ is a product of rings $S_ d$, $d = 0, \ldots , \dim (S)$ such that every maximal ideal $\mathfrak m$ of $S_ d$ has height $d$.

**Proof.**
The equivalence of the two statements follows from Lemma 10.24.3. Let $\mathfrak m \subset S$ be a maximal ideal. Every maximal chain of primes in $S_{\mathfrak m}$ has the same length equal to $\dim (S_{\mathfrak m})$, see Lemma 10.104.3. Hence, the dimension of the irreducible components passing through the point corresponding to $\mathfrak m$ all have dimension equal to $\dim (S_{\mathfrak m})$, see Lemma 10.114.4. Since $\mathop{\mathrm{Spec}}(S)$ is a Jacobson topological space the intersection of any two irreducible components of it contains a closed point if nonempty, see Lemmas 10.35.2 and 10.35.4. Thus we have shown that any two irreducible components that meet have the same dimension. The lemma follows easily from this, and the fact that $\mathop{\mathrm{Spec}}(S)$ has a finite number of irreducible components (see Lemmas 10.31.3 and 10.31.5).
$\square$

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