Lemma 10.40.6. Let $R \to R'$ be a ring map and let $M$ be a finite $R$-module. Then $\text{Supp}(M \otimes _ R R')$ is the inverse image of $\text{Supp}(M)$.
Proof. Let $\mathfrak p \in \text{Supp}(M)$. By Nakayama's lemma (Lemma 10.20.1) we see that
\[ M \otimes _ R \kappa (\mathfrak p) = M_\mathfrak p/\mathfrak p M_\mathfrak p \]
is a nonzero $\kappa (\mathfrak p)$ vector space. Hence for every prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$ we see that
\[ (M \otimes _ R R')_{\mathfrak p'}/\mathfrak p' (M \otimes _ R R')_{\mathfrak p'} = (M \otimes _ R R') \otimes _{R'} \kappa (\mathfrak p') = M \otimes _ R \kappa (\mathfrak p) \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p') \]
is nonzero. This implies $\mathfrak p' \in \text{Supp}(M \otimes _ R R')$. For the converse, if $\mathfrak p' \subset R'$ is a prime lying over an arbitrary prime $\mathfrak p \subset R$, then
\[ (M \otimes _ R R')_{\mathfrak p'} = M_\mathfrak p \otimes _{R_\mathfrak p} R'_{\mathfrak p'}. \]
Hence if $\mathfrak p' \in \text{Supp}(M \otimes _ R R')$ lies over the prime $\mathfrak p \subset R$, then $\mathfrak p \in \text{Supp}(M)$. $\square$
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