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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