Lemma 47.26.3. Let $(A, \mathfrak m, \kappa ) \to (B, \mathfrak n, l)$ be a flat local homorphism of Noetherian rings such that $\mathfrak n = \mathfrak m B$. If $E$ is the injective hull of $\kappa$, then $E \otimes _ A B$ is the injective hull of $l$.

Proof. Write $E = \bigcup E_ n$ as in Lemma 47.7.3. It suffices to show that $E_ n \otimes _{A/\mathfrak m^ n} B/\mathfrak n^ n$ is the injective hull of $l$ over $B/\mathfrak n$. This reduces us to the case where $A$ and $B$ are Artinian local. Observe that $\text{length}_ A(A) = \text{length}_ B(B)$ and $\text{length}_ A(E) = \text{length}_ B(E \otimes _ A B)$ by Algebra, Lemma 10.51.13. By Lemma 47.6.1 we have $\text{length}_ A(E) = \text{length}_ A(A)$ and $\text{length}_ B(E') = \text{length}_ B(B)$ where $E'$ is the injective hull of $l$ over $B$. We conclude $\text{length}_ B(E') = \text{length}_ B(E \otimes _ A B)$. Observe that

$\dim _ l((E \otimes _ A B)[\mathfrak n]) = \dim _ l(E[\mathfrak m] \otimes _ A B) = \dim _\kappa (E[\mathfrak m]) = 1$

where we have used flatness of $A \to B$ and $\mathfrak n = \mathfrak mB$. Thus there is an injective $B$-module map $E \otimes _ A B \to E'$ by Lemma 47.7.2. By equality of lengths shown above this is an isomorphism. $\square$

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