Lemma 15.6.5. In the situation of Lemma 15.6.4 for a $B'$-module $L'$ the adjunction map

$L' \longrightarrow (L' \otimes _{B'} B) \times _{(L' \otimes _{B'} A)} (L' \otimes _{B'} A')$

is surjective but in general not injective.

Proof. As in the proof of Lemma 15.6.4 let $J \subset B'$ be the kernel of the map $B' \to B$. Then $L' \otimes _{B'} B = L'/JL'$. Hence to prove surjectivity it suffices to show that elements of the form $(0, z)$ of the fibre product are in the image of the map of the lemma. The kernel of the map $L' \otimes _{B'} A' \to L' \otimes _{B'} A$ is the image of $L' \otimes _{B'} I \to L' \otimes _{B'} A'$. Since the map $J \to I$ induced by $B' \to A'$ is an isomorphism the composition

$L' \otimes _{B'} J \to L' \to (L' \otimes _{B'} B) \times _{(L' \otimes _{B'} A)} (L' \otimes _{B'} A')$

induces a surjection of $L' \otimes _{B'} J$ onto the set of elements of the form $(0, z)$. To see the map is not injective in general we present a simple example. Namely, take a field $k$, set $B' = k[x, y]/(xy)$, $A' = B'/(x)$, $B = B'/(y)$, $A = B'/(x, y)$ and $L' = B'/(x - y)$. In that case the class of $x$ in $L'$ is nonzero but is mapped to zero under the displayed arrow. $\square$

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