Lemma 15.6.7. Let $A, A', B, B', I, M, M', N, \varphi $ be as in Lemma 15.6.4. If $N$ finite over $B$ and $M'$ finite over $A'$, then $N' = N \times _{\varphi , M} M'$ is finite over $B'$.

**Proof.**
We will use the results of Lemma 15.6.4 without further mention. Choose generators $y_1, \ldots , y_ r$ of $N$ over $B$ and generators $x_1, \ldots , x_ s$ of $M'$ over $A'$. Using that $N = N' \otimes _{B'} B$ and $B' \to B$ is surjective we can find $u_1, \ldots , u_ r \in N'$ mapping to $y_1, \ldots , y_ r$ in $N$. Using that $M' = N' \otimes _{B'} A'$ we can find $v_1, \ldots , v_ t \in N'$ such that $x_ i = \sum v_ j \otimes a'_{ij}$ for some $a'_{ij} \in A'$. In particular we see that the images $\overline{v}_ j \in M'$ of the $v_ j$ generate $M'$ over $A'$. We claim that $u_1, \ldots , u_ r, v_1, \ldots , v_ t$ generate $N'$ as a $B'$-module. Namely, pick $\xi \in N'$. We first choose $b'_1, \ldots , b'_ r \in B'$ such that $\xi $ and $\sum b'_ i u_ i$ map to the same element of $N$. This is possible because $B' \to B$ is surjective and $y_1, \ldots , y_ r$ generate $N$ over $B$. The difference $\xi - \sum b'_ i u_ i$ is of the form $(0, \theta )$ for some $\theta $ in $IM'$. Say $\theta $ is $\sum t_ j\overline{v}_ j$ with $t_ j \in I$. As $J = \mathop{\mathrm{Ker}}(B' \to B)$ maps isomorphically to $I$ we can choose $s_ j \in J \subset B'$ mapping to $t_ j$. Because $N' = N \times _{\varphi , M} M'$ it follows that $\xi = \sum b'_ i u_ i + \sum s_ j v_ j$ as desired.
$\square$

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