Lemma 15.6.4. In Situation 15.6.1 the functor (15.6.3.1) has a right adjoint, namely the functor

$F : (N, M', \varphi ) \longmapsto N \times _{\varphi , M} M'$

where $M = M'/IM'$. Moreover, the composition of $F$ with (15.6.3.1) is the identity functor on $\text{Mod}_ B \times _{\text{Mod}_ A} \text{Mod}_{A'}$. In other words, setting $N' = N \times _{\varphi , M} M'$ we have $N' \otimes _{B'} B = N$ and $N' \otimes _{B'} A' = M'$.

Proof. The adjointness statement follows from the more general Lemma 15.5.4. To prove the final assertion, recall that $B' = B \times _ A A'$ and $N' = N \times _{\varphi , M} M'$ and extend these equalities to

$\vcenter { \xymatrix{ A & A' \ar[l] & I \ar[l] \\ B \ar[u] & B' \ar[l] \ar[u] & J \ar[l] \ar[u] } } \quad \text{and}\quad \vcenter { \xymatrix{ M & M' \ar[l] & K \ar[l] \\ N \ar[u]_\varphi & N' \ar[l] \ar[u] & L \ar[l] \ar[u] } }$

where $I, J, K, L$ are the kernels of the horizontal maps of the original diagrams. We present the proof as a sequence of observations:

1. $K = IM'$ (see statement lemma),

2. $B' \to B$ is surjective with kernel $J$ and $J \to I$ is bijective,

3. $N' \to N$ is surjective with kernel $L$ and $L \to K$ is bijective,

4. $JN' \subset L$,

5. $\mathop{\mathrm{Im}}(N \to M)$ generates $M$ as an $A$-module (because $N \otimes _ B A = M$),

6. $\mathop{\mathrm{Im}}(N' \to M')$ generates $M'$ as an $A'$-module (because it holds modulo $K$ and $L$ maps isomorphically to $K$),

7. $JN' = L$ (because $L \cong K = I M'$ is generated by images of elements $x n'$ with $x \in I$ and $n' \in N'$ by the previous statement),

8. $N' \otimes _{B'} B = N$ (because $N = N'/L$, $B = B'/J$, and the previous statement),

9. there is a map $\gamma : N' \otimes _{B'} A' \to M'$,

10. $\gamma$ is surjective (see above),

11. the kernel of the composition $N' \otimes _{B'} A' \to M' \to M$ is generated by elements $l \otimes 1$ and $n' \otimes x$ with $l \in K$, $n' \in N'$, $x \in I$ (because $M = N \otimes _ B A$ by assumption and because $N' \to N$ and $A' \to A$ are surjective with kernels $L$ and $I$),

12. any element of $N' \otimes _{B'} A'$ in the submodule generated by the elements $l \otimes 1$ and $n' \otimes x$ with $l \in L$, $n' \in N'$, $x \in I$ can be written as $l \otimes 1$ for some $l \in L$ (because $J$ maps isomorphically to $I$ we see that $n' \otimes x = n'x \otimes 1$ in $N' \otimes _{B'} A'$; similarly $x n' \otimes a' = n' \otimes xa' = n'(xa') \otimes 1$ in $N' \otimes _{B'} A'$ when $n' \in N'$, $x \in J$ and $a' \in A'$; since we have seen that $JN' = L$ this proves the assertion),

13. the kernel of $\gamma$ is zero (because by (10) and (11) any element of the kernel is of the form $l \otimes 1$ with $l \in L$ which is mapped to $l \in K \subset M'$ by $\gamma$).

This finishes the proof. $\square$

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