Lemma 15.5.4. Given a commutative diagram of rings

$\xymatrix{ R & R' \ar[l] \\ B \ar[u] & B' \ar[u] \ar[l] }$

the functor (15.5.3.1) has a right adjoint, namely the functor

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

(see proof for elucidation).

Proof. Given an object $(N, M', \varphi )$ of the category $\text{Mod}_ B \times _{\text{Mod}_ R} \text{Mod}_{R'}$ we set

$N \times _\varphi M' = \{ (n, m') \in N \times M' \mid \varphi (n \otimes 1) = m' \otimes 1\text{ in }M' \otimes _{R'} R\}$

viewed as a $B'$-module. The adjointness statement is that for a $B'$-module $L'$ and a triple $(N, M', \varphi )$ we have

$\mathop{\mathrm{Hom}}\nolimits _{B'}(L', N \times _\varphi M') = \mathop{\mathrm{Hom}}\nolimits _ B(L' \otimes _{B'} B, N) \times _{\mathop{\mathrm{Hom}}\nolimits _ R(L' \otimes _{B'} R, M' \otimes _{R'} R)} \mathop{\mathrm{Hom}}\nolimits _{R'}(L' \otimes _{B'} R', M')$

By Algebra, Lemma 10.14.3 the right hand side is equal to

$\mathop{\mathrm{Hom}}\nolimits _{B'}(L', N) \times _{\mathop{\mathrm{Hom}}\nolimits _{B'}(L', M' \otimes _{R'} R)} \mathop{\mathrm{Hom}}\nolimits _{B'}(L', M')$

Thus it is clear that for a pair $(g, f')$ of elements of this fibre product we get an $B'$-linear map $L' \to N \times _\varphi M'$, $l' \mapsto (g(l'), f'(l'))$. Conversely, given a $B'$ linear map $g' : L' \to N \times _\varphi M'$ we can set $g$ equal to the composition $L' \to N \times _\varphi M' \to N$ and $f'$ equal to the composition $L' \to N \times _\varphi M' \to M'$. These constructions are mutually inverse to each other and define the desired isomorphism. $\square$

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