Lemma 15.5.1. Let R be a ring. Let A \to B and C \to B be R-algebra maps. Assume
R is Noetherian,
A, B, C are of finite type over R,
A \to B is surjective, and
B is finite over C.
Then A \times _ B C is of finite type over R.
15.5 Fibre products of rings, I
Fibre products of rings have to do with pushouts of schemes. Some cases of pushouts of schemes are discussed in More on Morphisms, Section 37.14.
Proof. Set D = A \times _ B C. There is a commutative diagram
\xymatrix{ 0 & B \ar[l] & A \ar[l] & I \ar[l] & 0 \ar[l] \\ 0 & C \ar[l] \ar[u] & D \ar[l] \ar[u] & I \ar[l] \ar[u] & 0 \ar[l] }
with exact rows. Choose y_1, \ldots , y_ n \in B which are generators for B as a C-module. Choose x_ i \in A mapping to y_ i. Then 1, x_1, \ldots , x_ n are generators for A as a D-module. The map D \to A \times C is injective, and the ring A \times C is finite as a D-module (because it is the direct sum of the finite D-modules A and C). Hence the lemma follows from the Artin-Tate lemma (Algebra, Lemma 10.51.7). \square
Lemma 15.5.2. Let R be a Noetherian ring. Let I be a finite set. Suppose given a cartesian diagram
\xymatrix{ \prod B_ i & \prod A_ i \ar[l]^{\prod \varphi _ i} \\ Q \ar[u]^{\prod \psi _ i} & P \ar[u] \ar[l] }
with \psi _ i and \varphi _ i surjective, and Q, A_ i, B_ i of finite type over R. Then P is of finite type over R.
Proof. Follows from Lemma 15.5.1 and induction on the size of I. Namely, let I = I' \amalg \{ i_0\} . Let P' be the ring defined by the diagram of the lemma using I'. Then P' is of finite type by induction hypothesis. Finally, P sits in a fibre product diagram
\xymatrix{ B_{i_0} & A_{i_0} \ar[l] \\ P' \ar[u] & P \ar[u] \ar[l] & }
to which the lemma applies. \square
Lemma 15.5.3. Suppose given a cartesian diagram of rings
\xymatrix{ R & R' \ar[l]^ t \\ B \ar[u]_ s & B'\ar[u] \ar[l] }
i.e., B' = B \times _ R R'. If h \in B' corresponds to g \in B and f \in R' such that s(g) = t(f), then the diagram
\xymatrix{ R_{s(g)} = R_{t(f)} & (R')_ f \ar[l]^-t \\ B_ g \ar[u]_ s & (B')_ h \ar[u] \ar[l] }
is cartesian too.
Proof. The equality B' = B \times _ R R' tells us that
0 \to B' \to B \oplus R' \xrightarrow {s, -t} R
is an exact sequence of B'-modules. We have B_ g = B_ h, R'_ f = R'_ h, and R_{s(g)} = R_{t(f)} = R_ h as B'-modules. By exactness of localization (Algebra, Proposition 10.9.12) we find that
0 \to B'_ h \to B_ g \oplus R'_ f \xrightarrow {s, -t} R_{s(g)} = R_{t(f)}
is an exact sequence. This proves the lemma. \square
Consider a commutative diagram of rings
\xymatrix{ R & R' \ar[l] \\ B \ar[u] & B' \ar[u] \ar[l] }
Consider the functor (where the fibre product of categories is as constructed in Categories, Example 4.31.3)
15.5.3.1
\begin{equation} \label{more-algebra-equation-modules} \text{Mod}_{B'} \longrightarrow \text{Mod}_ B \times _{\text{Mod}_ R} \text{Mod}_{R'},\quad L' \longmapsto (L' \otimes _{B'} B, L' \otimes _{B'} R', can) \end{equation}
where can is the canonical identification L' \otimes _{B'} B \otimes _ B R = L' \otimes _{B'} R' \otimes _{R'} R. In the following we will write (N, M', \varphi ) for an object of the right hand side, i.e., N is a B-module, M' is an R'-module and \varphi : N \otimes _ B R \to M' \otimes _{R'} R is an isomorphism.
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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