The Stacks project

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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 36.14.

Lemma 15.5.1. Let $R$ be a ring. Let $A \to B$ and $C \to B$ be $R$-algebra maps. Assume

  1. $R$ is Noetherian,

  2. $A$, $B$, $C$ are of finite type over $R$,

  3. $A \to B$ is surjective, and

  4. $B$ is finite over $C$.

Then $A \times _ B C$ is of finite type over $R$.

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 10.50.7). $\square$

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 4.30.3)

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 10.13.3 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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