## 15.7 Fibre products of rings, III

In this section we discuss fibre products in the following situation.

Situation 15.7.1. Let $A, A', B, B', I$ be as in Situation 15.6.1. Let $B' \to D'$ be a ring map. Set $D = D' \otimes _{B'} B$, $C' = D' \otimes _{B'} A'$, and $C = D' \otimes _{B'} A$. This leads to a big commutative diagram

$\xymatrix{ C & & & C' \ar[lll] \\ & A \ar[ul] & A' \ar[l] \ar[ru] \\ & B \ar[u] \ar[ld] & B' \ar[l] \ar[u] \ar[rd] \\ D \ar[uuu] & & & D' \ar[lll] \ar[uuu] }$

of rings. Observe that we do not assume that the map $D' \to D \times _ C C'$ is an isomorphism1. In this situation we have the functor

15.7.1.1
\begin{equation} \label{more-algebra-equation-relative-functor} \text{Mod}_{D'} \longrightarrow \text{Mod}_ D \times _{\text{Mod}_ C} \text{Mod}_{C'},\quad L' \longmapsto (L' \otimes _{D'} D, L' \otimes _{D'} C', can) \end{equation}

analogous to (15.6.3.1). Note that $L' \otimes _{D'} D = L \otimes _{D'} (D' \otimes _{B'} B) = L \otimes _{B'} B$ and similarly $L' \otimes _{D'} C' = L \otimes _{D'} (D' \otimes _{B'} A') = L \otimes _{B'} A'$ hence the diagram

$\xymatrix{ \text{Mod}_{D'} \ar[r] \ar[d] & \text{Mod}_ D \times _{\text{Mod}_ C} \text{Mod}_{C'} \ar[d] \\ \text{Mod}_{B'} \ar[r] & \text{Mod}_ B \times _{\text{Mod}_ A} \text{Mod}_{A'} }$

is commutative. In the following we will write $(N, M', \varphi )$ for an object of $\text{Mod}_ D \times _{\text{Mod}_ C} \text{Mod}_{C'}$, i.e., $N$ is a $D$-module, $M'$ is an $C'$-module and $\varphi : N \otimes _ B A \to M' \otimes _{A'} A$ is an isomorphism of $C$-modules. However, it is often more convenient think of $\varphi$ as a $D$-linear map $\varphi : N \to M'/IM'$ which induces an isomorphism $N \otimes _ B A \to M' \otimes _{A'} A = M'/IM'$.

Lemma 15.7.2. In Situation 15.7.1 the functor (15.7.1.1) 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 (15.7.1.1) is the identity functor on $\text{Mod}_ D \times _{\text{Mod}_ C} \text{Mod}_{C'}$. In other words, setting $N' = N \times _{\varphi , M} M'$ we have $N' \otimes _{D'} D = N$ and $N' \otimes _{D'} C' = M'$.

Proof. The adjointness statement follows from the more general Lemma 15.5.4. The final assertion follows from the corresponding assertion of Lemma 15.6.4 because $N' \otimes _{D'} D = N' \otimes _{D'} D' \otimes _{B'} B = N' \otimes _{B'} B$ and $N' \otimes _{D'} C' = N' \otimes _{D'} D' \otimes _{B'} A' = N' \otimes _{B'} A'$. $\square$

Lemma 15.7.3. In Situation 15.7.1 the map $JD' \to IC'$ is surjective where $J = \mathop{\mathrm{Ker}}(B' \to B)$.

Proof. Since $C' = D' \otimes _{B'} A'$ we have that $IC'$ is the image of $D' \otimes _{B'} I = C' \otimes _{A'} I \to C'$. As the ring map $B' \to A'$ induces an isomorphism $J \to I$ the lemma follows. $\square$

Lemma 15.7.4. Let $A, A', B, B', C, C', D, D', I, M', M, N, \varphi$ be as in Lemma 15.7.2. If $N$ finite over $D$ and $M'$ finite over $C'$, then $N' = N \times _{\varphi , M} M'$ is finite over $D'$.

Proof. Recall that $D' \to D \times _ C C'$ is surjective by Lemma 15.6.5. Observe that $N' = N \times _{\varphi , M} M'$ is a module over $D \times _ C C'$. We can apply Lemma 15.6.7 to the data $C, C', D, D', IC', M', M, N, \varphi$ to see that $N' = N \times _{\varphi , M} M'$ is finite over $D \times _ C C'$. Thus it is finite over $D'$. $\square$

Lemma 15.7.5. With $A, A', B, B', C, C', D, D', I$ as in Situation 15.7.1.

1. Let $(N, M', \varphi )$ be an object of $\text{Mod}_ D \times _{\text{Mod}_ C} \text{Mod}_{C'}$. If $M'$ is flat over $A'$ and $N$ is flat over $B$, then $N' = N \times _{\varphi , M} M'$ is flat over $B'$.

2. If $L'$ is a $D'$-module flat over $B'$, then $L' = (L \otimes _{D'} D) \times _{(L \otimes _{D'} C)} (L \otimes _{D'} C')$.

3. The category of $D'$-modules flat over $B'$ is equivalent to the categories of objects $(N, M', \varphi )$ of $\text{Mod}_ D \times _{\text{Mod}_ C} \text{Mod}_{C'}$ with $N$ flat over $B$ and $M'$ flat over $A'$.

Proof. Part (1) follows from part (1) of Lemma 15.6.8.

Part (2) follows from part (2) of Lemma 15.6.8 using that $L' \otimes _{D'} D = L' \otimes _{B'} B$, $L' \otimes _{D'} C' = L' \otimes _{B'} A'$, and $L' \otimes _{D'} C = L' \otimes _{B'} A$, see discussion in Situation 15.7.1.

Part (3) is an immediate consequence of (1) and (2). $\square$

The following lemma is a good deal more interesting than its counter part in the absolute case (Lemma 15.6.9), although the proof is essentially the same.

Lemma 15.7.6. Let $A, A', B, B', C, C', D, D', I, M', M, N, \varphi$ be as in Lemma 15.7.2. If

1. $N$ is finitely presented over $D$ and flat over $B$,

2. $M'$ finitely presented over $C'$ and flat over $A'$, and

3. the ring map $B' \to D'$ factors as $B' \to D'' \to D'$ with $B' \to D''$ flat and $D'' \to D'$ of finite presentation,

then $N' = N \times _ M M'$ is finitely presented over $D'$.

Proof. Choose a surjection $D''' = D''[x_1, \ldots , x_ n] \to D'$ with finitely generated kernel $J$. By Algebra, Lemma 10.36.23 it suffices to show that $N'$ is finitely presented as a $D'''$-module. Moreover, $D''' \otimes _{B'} B \to D' \otimes _{B'} B = D$ and $D''' \otimes _{B'} A' \to D' \otimes _{B'} A' = C'$ are surjections whose kernels are generated by the image of $J$, hence $N$ is a finitely presented $D''' \otimes _{B'} B$-module and $M'$ is a finitely presented $D''' \otimes _{B'} A'$-module by Algebra, Lemma 10.36.23 again. Thus we may replace $D'$ by $D'''$ and $D$ by $D''' \otimes _{B'} B$, etc. Since $D'''$ is flat over $B'$, it follows that we may assume that $B' \to D'$ is flat.

Assume $B' \to D'$ is flat. By Lemma 15.7.4 the module $N'$ is finite over $D'$. Choose a surjection $(D')^{\oplus n} \to N'$ with kernel $K'$. By base change we obtain maps $D^{\oplus n} \to N$, $(C')^{\oplus n} \to M'$, and $C^{\oplus n} \to M$ with kernels $K_ D$, $K_{C'}$, and $K_ C$. There is a canonical map

$K' \longrightarrow K_ D \times _{K_ C} K_{C'}$

On the other hand, since $N' = N \times _ M M'$ and $D' = D \times _ C C'$ (by Lemma 15.6.8; applied to the flat $B'$-module $D'$) there is also a canonical map $K_ D \times _{K_ C} K_{C'} \to K'$ inverse to the displayed arrow. Hence the displayed map is an isomorphism. By Algebra, Lemma 10.5.3 the modules $K_ D$ and $K_{C'}$ are finite. We conclude from Lemma 15.7.4 that $K'$ is a finite $D'$-module provided that $K_ D \to K_ C$ and $K_{C'} \to K_ C$ induce isomorphisms $K_ D \otimes _ B A = K_ C = K_{C'} \otimes _{A'} A$. This is true because the flatness assumptions implies the sequences

$0 \to K_ D \to D^{\oplus n} \to N \to 0 \quad \text{and}\quad 0 \to K_{C'} \to (C')^{\oplus n} \to M' \to 0$

stay exact upon tensoring, see Algebra, Lemma 10.39.12. $\square$

Lemma 15.7.7. Let $A, A', B, B', I$ be as in Situation 15.6.1. Let $(D, C', \varphi )$ be a system consisting of an $B$-algebra $D$, a $A'$-algebra $C'$ and an isomorphism $D \otimes _ B A \to C'/IC' = C$. Set $D' = D \times _ C C'$ (as in Lemma 15.6.4). Then

1. $B' \to D'$ is finite type if and only if $B \to D$ and $A' \to C'$ are finite type,

2. $B' \to D'$ is flat if and only if $B \to D$ and $A' \to C'$ are flat,

3. $B' \to D'$ is flat and of finite presentation if and only if $B \to D$ and $A' \to C'$ are flat and of finite presentation,

4. $B' \to D'$ is smooth if and only if $B \to D$ and $A' \to C'$ are smooth,

5. $B' \to D'$ is étale if and only if $B \to D$ and $A' \to C'$ are étale.

Moreover, if $D'$ is a flat $B'$-algebra, then $D' \to (D' \otimes _{B'} B) \times _{(D' \otimes _{B'} A)} (D' \otimes _{B'} A')$ is an isomorphism. In this way the category of flat $B'$-algebras is equivalent to the categories of systems $(D, C', \varphi )$ as above with $D$ flat over $B$ and $C'$ flat over $A'$.

Proof. The implication “$\Rightarrow$” follows from Algebra, Lemmas 10.14.2, 10.39.7, 10.137.4, and 10.143.3 because we have $D' \otimes _{B'} B = D$ and $D' \otimes _{B'} A' = C'$ by Lemma 15.6.4. Thus it suffices to prove the implications in the other direction.

Ad (1). Assume $D$ of finite type over $B$ and $C'$ of finite type over $A'$. We will use the results of Lemma 15.6.4 without further mention. Choose generators $x_1, \ldots , x_ r$ of $D$ over $B$ and generators $y_1, \ldots , y_ s$ of $C'$ over $A'$. Using that $D = D' \otimes _{B'} B$ and $B' \to B$ is surjective we can find $u_1, \ldots , u_ r \in D'$ mapping to $x_1, \ldots , x_ r$ in $D$. Using that $C' = D' \otimes _{B'} A'$ we can find $v_1, \ldots , v_ t \in D'$ such that $y_ i = \sum v_ j \otimes a'_{ij}$ for some $a'_{ij} \in A'$. In particular, the images of $v_ j$ in $C'$ generate $C'$ as an $A'$-algebra. Set $N = r + t$ and consider the cube of rings

$\xymatrix{ A[x_1, \ldots , x_ N] & & A'[x_1, \ldots , x_ N] \ar[ll] \\ & A \ar[lu] & & A' \ar[ll] \ar[lu] \\ B[x_1, \ldots , x_ N] \ar[uu] & & B'[x_1, \ldots , x_ N] \ar[uu] \ar[ll] \\ & B \ar[uu] \ar[lu] & & B' \ar[ll] \ar[uu] \ar[lu] }$

Observe that the back square is cartesian as well. Consider the ring map

$B'[x_1, \ldots , x_ N] \to D',\quad x_ i \mapsto u_ i \quad \text{and}\quad x_{r + j} \mapsto v_ j.$

Then we see that the induced maps $B[x_1, \ldots , x_ N] \to D$ and $A'[x_1, \ldots , x_ N] \to C'$ are surjective, in particular finite. We conclude from Lemma 15.7.4 that $B'[x_1, \ldots , x_ N] \to D'$ is finite, which implies that $D'$ is of finite type over $B'$ for example by Algebra, Lemma 10.6.2.

Ad (2). The implication “$\Leftarrow$” follows from Lemma 15.7.5. Moreover, the final statement follows from the final statement of Lemma 15.7.5.

Ad (3). Assume $B \to D$ and $A' \to C'$ are flat and of finite presentation. The flatness of $B' \to D'$ we've seen in (2). We know $B' \to D'$ is of finite type by (1). Choose a surjection $B'[x_1, \ldots , x_ N] \to D'$. By Algebra, Lemma 10.6.3 the ring $D$ is of finite presentation as a $B[x_1, \ldots , x_ N]$-module and the ring $C'$ is of finite presentation as a $A'[x_1, \ldots , x_ N]$-module. By Lemma 15.7.6 we see that $D'$ is of finite presentation as a $B'[x_1, \ldots , x_ N]$-module, i.e., $B' \to D'$ is of finite presentation.

Ad (4). Assume $B \to D$ and $A' \to C'$ smooth. By (3) we see that $B' \to D'$ is flat and of finite presentation. By Algebra, Lemma 10.137.17 it suffices to check that $D' \otimes _{B'} k$ is smooth for any field $k$ over $B'$. If the composition $J \to B' \to k$ is zero, then $B' \to k$ factors as $B' \to B \to k$ and we see that

$D' \otimes _{B'} k = D' \otimes _{B'} B \otimes _ B k = D \otimes _ B k$

is smooth as $B \to D$ is smooth. If the composition $J \to B' \to k$ is nonzero, then there exists an $h \in J$ which does not map to zero in $k$. Then $B' \to k$ factors as $B' \to B'_ h \to k$. Observe that $h$ maps to zero in $B$, hence $B_ h = 0$. Thus by Lemma 15.5.3 we have $B'_ h = A'_ h$ and we get

$D' \otimes _{B'} k = D' \otimes _{B'} B'_ h \otimes _{B'_ h} k = C'_ h \otimes _{A'_ h} k$

is smooth as $A' \to C'$ is smooth.

Ad (5). Assume $B \to D$ and $A' \to C'$ are étale. By (4) we see that $B' \to D'$ is smooth. As we can read off whether or not a smooth map is étale from the dimension of fibres we see that (5) holds (argue as in the proof of (4) to identify fibres – some details omitted). $\square$

Remark 15.7.8. In Situation 15.7.1. Assume $B' \to D'$ is of finite presentation and suppose we are given a $D'$-module $L'$. We claim there is a bijective correspondence between

1. surjections of $D'$-modules $L' \to Q'$ with $Q'$ of finite presentation over $D'$ and flat over $B'$, and

2. pairs of surjections of modules $(L' \otimes _{D'} D \to Q_1, L' \otimes _{D'} C' \to Q_2)$ with

1. $Q_1$ of finite presentation over $D$ and flat over $B$,

2. $Q_2$ of finite presentation over $C'$ and flat over $A'$,

3. $Q_1 \otimes _ D C = Q_2 \otimes _{C'} C$ as quotients of $L' \otimes _{D'} C$.

The correspondence between these is given by $Q \mapsto (Q_1, Q_2)$ with $Q_1 = Q \otimes _{D'} D$ and $Q_2 = Q \otimes _{D'} C'$. And for the converse we use $Q = Q_1 \times _{Q_{12}} Q_2$ where $Q_{12}$ the common quotient $Q_1 \otimes _ D C = Q_2 \otimes _{C'} C$ of $L' \otimes _{D'} C$. As quotient map we use

$L' \longrightarrow (L' \otimes _{D'} D) \times _{(L' \otimes _{D'} C)} (L' \otimes _{D'} C') \longrightarrow Q_1 \times _{Q_{12}} Q_2 = Q$

where the first arrow is surjective by Lemma 15.6.5 and the second by Lemma 15.6.6. The claim follows by Lemmas 15.7.5 and 15.7.6.

 But $D' \to D \times _ C C'$ is surjective by Lemma 15.6.5.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).