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 10.13.2, 10.38.7, 10.135.4, and 10.141.3 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 10.135.16 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). $\square$

Comment #1802 by Matthieu Romagny on

Rtpo in statement of Lemma: $C'/IC'$

Comment #1803 by Matthieu Romagny on

Comment #1809 by Matthieu Romagny on

In the proof of (3) I don't see why "the ring C′ is of finite presentation as a $A'[x_1,\dots,x_N]$-module". This would follow if $D'\to C'$ was surjective which has no reason to hold in general. Is there an argument I am missing ? Or should we incorporate generators for $C'$ as an $A'$-algebra in the initial choice of surjection $B'[x_1,\dots,x_N]\to D'$ ?

Comment #1827 by on

@#1802 and @#1803 OK, I fixed the Rtpo $=$ typo.

@#1809. This lemma is one of the most ugly things ever. But I think the argument is correct. Namely, we have that $B'[x_i] \to D'$ is surjective and we have that $C' = D' \otimes_{B'} A'$ hence also $A'[x_i] \to C'$ is surjective by right exactness of tensor product. Now we are assuming $C'$ is of finite presentation as an $A'$-algebra, hence it is of finite presentation of an $A'[x_i]$-algebra (by the algebra lemma quoted), and then we find it is of finite presentation as a module because the map is surjective. OK?

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