The Stacks project

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). $\square$


Comments (4)

Comment #1802 by Matthieu Romagny on

Rtpo in statement of Lemma:

Comment #1803 by Matthieu Romagny on

Whoops, typo in my comment : read "typo" instead of "Rtpo"!

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 -module". This would follow if was surjective which has no reason to hold in general. Is there an argument I am missing ? Or should we incorporate generators for as an -algebra in the initial choice of surjection ?

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 is surjective and we have that hence also is surjective by right exactness of tensor product. Now we are assuming is of finite presentation as an -algebra, hence it is of finite presentation of an -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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