Lemma 35.14.1. Let $R \to A \to B$ be ring maps. Assume $R \to B$ is of finite presentation and $A \to B$ faithfully flat and of finite presentation. Then $R \to A$ is of finite presentation.

## 35.14 Descent of finiteness and smoothness properties of morphisms

In this section we show that several properties of morphisms (being smooth, locally of finite presentation, and so on) descend under faithfully flat morphisms. We start with an algebraic version. (The “Noetherian” reader should consult Lemma 35.14.2 instead of the next lemma.)

**Proof.**
Consider the algebra $C = B \otimes _ A B$ together with the pair of maps $p, q : B \to C$ given by $p(b) = b \otimes 1$ and $q(b) = 1 \otimes b$. Of course the two compositions $A \to B \to C$ are the same. Note that as $p : B \to C$ is flat and of finite presentation (base change of $A \to B$), the ring map $R \to C$ is of finite presentation (as the composite of $R \to B \to C$).

We are going to use the criterion Algebra, Lemma 10.127.3 to show that $R \to A$ is of finite presentation. Let $S$ be any $R$-algebra, and suppose that $S = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } S_\lambda $ is written as a directed colimit of $R$-algebras. Let $A \to S$ be an $R$-algebra homomorphism. We have to show that $A \to S$ factors through one of the $S_\lambda $. Consider the rings $B' = S \otimes _ A B$ and $C' = S \otimes _ A C = B' \otimes _ S B'$. As $B$ is faithfully flat of finite presentation over $A$, also $B'$ is faithfully flat of finite presentation over $S$. By Algebra, Lemma 10.168.1 part (2) applied to the pair $(S \to B', B')$ and the system $(S_\lambda )$ there exists a $\lambda _0 \in \Lambda $ and a flat, finitely presented $S_{\lambda _0}$-algebra $B_{\lambda _0}$ such that $B' = S \otimes _{S_{\lambda _0}} B_{\lambda _0}$. For $\lambda \geq \lambda _0$ set $B_\lambda = S_\lambda \otimes _{S_{\lambda _0}} B_{\lambda _0}$ and $C_\lambda = B_\lambda \otimes _{S_\lambda } B_\lambda $.

We interrupt the flow of the argument to show that $S_\lambda \to B_\lambda $ is faithfully flat for $\lambda $ large enough. (This should really be a separate lemma somewhere else, maybe in the chapter on limits.) Since $\mathop{\mathrm{Spec}}(B_{\lambda _0}) \to \mathop{\mathrm{Spec}}(S_{\lambda _0})$ is flat and of finite presentation it is open (see Morphisms, Lemma 29.25.10). Let $I \subset S_{\lambda _0}$ be an ideal such that $V(I) \subset \mathop{\mathrm{Spec}}(S_{\lambda _0})$ is the complement of the image. Note that formation of the image commutes with base change. Hence, since $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(S)$ is surjective, and $B' = B_{\lambda _0} \otimes _{S_{\lambda _0}} S$ we see that $IS = S$. Thus for some $\lambda \geq \lambda _0$ we have $IS_{\lambda } = S_\lambda $. For this and all greater $\lambda $ the morphism $\mathop{\mathrm{Spec}}(B_\lambda ) \to \mathop{\mathrm{Spec}}(S_\lambda )$ is surjective.

By analogy with the notation in the first paragraph of the proof denote $p_\lambda , q_\lambda : B_\lambda \to C_\lambda $ the two canonical maps. Then $B' = \mathop{\mathrm{colim}}\nolimits _{\lambda \geq \lambda _0} B_\lambda $ and $C' = \mathop{\mathrm{colim}}\nolimits _{\lambda \geq \lambda _0} C_\lambda $. Since $B$ and $C$ are finitely presented over $R$ there exist (by Algebra, Lemma 10.127.3 applied several times) a $\lambda \geq \lambda _0$ and an $R$-algebra maps $B \to B_\lambda $, $C \to C_\lambda $ such that the diagram

is commutative. OK, and this means that $A \to B \to B_\lambda $ maps into the equalizer of $p_\lambda $ and $q_\lambda $. By Lemma 35.3.6 we see that $S_\lambda $ is the equalizer of $p_\lambda $ and $q_\lambda $. Thus we get the desired ring map $A \to S_\lambda $ and we win. $\square$

Here is an easier version of this dealing with the property of being of finite type.

Lemma 35.14.2. Let $R \to A \to B$ be ring maps. Assume $R \to B$ is of finite type and $A \to B$ faithfully flat and of finite presentation. Then $R \to A$ is of finite type.

**Proof.**
By Algebra, Lemma 10.168.2 there exists a commutative diagram

with $R \to A_0$ of finite presentation, $A_0 \to B_0$ faithfully flat of finite presentation and $B = A \otimes _{A_0} B_0$. Since $R \to B$ is of finite type by assumption, we may add some elements to $A_0$ and assume that the map $B_0 \to B$ is surjective! In this case, since $A_0 \to B_0$ is faithfully flat, we see that as

is surjective, also $A_0 \to A$ is surjective. Hence we win. $\square$

Lemma 35.14.3. Let

be a commutative diagram of morphisms of schemes. Assume that $f$ is surjective, flat and locally of finite presentation and assume that $p$ is locally of finite presentation (resp. locally of finite type). Then $q$ is locally of finite presentation (resp. locally of finite type).

**Proof.**
The problem is local on $S$ and $Y$. Hence we may assume that $S$ and $Y$ are affine. Since $f$ is flat and locally of finite presentation, we see that $f$ is open (Morphisms, Lemma 29.25.10). Hence, since $Y$ is quasi-compact, there exist finitely many affine opens $X_ i \subset X$ such that $Y = \bigcup f(X_ i)$. Clearly we may replace $X$ by $\coprod X_ i$, and hence we may assume $X$ is affine as well. In this case the lemma is equivalent to Lemma 35.14.1 (resp. Lemma 35.14.2) above.
$\square$

We use this to improve some of the results on morphisms obtained earlier.

Lemma 35.14.4. Let

be a commutative diagram of morphisms of schemes. Assume that

$f$ is surjective, and syntomic (resp. smooth, resp. étale),

$p$ is syntomic (resp. smooth, resp. étale).

Then $q$ is syntomic (resp. smooth, resp. étale).

**Proof.**
Combine Morphisms, Lemmas 29.30.16, 29.34.19, and 29.36.19 with Lemma 35.14.3 above.
$\square$

Actually we can strengthen this result as follows.

Lemma 35.14.5. Let

be a commutative diagram of morphisms of schemes. Assume that

$f$ is surjective, flat, and locally of finite presentation,

$p$ is smooth (resp. étale).

Then $q$ is smooth (resp. étale).

**Proof.**
Assume (1) and that $p$ is smooth. By Lemma 35.14.3 we see that $q$ is locally of finite presentation. By Morphisms, Lemma 29.25.13 we see that $q$ is flat. Hence now it suffices to show that the fibres of $q$ are smooth, see Morphisms, Lemma 29.34.3. Apply Varieties, Lemma 33.25.9 to the flat surjective morphisms $X_ s \to Y_ s$ for $s \in S$ to conclude. We omit the proof of the étale case.
$\square$

Remark 35.14.6. With the assumptions (1) and $p$ smooth in Lemma 35.14.5 it is not automatically the case that $X \to Y$ is smooth. A counter example is $S = \mathop{\mathrm{Spec}}(k)$, $X = \mathop{\mathrm{Spec}}(k[s])$, $Y = \mathop{\mathrm{Spec}}(k[t])$ and $f$ given by $t \mapsto s^2$. But see also Lemma 35.14.7 for some information on the structure of $f$.

Lemma 35.14.7. Let

be a commutative diagram of morphisms of schemes. Assume that

$f$ is surjective, flat, and locally of finite presentation,

$p$ is syntomic.

Then both $q$ and $f$ are syntomic.

**Proof.**
By Lemma 35.14.3 we see that $q$ is of finite presentation. By Morphisms, Lemma 29.25.13 we see that $q$ is flat. By Morphisms, Lemma 29.30.10 it now suffices to show that the local rings of the fibres of $Y \to S$ and the fibres of $X \to Y$ are local complete intersection rings. To do this we may take the fibre of $X \to Y \to S$ at a point $s \in S$, i.e., we may assume $S$ is the spectrum of a field. Pick a point $x \in X$ with image $y \in Y$ and consider the ring map

This is a flat local homomorphism of local Noetherian rings. The local ring $\mathcal{O}_{X, x}$ is a complete intersection. Thus may use Avramov's result, see Divided Power Algebra, Lemma 23.8.9, to conclude that both $\mathcal{O}_{Y, y}$ and $\mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x}$ are complete intersection rings. $\square$

The following type of lemma is occasionally useful.

Lemma 35.14.8. Let $X \to Y \to Z$ be morphism of schemes. Let $P$ be one of the following properties of morphisms of schemes: flat, locally finite type, locally finite presentation. Assume that $X \to Z$ has $P$ and that $\{ X \to Y\} $ can be refined by an fppf covering of $Y$. Then $Y \to Z$ is $P$.

**Proof.**
Let $\mathop{\mathrm{Spec}}(C) \subset Z$ be an affine open and let $\mathop{\mathrm{Spec}}(B) \subset Y$ be an affine open which maps into $\mathop{\mathrm{Spec}}(C)$. The assumption on $X \to Y$ implies we can find a standard affine fppf covering $\{ \mathop{\mathrm{Spec}}(B_ j) \to \mathop{\mathrm{Spec}}(B)\} $ and lifts $x_ j : \mathop{\mathrm{Spec}}(B_ j) \to X$. Since $\mathop{\mathrm{Spec}}(B_ j)$ is quasi-compact we can find finitely many affine opens $\mathop{\mathrm{Spec}}(A_ i) \subset X$ lying over $\mathop{\mathrm{Spec}}(B)$ such that the image of each $x_ j$ is contained in the union $\bigcup \mathop{\mathrm{Spec}}(A_ i)$. Hence after replacing each $\mathop{\mathrm{Spec}}(B_ j)$ by a standard affine Zariski coverings of itself we may assume we have a standard affine fppf covering $\{ \mathop{\mathrm{Spec}}(B_ i) \to \mathop{\mathrm{Spec}}(B)\} $ such that each $\mathop{\mathrm{Spec}}(B_ i) \to Y$ factors through an affine open $\mathop{\mathrm{Spec}}(A_ i) \subset X$ lying over $\mathop{\mathrm{Spec}}(B)$. In other words, we have ring maps $C \to B \to A_ i \to B_ i$ for each $i$. Note that we can also consider

and that the ring map $B \to \prod B_ i$ is faithfully flat and of finite presentation.

The case $P = flat$. In this case we know that $C \to A$ is flat and we have to prove that $C \to B$ is flat. Suppose that $N \to N' \to N''$ is an exact sequence of $C$-modules. We want to show that $N \otimes _ C B \to N' \otimes _ C B \to N'' \otimes _ C B$ is exact. Let $H$ be its cohomology and let $H'$ be the cohomology of $N \otimes _ C B' \to N' \otimes _ C B' \to N'' \otimes _ C B'$. As $B \to B'$ is flat we know that $H' = H \otimes _ B B'$. On the other hand $N \otimes _ C A \to N' \otimes _ C A \to N'' \otimes _ C A$ is exact hence has zero cohomology. Hence the map $H \to H'$ is zero (as it factors through the zero module). Thus $H' = 0$. As $B \to B'$ is faithfully flat we conclude that $H = 0$ as desired.

The case $P = locally\ finite\ type$. In this case we know that $C \to A$ is of finite type and we have to prove that $C \to B$ is of finite type. Because $B \to B'$ is of finite presentation (hence of finite type) we see that $A \to B'$ is of finite type, see Algebra, Lemma 10.6.2. Therefore $C \to B'$ is of finite type and we conclude by Lemma 35.14.2.

The case $P = locally\ finite\ presentation$. In this case we know that $C \to A$ is of finite presentation and we have to prove that $C \to B$ is of finite presentation. Because $B \to B'$ is of finite presentation and $B \to A$ of finite type we see that $A \to B'$ is of finite presentation, see Algebra, Lemma 10.6.2. Therefore $C \to B'$ is of finite presentation and we conclude by Lemma 35.14.1. $\square$

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