The Stacks project

Proof. We write everything out completely.

Proof of (1). Let $f : X \to Y$ be a morphism of schemes. Let $\{ X_ i \to X\} _{i \in I}$ be an étale covering of $X$. If each composition $h_ i : X_ i \to Y$ has $\mathcal{P}$, then for each $x \in X$ we can find an $i \in I$ and a point $x_ i \in X_ i$ mapping to $x$. Then $(X_ i, x_ i) \to (X, x)$ is an étale morphism of germs, and $\text{id}_ Y : Y \to Y$ is an étale morphism, and $h_ i$ is as in part (3) of Definition 35.32.3. Thus we see that $f$ has $\mathcal{P}$. Conversely, if $f$ has $\mathcal{P}$ then each $X_ i \to Y$ has $\mathcal{P}$ by Definition 35.32.3 part (1).

Proof of (2). Let $f : X \to Y$ be a morphism of schemes. Let $\{ Y_ i \to Y\} _{i \in I}$ be an étale covering of $Y$. Write $X_ i = Y_ i \times _ Y X$ and $h_ i : X_ i \to Y_ i$ for the base change of $f$. If each $h_ i : X_ i \to Y_ i$ has $\mathcal{P}$, then for each $x \in X$ we pick an $i \in I$ and a point $x_ i \in X_ i$ mapping to $x$. Then $(X_ i, x_ i) \to (X, x)$ is an étale morphism of germs, $Y_ i \to Y$ is étale, and $h_ i$ is as in part (3) of Definition 35.32.3. Thus we see that $f$ has $\mathcal{P}$. Conversely, if $f$ has $\mathcal{P}$, then each $X_ i \to Y_ i$ has $\mathcal{P}$ by Definition 35.32.3 part (2).

Proof of (3). Assume $f : X \to Y$ has $\mathcal{P}$ and $g : Y \to Z$ is étale. For every $x \in X$ we can think of $(X, x) \to (X, x)$ as an étale morphism of germs, $Y \to Z$ is an étale morphism, and $h = f$ is as in part (3) of Definition 35.32.3. Thus we see that $g \circ f$ has $\mathcal{P}$.

Proof of (4). Let $f : X \to Y$ be a morphism and $g : Y \to Z$ étale such that $g \circ f$ has $\mathcal{P}$. Then by Definition 35.32.3 part (2) we see that $\text{pr}_ Y : Y \times _ Z X \to Y$ has $\mathcal{P}$. But the morphism $(f, 1) : X \to Y \times _ Z X$ is étale as a section to the étale projection $\text{pr}_ X : Y \times _ Z X \to X$, see Morphisms, Lemma 29.36.18. Hence $f = \text{pr}_ Y \circ (f, 1)$ has $\mathcal{P}$ by Definition 35.32.3 part (1). $\square$

Proof. The equivalence of (a) and (b) is part of Definition 35.32.3. The equivalence of (a) and (e) is Lemma 35.32.4 part (2). The equivalence of (a) and (f) is Lemma 35.32.4 part (1). As (a) is now equivalent to (e) and (f) it follows that (a) equivalent to (g).

It is clear that (c) implies (a). If (a) holds, then for any diagram as in (c) the morphism $f \circ a$ has $\mathcal{P}$ by Definition 35.32.3 part (1), whereupon $h$ has $\mathcal{P}$ by Lemma 35.32.4 part (4). Thus (a) and (c) are equivalent. It is clear that (c) implies (d). To see that (d) implies (a) assume we have a diagram as in (c) with $a : U \to X$ surjective and $h$ having $\mathcal{P}$. Then $b \circ h$ has $\mathcal{P}$ by Lemma 35.32.4 part (3). Since $\{ a : U \to X\} $ is an étale covering we conclude that $f$ has $\mathcal{P}$ by Lemma 35.32.4 part (1). $\square$

Proof. Let $\mathcal{P}$ be a property of morphisms of schemes which satisfies conditions (1), (2) and (3) of the lemma. By Lemma 35.26.2 we see that $\mathcal{P}$ is stable under precomposing with étale morphisms. By Lemma 35.22.2 we see that $\mathcal{P}$ is stable under étale base change. Hence it suffices to prove part (3) of Definition 35.32.3 holds.

More precisely, suppose that $f : X \to Y$ is a morphism of schemes which satisfies Definition 35.32.3 part (3)(b). In other words, for every $x \in X$ there exists an étale morphism $a_ x : U_ x \to X$, a point $u_ x \in U_ x$ mapping to $x$, an étale morphism $b_ x : V_ x \to Y$, and a morphism $h_ x : U_ x \to V_ x$ such that $f \circ a_ x = b_ x \circ h_ x$ and $h_ x$ has $\mathcal{P}$. The proof of the lemma is complete once we show that $f$ has $\mathcal{P}$. Set $U = \coprod U_ x$, $a = \coprod a_ x$, $V = \coprod V_ x$, $b = \coprod b_ x$, and $h = \coprod h_ x$. We obtain a commutative diagram

with $a$, $b$ étale, $a$ surjective. Note that $h$ has $\mathcal{P}$ as each $h_ x$ does and $\mathcal{P}$ is étale local on the target. Because $a$ is surjective and $\mathcal{P}$ is étale local on the source, it suffices to prove that $b \circ h$ has $\mathcal{P}$. This reduces the lemma to proving that $\mathcal{P}$ is stable under postcomposing with an étale morphism.

During the rest of the proof we let $f : X \to Y$ be a morphism with property $\mathcal{P}$ and $g : Y \to Z$ is an étale morphism. Consider the following statements:

1. With no additional assumptions $g \circ f$ has property $\mathcal{P}$.

2. Whenever $Z$ is affine $g \circ f$ has property $\mathcal{P}$.

3. Whenever $X$ and $Z$ are affine $g \circ f$ has property $\mathcal{P}$.

4. Whenever $X$, $Y$, and $Z$ are affine $g \circ f$ has property $\mathcal{P}$.

Once we have proved (-) the proof of the lemma will be complete.

Claim 1: (AAA) $\Rightarrow $ (AA). Namely, let $f : X \to Y$, $g : Y \to Z$ be as above with $X$, $Z$ affine. As $X$ is affine hence quasi-compact we can find finitely many affine open $Y_ i \subset Y$, $i = 1, \ldots , n$ such that $X = \bigcup _{i = 1, \ldots , n} f^{-1}(Y_ i)$. Set $X_ i = f^{-1}(Y_ i)$. By Lemma 35.22.2 each of the morphisms $X_ i \to Y_ i$ has $\mathcal{P}$. Hence $\coprod _{i = 1, \ldots , n} X_ i \to \coprod _{i = 1, \ldots , n} Y_ i$ has $\mathcal{P}$ as $\mathcal{P}$ is étale local on the target. By (AAA) applied to $\coprod _{i = 1, \ldots , n} X_ i \to \coprod _{i = 1, \ldots , n} Y_ i$ and the étale morphism $\coprod _{i = 1, \ldots , n} Y_ i \to Z$ we see that $\coprod _{i = 1, \ldots , n} X_ i \to Z$ has $\mathcal{P}$. Now $\{ \coprod _{i = 1, \ldots , n} X_ i \to X\} $ is an étale covering, hence as $\mathcal{P}$ is étale local on the source we conclude that $X \to Z$ has $\mathcal{P}$ as desired.

Claim 2: (AAA) $\Rightarrow $ (A). Namely, let $f : X \to Y$, $g : Y \to Z$ be as above with $Z$ affine. Choose an affine open covering $X = \bigcup X_ i$. As $\mathcal{P}$ is étale local on the source we see that each $f|_{X_ i} : X_ i \to Y$ has $\mathcal{P}$. By (AA), which follows from (AAA) according to Claim 1, we see that $X_ i \to Z$ has $\mathcal{P}$ for each $i$. Since $\{ X_ i \to X\} $ is an étale covering and $\mathcal{P}$ is étale local on the source we conclude that $X \to Z$ has $\mathcal{P}$.

Claim 3: (AAA) $\Rightarrow $ (-). Namely, let $f : X \to Y$, $g : Y \to Z$ be as above. Choose an affine open covering $Z = \bigcup Z_ i$. Set $Y_ i = g^{-1}(Z_ i)$ and $X_ i = f^{-1}(Y_ i)$. By Lemma 35.22.2 each of the morphisms $X_ i \to Y_ i$ has $\mathcal{P}$. By (A), which follows from (AAA) according to Claim 2, we see that $X_ i \to Z_ i$ has $\mathcal{P}$ for each $i$. Since $\mathcal{P}$ is local on the target and $X_ i = (g \circ f)^{-1}(Z_ i)$ we conclude that $X \to Z$ has $\mathcal{P}$.

Thus to prove the lemma it suffices to prove (AAA). Let $f : X \to Y$ and $g : Y \to Z$ be as above $X, Y, Z$ affine. Note that an étale morphism of affines has universally bounded fibres, see Morphisms, Lemma 29.36.6 and Lemma 29.57.9. Hence we can do induction on the integer $n$ bounding the degree of the fibres of $Y \to Z$. See Morphisms, Lemma 29.57.8 for a description of this integer in the case of an étale morphism. If $n = 1$, then $Y \to Z$ is an open immersion, see Lemma 35.25.2, and the result follows from assumption (3) of the lemma. Assume $n > 1$.

Consider the following commutative diagram

Note that we have a decomposition into open and closed subschemes $Y \times _ Z Y = \Delta _{Y/Z}(Y) \amalg Y'$, see Morphisms, Lemma 29.35.13. As a base change the degrees of the fibres of the second projection $\text{pr} : Y \times _ Z Y \to Y$ are bounded by $n$, see Morphisms, Lemma 29.57.5. On the other hand, $\text{pr}|_{\Delta (Y)} : \Delta (Y) \to Y$ is an isomorphism and every fibre has exactly one point. Thus, on applying Morphisms, Lemma 29.57.8 we conclude the degrees of the fibres of the restriction $\text{pr}|_{Y'} : Y' \to Y$ are bounded by $n - 1$. Set $X' = f_ Y^{-1}(Y')$. Picture

As $\mathcal{P}$ is étale local on the target and hence stable under étale base change (see Lemma 35.22.2) we see that $f_ Y$ has $\mathcal{P}$. Hence, as $\mathcal{P}$ is étale local on the source, $f' = f_ Y|_{X'}$ has $\mathcal{P}$. By induction hypothesis we see that $X' \to Y$ has $\mathcal{P}$. As $\mathcal{P}$ is local on the source, and $\{ X \to X \times _ Z Y, X' \to X \times _ Y Z\} $ is an étale covering, we conclude that $\text{pr} \circ f_ Y$ has $\mathcal{P}$. Note that $g \circ f$ can be viewed as a morphism $g \circ f : X \to g(Y)$. As $\text{pr} \circ f_ Y$ is the pullback of $g \circ f : X \to g(Y)$ via the étale covering $\{ Y \to g(Y)\} $, and as $\mathcal{P}$ is étale local on the target, we conclude that $g \circ f : X \to g(Y)$ has property $\mathcal{P}$. Finally, applying assumption (3) of the lemma once more we conclude that $g \circ f : X \to Z$ has property $\mathcal{P}$. $\square$

Proof. Lemma 35.26.3 applies since $\mathcal{P}$ is étale local on the source by Lemma 35.32.4.

Assume $x \in W(f)$. Let $U' \subset X'$ and $V' \subset Y'$ be open neighbourhoods of $x'$ and $y'$ such that $f'(U') \subset V'$, $g'(U') \subset W(f)$ and $g'|_{U'}$ and $g|_{V'}$ are étale. Then $f \circ g'|_{U'} = g \circ f'|_{U'}$ has $\mathcal{P}$ by property (1) of Definition 35.32.3. Then $f'|_{U'} : U' \to V'$ has property $\mathcal{P}$ by (4) of Lemma 35.32.4. Then by (3) of Lemma 35.32.4 we conclude that $f'_{U'} : U' \to Y'$ has $\mathcal{P}$. Hence $U' \subset W(f')$ by definition. Hence $x' \in W(f')$.

Assume $x' \in W(f')$. Let $U' \subset X'$ and $V' \subset Y'$ be open neighbourhoods of $x'$ and $y'$ such that $f'(U') \subset V'$, $U' \subset W(f')$ and $g'|_{U'}$ and $g|_{V'}$ are étale. Then $U' \to Y'$ has $\mathcal{P}$ by definition of $W(f')$. Then $U' \to V'$ has $\mathcal{P}$ by (4) of Lemma 35.32.4. Then $U' \to Y$ has $\mathcal{P}$ by (3) of Lemma 35.32.4. Let $U \subset X$ be the image of the étale (hence open) morphism $g'|_ U' : U' \to X$. Then $\{ U' \to U\} $ is an étale covering and we conclude that $U \to Y$ has $\mathcal{P}$ by (1) of Lemma 35.32.4. Thus $U \subset W(f)$ by definition. Hence $x \in W(f)$. $\square$

Proof. We may replace $k$ by any extension field in order to prove this. Let $Z$ be an irreducible component of $\mathbf{A}^ n_ k \setminus W$. Assume $\dim (Z) \geq 1$, to get a contradiction. Then there exists an extension field $k'/k$ and a $k'$-valued point $\xi = (\xi _1, \ldots , \xi _ n) \in (k')^ n$ of $Z_{k'} \subset \mathbf{A}^ n_{k'}$ such that at least one of $x_1, \ldots , x_ n$ is transcendental over the prime field. Claim: the orbit of $\xi $ under the group generated by the transformations $T_{i, j, d}$ is Zariski dense in $\mathbf{A}^ n_{k'}$. The claim will give the desired contradiction.

If the characteristic of $k'$ is zero, then already the operators $T_{i, j, 0}$ will be enough since these transform $\xi $ into the points

for arbitrary $(a_1, \ldots , a_ n) \in \mathbf{Z}_{\geq 0}^ n$. If the characteristic is $p > 0$, we may assume after renumbering that $\xi _ n$ is transcendental over $\mathbf{F}_ p$. By successively applying the operators $T_{i, n, d}$ for $i < n$ we see the orbit of $\xi $ contains the elements

for arbitrary $(P_1, \ldots , P_{n - 1}) \in \mathbf{F}_ p[t]$. Thus the Zariski closure of the orbit contains the coordinate hyperplane $x_ n = \xi _ n$. Repeating the argument with a different coordinate, we conclude that the Zariski closure contains $x_ i = \xi _ i + P(\xi _ n)$ for any $P \in \mathbf{F}_ p[t]$ such that $\xi _ i + P(\xi _ n)$ is transcendental over $\mathbf{F}_ p$. Since there are infinitely many such $P$ the claim follows.

Of course the argument in the preceding paragraph also applies if $Z = \{ z\} $ has dimension $0$ and the coordinates of $z$ in $\kappa (z)$ are not algebraic over $\mathbf{F}_ p$. The lemma follows. $\square$

Proof. Since $\mathcal{P}$ is étale local on the source we see that $x \in W(f)$ if and only if the image of $x$ in $X \times _ Y Y'$ is in $W(X \times _ Y Y' \to Y')$. Hence we may assume the diagram in the lemma is cartesian.

Assume $x \in W(f)$. Since $\mathcal{P}$ is smooth local on the target we see that $(g')^{-1}W(f) = W(f) \times _ Y Y' \to Y'$ has $\mathcal{P}$. Hence $(g')^{-1}W(f) \subset W(f')$. We conclude $x' \in W(f')$.

Assume $x' \in W(f')$. For any open neighbourhood $V' \subset Y'$ of $y'$ we may replace $Y'$ by $V'$ and $X'$ by $U' = (f')^{-1}V'$ because $V' \to Y'$ is smooth and hence the base change $W(f') \cap U' \to V'$ of $W(f') \to Y'$ has property $\mathcal{P}$. Thus we may assume there exists an étale morphism $Y' \to \mathbf{A}^ n_ Y$ over $Y$, see Morphisms, Lemma 29.36.20. Picture

By Lemma 35.32.6 (and because étale coverings are smooth coverings) we see that $\mathcal{P}$ is étale local on the source-and-target. By Lemma 35.32.9 we see that $W(f')$ is the inverse image of the open $W(f_ n) \subset \mathbf{A}^ n_ X$. In particular $W(f_ n)$ contains a point lying over $x$. After replacing $X$ by the image of $W(f_ n)$ (which is open) we may assume $W(f_ n) \to X$ is surjective. Claim: $W(f_ n) = \mathbf{A}^ n_ X$. The claim implies $f$ has $\mathcal{P}$ as $\mathcal{P}$ is local in the smooth topology and $\{ \mathbf{A}^ n_ Y \to Y\} $ is a smooth covering.

Essentially, the claim follows as $W(f_ n) \subset \mathbf{A}^ n_ X$ is a “translation invariant” open which meets every fibre of $\mathbf{A}^ n_ X \to X$. However, to produce an argument along these lines one has to do étale localization on $Y$ to produce enough translations and it becomes a bit annoying. Instead we use the automorphisms of Lemma 35.32.10 and étale morphisms of affine spaces. We may assume $n \geq 2$. Namely, if $n = 0$, then we are done. If $n = 1$, then we consider the diagram

We have $p^{-1}(W(f_1)) \subset W(f_2)$ (see first paragraph of the proof). Thus $W(f_2) \to X$ is still surjective and we may work with $f_2$. Assume $n \geq 2$.

For any $1 \leq i, j \leq n$ with $i \not= j$ and $d \geq 0$ denote $T_{i, j, d}$ the automorphism of $\mathbf{A}^ n$ defined in Lemma 35.32.10. Then we get a commutative diagram

whose vertical arrows are isomorphisms. We conclude that $T_{i, j, d}(W(f_ n)) = W(f_ n)$. Applying Lemma 35.32.10 we conclude for any $x \in X$ the fibre $W(f_ n)_ x \subset \mathbf{A}^ n_ x$ is either $\mathbf{A}^ n_ x$ (this is what we want) or $\kappa (x)$ has characteristic $p > 0$ and $W(f_ n)_ x$ is the complement of a finite set $Z_ x \subset \mathbf{A}^ n_ x$ of closed points. The second possibility cannot occur. Namely, consider the morphism $T_ p : \mathbf{A}^ n \to \mathbf{A}^ n$ given by

As above we get a commutative diagram

The morphism $T_ p : \mathbf{A}^ n_ X \to \mathbf{A}^ n_ X$ is étale at every point lying over $x$ and the morphism $T_ p : \mathbf{A}^ n_ Y \to \mathbf{A}^ n_ Y$ is étale at every point lying over the image of $x$ in $Y$. (Details omitted; hint: compute the derivatives.) We conclude that

by Lemma 35.32.9 (we've already seen $\mathcal{P}$ is étale local on the source-and-target). Since $T_ p : \mathbf{A}^ n_ x \to \mathbf{A}^ n_ x$ is finite étale of degree $p^ n > 1$ we see that if $Z_ x$ is not empty then it contains $T_ p^{-1}(Z_ x)$ which is bigger. This contradiction finishes the proof. $\square$