The Stacks project

Proof. First assume that $A = \mathbf{Z}[f_1, f_2]$ is the polynomial ring. In this case our almost blow up square is the blowing up of $X = \mathop{\mathrm{Spec}}(A)$ in the closed subscheme $Z$ and in fact $X' \subset \mathbf{P}^1_ X$ is an effective Cartier divisor cut out by the global section $f_2T_0 - f_1 T_1$ of $\mathcal{O}_{\mathbf{P}^1_ X}(1)$. Thus we have a resolution

Using the description of the cohomology given in Cohomology of Schemes, Section 30.8 it follows that in this case $\Gamma (X, \mathcal{O}_ X) \to \Gamma (X', \mathcal{O}_{X'})$ is an isomorphism and $H^1(X', \mathcal{O}_{X'}) = 0$.

Next, we observe that any diagram as in Example 38.37.11 is the base change of the diagram in the previous paragraph by the ring map $\mathbf{Z}[f_1, f_2] \to A$. Hence by More on Morphisms, Lemmas 37.72.1, 37.72.2, and 37.72.4 we conclude that $H^1(X', \mathcal{O}_{X'})$ is zero in general and the surjectivity of the map $H^0(X, \mathcal{O}_ X) \to H^0(X', \mathcal{O}_{X'})$ in general.

Next, in the general case, let us study the kernel. If $a \in A$ maps to zero, then looking on affine charts we see that

for some $r \geq 0$ and $a_0, \ldots , a_ r \in A$ and similarly

for some $s \geq 0$ and $b_0, \ldots , b_ s \in A$. This means we have

If $(a', r', a'_ i, s', b'_ j)$ is a second such system, then we have

as desired. $\square$

Proof. Denote $\mathcal{F}$ the Zariski sheafification of the functor

For quasi-compact and quasi-separated schemes $X$ we have $\mathcal{F}(X) = \mathop{\mathrm{colim}}\nolimits _ F \Gamma (X, \mathcal{O}_ X)$. by Sheaves, Lemma 6.29.1 and the fact that $\mathcal{O}$ is a sheaf for the Zariski topology. Thus it suffices to show that $\mathcal{F}$ is a h sheaf. To prove this we check conditions (1), (2), (3), and (4) of Lemma 38.37.12. Condition (1) holds because we performed an (almost unnecessary) Zariski sheafification. Condition (2) holds because $\mathcal{O}$ is an fppf sheaf (Descent, Lemma 35.8.1) and if $A$ is the equalizer of two maps $B \to C$ of $\mathbf{F}_ p$-algebras, then $\mathop{\mathrm{colim}}\nolimits _ F A$ is the equalizer of the two maps $\mathop{\mathrm{colim}}\nolimits _ F B \to \mathop{\mathrm{colim}}\nolimits _ F C$.

We check condition (3). Let $A, f, J$ be as in Example 38.37.10. We have to show that

This reduces to the following algebra question: suppose $a', a'' \in A$ are such that $F^ n(a' - a'') \in fA + J$. Find $a \in A$ and $m \geq 0$ such that $a - F^ m(a') \in J$ and $a - F^ m(a'') \in fA$ and show that the pair $(a, m)$ is uniquely determined up to a replacement of the form $(a, m) \mapsto (F(a), m + 1)$. To do this just write $F^ n(a' - a'') = f h + g$ with $h \in A$ and $g \in J$ and set $a = F^ n(a') - g = F^ n(a'') + fh$ and set $m = n$. To see uniqueness, suppose $(a_1, m_1)$ is a second solution. By a replacement of the form given above we may assume $m = m_1$. Then we see that $a - a_1 \in J$ and $a - a_1 \in fA$. Since $J$ is annihilated by a power of $f$ we see that $a - a_1$ is a nilpotent element. Hence $F^ k(a - a_1)$ is zero for some large $k$. Thus after doing more replacements we get $a = a_1$.

We check condition (4). Let $X, X', Z, E$ be as in Example 38.37.11. By Lemma 38.38.1 we see that

is bijective. Since $E = \mathbf{P}^1_ Z$ in this case we also see that $\mathcal{F}(Z) \to \mathcal{F}(E)$ is bijective. Thus the conclusion holds in this case as well. $\square$

Proof. To prove $\mathcal{F}$ is a sheaf, let's check conditions (1), (2), (3), and (4) of Lemma 38.37.12. Condition (1) holds because limits of sheaves are sheaves and $\mathcal{O}$ is a Zariski sheaf. Condition (2) holds because $\mathcal{O}$ is an fppf sheaf (Descent, Lemma 35.8.1) and if $A$ is the equalizer of two maps $B \to C$ of $\mathbf{F}_ p$-algebras, then $\mathop{\mathrm{lim}}\nolimits _ F A$ is the equalizer of the two maps $\mathop{\mathrm{lim}}\nolimits _ F B \to \mathop{\mathrm{lim}}\nolimits _ F C$.

We check condition (3). Let $A, f, J$ be as in Example 38.37.10. We have to show that

is bijective. Since $J$ is annihilated by a power of $f$ we see that $\mathfrak a = fA \cap J$ is a nilpotent ideal, i.e., there exists an $n$ such that $\mathfrak a^ n = 0$. It is straightforward to verify that in this case $\mathop{\mathrm{lim}}\nolimits _ F A \to \mathop{\mathrm{lim}}\nolimits _ F A/\mathfrak a$ is bijective.

We check condition (4). Let $X, X', Z, E$ be as in Example 38.37.11. By Lemma 38.38.1 and the same argument as above we see that

is bijective. Since $E = \mathbf{P}^1_ Z$ in this case we also see that $\mathcal{F}(Z) \to \mathcal{F}(E)$ is bijective. Thus the conclusion holds in this case as well. $\square$

Proof. To prove $\mathcal{F}$ is a sheaf, let's check conditions (1) and (2) of Topologies, Lemma 34.8.15. Condition (1) holds because formation of $X^{awn}$ commutes with open coverings, see Morphisms, Lemma 29.47.7 and its proof.

Let $\pi : Y \to X$ be a surjective proper morphism. We have to show that the equalizer of the two maps

is equal to $\Gamma (X^{awn}, \mathcal{O}_{X^{awn}})$. Let $f$ be an element of this equalizer. Then we consider the morphism

Since $Y^{awn} \to X$ is universally closed, the scheme theoretic image $Z$ of $f$ is a closed subscheme of $\mathbf{A}^1_ X$ proper over $X$ and $f : Y^{awn} \to Z$ is surjective. See Morphisms, Lemma 29.41.10. Thus $Z \to X$ is finite (Morphisms, Lemma 29.44.11) and surjective.

Let $k$ be a field and let $z_1, z_2 : \mathop{\mathrm{Spec}}(k) \to Z$ be two morphisms equalized by $Z \to X$. We claim that $z_1 = z_2$. It suffices to show the images $\lambda _ i = z_ i^*f \in k$ agree (as the structure sheaf of $Z$ is generated by $f$ over the structure sheaf of $X$). To see this we choose a field extension $K/k$ and morphisms $y_1, y_2 : \mathop{\mathrm{Spec}}(K) \to Y^{awn}$ such that $z_ i \circ (\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(k)) = f \circ y_ i$. This is possible by the surjectivity of the map $Y^{awn} \to Z$. Choose an algebraically closed extension $\Omega /k$ of very large cardinality. For any $k$-algebra maps $\sigma _ i : K \to \Omega$ we obtain

Since the canonical morphism $(Y \times _ X Y)^{awn} \to Y^{awn} \times _ X Y^{awn}$ is a universal homeomorphism and since $\Omega$ is algebraically closed, we can lift the composition above uniquely to a morphism $\mathop{\mathrm{Spec}}(\Omega ) \to (Y \times _ X Y)^{awn}$. Since $f$ is in the equalizer above, this proves that $\sigma _1(\lambda _1) = \sigma _2(\lambda _2)$. An easy lemma about field extensions shows that this implies $\lambda _1 = \lambda _2$; details omitted.

We conclude that $Z \to X$ is universally injective, i.e., $Z \to X$ is injective on points and induces purely inseparated residue field extensions (Morphisms, Lemma 29.10.2). All in all we conclude that $Z \to X$ is a universal homeomorphism, see Morphisms, Lemma 29.45.5.

Let $g : X^{awn} \to Z$ be the map obtained from the universal property of $X^{awn}$. Then $Y^{awn} \to X^{awn} \to Z$ and $f : Y^{awn} \to Z$ are two morphisms over $X$. By the universal property of $Y^{awn} \to Y$ the two corresponding morphisms $Y^{awn} \to Y \times _ X Z$ over $Y$ have to be equal. This implies that $g \circ \pi ^{wan} = f$ as morphisms into $\mathbf{A}^1_ X$ and we conclude that $g \in \Gamma (X^{awn}, \mathcal{O}_{X^{awn}})$ is the element we were looking for. $\square$

Proof. In Lemma 38.38.4 we have seen that the rule $\mathcal{F}$ of the lemma defines a sheaf in the ph topology and hence a fortiori a sheaf for the h topology. Clearly, there is a canonical map of presheaves of rings $\mathcal{O} \to \mathcal{F}$. To finish the proof, it suffices to show

1. if $f \in \mathcal{O}(X)$ maps to zero in $\mathcal{F}(X)$, then there is a h covering $\{ X_ i \to X\}$ such that $f|_{X_ i} = 0$, and

2. given $f \in \mathcal{F}(X)$ there is a h covering $\{ X_ i \to X\}$ such that $f|_{X_ i}$ is the image of $f_ i \in \mathcal{O}(X_ i)$.

Let $f$ be as in (1). Then $f|_{X^{awn}} = 0$. This means that $f$ is locally nilpotent. Thus if $X' \subset X$ is the closed subscheme cut out by $f$, then $X' \to X$ is a surjective closed immersion of finite presentation. Hence $\{ X' \to X\}$ is the desired h covering. Let $f$ be as in (2). After replacing $X$ by the members of an affine open covering we may assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. Then $f \in A^{awn}$, see Morphisms, Lemma 29.47.6. By Morphisms, Lemma 29.46.11 we can find a ring map $A \to B$ of finite presentation such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism and such that $f$ is the image of an element $b \in B$ under the canonical map $B \to A^{awn}$. Then $\{ \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)\}$ is an h covering and we conclude. The statement about the ph topology follows in the same manner (or it can be deduced from the statement for the h topology). $\square$

Proof. It is immediate from condition (2)(b) in Morphisms, Definition 29.47.1 that if $A$ is absolutely weakly normal, then it is perfect.

Assume $A$ is perfect. Suppose $x, y \in A$ with $x^3 = y^2$. If $p > 3$ then we can write $p = 2n + 3m$ for some $n, m > 0$. Choose $a, b \in A$ with $a^ p = x$ and $b^ p = y$. Setting $c = a^ n b^ m$ we have

and hence $c^2 = x$. Similarly $c^3 = y$. If $p = 2$, then write $x = a^2$ to get $a^6 = y^2$ which implies $a^3 = y$. If $p = 3$, then write $y = a^3$ to get $x^3 = a^6$ which implies $x = a^2$.

Suppose $x, y \in A$ with $\ell ^\ell x = y^\ell$ for some prime number $\ell$. If $\ell \not= p$, then $a = y/\ell$ satisfies $a^\ell = x$ and $\ell a = y$. If $\ell = p$, then $y = 0$ and $x = a^ p$ for some $a$. $\square$

Proof. Proof of (1). Observe that $A \to \mathop{\mathrm{colim}}\nolimits _ F A$ induces a universal homeomorphism on spectra by Algebra, Lemma 10.46.7. Thus it suffices to show that $B = \mathop{\mathrm{colim}}\nolimits _ F A$ is absolutely weakly normal, see Morphisms, Lemma 29.47.6. Note that the ring map $F : B \to B$ is an automorphism, in other words, $B$ is a perfect ring. Hence Lemma 38.38.6 applies.

Proof of (2). This follows from (1) and Lemmas 38.38.2 and 38.38.5 by looking affine locally. $\square$