Higher pushforwards along finite morphisms

Posted on 12 January 2018 by Remy

This post is about one of my favourite answers I have given on MathOverflow, although it seems to have gone by mostly unnoticed. In the post, Qixiao asks (essentially) the following:

Question. If $f \colon X \to Y$ is a finite morphism of schemes, is the pushforward $f_* \colon \Sh(X) \to \Sh(Y)$ exact?

Note that this is true on the subcategory of quasicoherent sheaves because affine morphisms have no quasicoherent higher pushforwards. Also, in the étale topology the pushforward along a finite morphism is exact on the category of all abelian sheaves; see e.g. Tag 03QP.

However, we show that the answer to the question above is negative.

Example. Let $Y$ be the spectrum of a DVR $(R,\mathfrak m)$ , let $R \to S$ be a finite extension of domains such that $S$ has exactly two primes $\mathfrak p, \mathfrak q$ above $\mathfrak m$ , and let $X = \Spec S$ . For example, $R = \Z_{(5)}$ and $S = \Z_{(5)}[i]$ , or $R = k[x]_{(x)}$ and $S = k[x]_{(x)}[\sqrt{x+1}]$ if you prefer a more geometric example.

By my previous post, the global sections functor $\Gamma \colon \Sh(Y) \to \Ab$ is exact. If the same were true for $f_* \colon \Sh(X) \to \Sh(Y)$ , then the global sections functor on $X$ would be exact as well. Thus, it suffices to prove that this is not the case, i.e. to produce a surjection $\mathscr F \to \mathscr G$ of sheaves on $X$ such that the map on global sections is not surjective.

The topological space of $X$ consists of closed points $x,y$ and a generic point $\eta$ . Let $U = \{\eta\}$ and $Z = U^{\operatorname{c}} = \{x,y\}$ ; then $U$ is open and $Z$ is closed. Hence, for any sheaf $\mathscr F$ on $X$ , we have a short exact sequence (see e.g. Tag 02UT)

$0 \to j_! (\mathscr F|_U) \to \mathscr F \to i_* (\mathscr F|_Z) \to 0,$

where $j \colon U \to X$ and $i \colon Z \to X$ are the inclusions. Let $\mathscr F$ be the constant sheaf $\Z$ ; then the same goes for $\mathscr F|_U$ and $\mathscr F|_Z$ . Then the map

$H^0(X,\mathscr F) \to H^0(X,i_*(\mathscr F|_Z)) = H^0(Z,\mathscr F|_Z)$

is given by the diagonal map $\Z \to \Z \oplus \Z$ , since $X$ is connected by $Z$ has two connected components. This is visibly not surjective. $\qedsymbol$

Cohomology of a local scheme

Posted on 6 January 2018 by Remy

This is a continuation of my previous post on local schemes. Here is a ridiculous lemma.

Lemma. Let $(X,x)$ be a local scheme, and let $\mathscr F$ be any abelian sheaf on $X$ . Then $H^i(X,\mathscr F) = 0$ for all $i > 0$ .

Proof. It suffices to show that the global sections functor $\Gamma \colon \Sh(X) \to \Ab$ is exact. Let $\mathscr F \to \mathscr G$ be a surjection of abelian sheaves on $X$ , and let $s \in \mathscr G(X)$ be a global section. Then $s$ can be lifted to a section of $\mathscr F$ in an open neighbourhood $U$ of $x$ . But the only open neighbourhood of $x$ is $X$ . Thus, $s$ can be lifted to a section of $\mathscr F(X)$ . $\qedsymbol$

What’s going on is that the functors $\mathscr F \mapsto \Gamma(X,\mathscr F)$ and $\mathscr F \mapsto \mathscr F_x$ are naturally isomorphic, due to the absence of open neighbourhoods of $x$ .

Remark. It seems believable that there are suitable site-theoretic versions of this lemma as well. For example, a strictly Henselian local ring has no higher cohomology in the étale topology. The argument is essentially the same: every open neighbourhood of the closed point has a section; see e.g. the proof of Tag 03QO.

Local schemes

Posted on 29 December 2017 by Remy

Consider the following definition. It seems to be standard, although I have not found a place where it is actually spelled out in this way.

Definition. A pointed scheme $(X,x)$ is local if $x$ is contained in every nonempty closed subset of $X$ .

Example. If $(A,\mathfrak m)$ is a local ring, then $(\Spec A,\mathfrak m)$ is a local scheme. Indeed, $\mathfrak m$ is contained in every nonempty closed subset $V(I) \subseteq X$ , because every strict ideal $I \subsetneq A$ is contained in $\mathfrak m$ .

We prove that this is actually the only example.

Lemma. Let $(X,x)$ be a local scheme. Then $X$ is affine, and $A = \Gamma(X,\mathcal O_X)$ is a local ring whose maximal ideal corresponds to the point $x \in X = \Spec A$ .

Proof. Let $U$ be an affine open neighbourhood of $x$ . Then the complement $V$ is a closed set not containing $x$ , hence $V = \varnothing$ . Thus, $X = U$ is affine. Let $A = \Gamma(X,\mathcal O_X)$ . Let $\mathfrak m$ be a maximal ideal of $A$ ; then $V(\mathfrak m) = \{\mathfrak m\}$ . Since this contains $x$ , we must have $x = \mathfrak m$ , i.e. $x$ corresponds to the (necessarily unique) maximal ideal $\mathfrak m \subseteq A$ . $\qedsymbol$

A fun example of a representable functor

Posted on 3 December 2017 by Remy

This post is about representable functors:

Definition. Let $F \colon \mathscr C \to \Set$ be a functor. Then $F$ is representable if it is isomorphic to $\Hom(A,-)$ for some $A \in \ob \mathscr C$ . In this case, we say that $A$ represents $F$ .

Exercise. If such $A$ exists, then it is unique up to unique isomorphism.

Really one should encode the isomorphism $\Hom(A,-) \stackrel\sim\to F$ as well, but this is often dropped from the notation. By the Yoneda lemma, every natural transformation $\Hom(A,-) \to F$ is uniquely determined by the element of $F(A)$ corresponding to the identity of $A$ .

When $\Hom(A,-) \to F$ is a natural isomorphism, the corresponding element $a \in F(A)$ is called the universal object of $F$ . It has the property that for every $B \in \mathscr C$ and any $b \in F(B)$ , there exists a unique morphism $f \colon A \to B$ such that $(Ff)(a) = b$ .

Example. The forgetful functor $\Ab \to \Set$ is represented by $\Z$ . Indeed, the natural map

$\begin{align*} \Hom(\Z,M) &\to M\\ f &\mapsto f(1) \end{align*}$

is an isomorphism. The universal element is $1 \in \Z$ .

Example. Similarly, the forgetful functor $\Ring \to \Set$ is represented by $\Z[x]$ . The universal element is $x$ .

A fun exercise (for the rest of your life!) is to see whether functors you encounter in your work are representable. See for example this post about some more geometric examples.

The main example for today is the following:

Lemma. The functor $\Top\op \to \Set$ that associates to a topological space $(X,\mathcal T_X)$ its topology $\mathcal T_X$ is representable.

Proof. Consider the topological space $Y = \{0,1\}$ with topology $\{\varnothing, \{1\},\{0,1\}\}$ . Then there is a natural map

$\begin{align*} \Hom(X,Y) &\to \mathcal T_X\\ f &\mapsto f^{-1}(\{1\}). \end{align*}$

Conversely, given an open set $U$ , we can associate the characteristic function $\mathbb I_U$ . This gives an inverse of the map above. $\qedsymbol$

The space $Y$ we constructed is called the Sierpiński space. The universal open set is $\{1\}$ .

Remark. The space $Y^I$ represents the data of open sets $U_i$ for $i \in I$ : for any continuous map $f \colon X \to Y^I$ , we have $U_i = f^{-1}(Y_i)$ , where $Y_i = \pi_i^{-1}(\{1\}) \subseteq Y^I$ . If $Z_i$ denotes the complementary open, then the $U_i$ form a cover of $X$ if and only if $\bigcap_{i \in I} Z_i = \varnothing$ . This corresponds to the statement that $f$ lands in $Y^I\setminus\{(0,0,\ldots)\}$ .

Thus, the open cover $Y^I\setminus\{0\} = \bigcup_{i \in I} Y_i$ is the universal open cover, i.e. for every open covering $X = \bigcup U_i$ there exists a unique continuous map $f \colon X \to Y^I\setminus\{0\}$ such that $U_i = f^{-1}(Y_i)$ .

A weird Dedekind domain

Posted on 28 November 2017 by Remy

Here is a random question that I was wondering about at some point (just out of curiosity):

Question. Does there exist a Dedekind domain $R$ that has infinitely many points whose residue field has characteristic $0$ and finitely many points (at least one) whose residue field has positive characteristic?

I don’t fully recall why this question came up, but it had something to do with a similar property that was satisfied by an object involved in the definition of the Fargues–Fontaine curve. However, we don’t need such deep theory to discuss this elementary commutative algebra question.

Lemma. Such a Dedekind domain exists.

Proof. We will construct $R$ as a localisation of $\mathbb Z[x]$ . Recall that prime ideals of $\mathbb Z[x]$ come in four types:

The generic point $(0)$ , of height 0;
Height 1 primes $(p)$ for every prime $p \in \mathbb Z$ ;
Height 1 primes $(f)$ for every irreducible polynomial $f \in \mathbb Z[x]$ ;
Height 2 closed points $(p,f)$ for $p \in \mathbb Z$ a prime and $f \in \mathbb Z[x]$ a polynomial whose reduction $\bar f \in \mathbb F_p[x]$ is irreducible.

We first localise at $(p,x)$ ; then the only primes we have left are the ones contained in $(p,x)$ . This is the generic point, the height 1 prime $(p)$ , the height 1 primes $(f)$ where $f \in \mathbb Z[x]$ is an irreducible polynomial whose constant coefficient is divisible by $p$ (e.g. $(x), (x-p), \ldots$ ), and exactly one height 2 prime $(p,x)$ .

Next, we invert $x$ ; denoting the resulting ring by $R$ . This gets rid of all prime ideals containing $x$ , which are $(p,x)$ and $(x)$ . In particular, there are no more height 2 primes, so $R$ is 1-dimensional. It is a normal Noetherian domain because it a localisation of a normal Noetherian domain. Therefore, $R$ is a Dedekind domain.

The primes of $R$ are $(0)$ with residue field $\mathbb Q(x)$ ; the prime $(p)$ with residue field $\mathbb F_p(x)$ ; and the prime ideals $(f)$ with $f \neq x$ a polynomial whose constant coefficient is divisible by $p$ , whose residue field is a finite extension of $\mathbb Q$ . $\qedsymbol$

Remark. The ring $R$ we constructed is essentially of finite type over $\mathbb Z$ (a localisation of a finite type $\mathbb Z$ -algebra). There are no examples of finite type over $\mathbb Z$ , because by Chevalley’s theorem the image of $\operatorname{Spec} R \to \operatorname{Spec} \mathbb Z$ would be constructible. However, no set of the form $\{(0)\} \cup S$ for $S \subseteq \operatorname{Spec} \mathbb Z$ finite is constructible. (Alternatively, the weak Nullstellensatz implies that every closed point of a finite type $\mathbb Z$ -algebra has residue characteristic $p > 0$ .)

I was a little surprised that we can make examples when we drop the finite type assumption. I don’t know if this type of ring has ever been used for anything.

Cardinality of fraction field

Posted on 24 June 2017 by Remy

This is a ridiculous lemma that I came up with.

Lemma. Let $R$ be a (commutative) ring, and let $K$ be its total ring of fractions. Then $R$ and $K$ have the same cardinality.

Proof. If $R$ is finite, my previous post shows that $R = K$ . If $R$ is infinite, then $K$ is a subquotient of $R^2$ , hence $|K| \leq |R^2| = |R|$ . But $R$ injects into $K$ , so $|R| \leq |K|$ . $\qedsymbol$

Corollary. If $R$ is a domain, then $|\operatorname{Frac}(R)| = |R|$ .

Proof. This is a special case of the lemma. $\qedsymbol$

Finite domains are fields

Posted on 23 June 2017 by Remy

This is one of the classics.

Lemma. Let $R$ be a finite commutative ring. Then every element is either a unit or a zero-divisor.

Proof. If $x \in R$ is not a zero-divisor, then the map $x \colon R \to R$ is injective. Since $R$ is finite, it is also surjective, so there exists $y \in R$ with $xy = 1$ . $\qedsymbol$

Corollary 1. Let $R$ be a finite commutative ring. Then $R$ is its own total ring of fractions.

Proof. The total ring of fractions is the ring $R[S^{-1}]$ , where $S$ is the set of non-zerodivisors. But that set consists of units by the lemma above, so inverting them doesn’t change $R$ . $\qedsymbol$

Corollary 2. Let $R$ be a finite domain. Then $R$ is a field.

Proof. In this case, the total ring of fractions is the fraction field. Therefore, $R$ is its own fraction field by Corollary 1. $\qedsymbol$

Étale and universally injective

Posted on 28 June 2015 by Remy

This is the second in a three-part post about a proof that I contributed to the Stacks project. The result was already there, but I found a slightly easier proof. This post contains the actual result, and the next post will contain a fun application.

Remark. Recall that a morphism $f \colon X \ra Y$ of schemes is étale if $f$ is flat and locally of finite presentation, and $\Omega_{X/Y} = 0$ .

Lemma. Suppose $f \colon X \ra S$ is étale and universally injective. Then $f$ is an open immersion.

Proof. Since étale morphisms are open and surjectivity is stable under base change, we may assume that $f$ is universally bijective; then we have to prove that $f$ is an isomorphism. Since étale morphisms are open, $f$ is in fact a universal homeomorphism. By Tag 04DE, $f$ is affine.

The question is local on $S$ , so we may assume $S$ is affine, and hence so is $X$ . Say $f$ is induced by $g \colon A \ra B$ . Now $f$ is proper and affine, hence finite. Moreover, since $B$ is finitely presented and finite as $A$ -algebra, and $B$ is a finitely presented $B$ -module, it is also a finitely presented $A$ -module (Tag 0564).

Now $B$ is flat of finite presentation over $A$ , hence locally free (actually, we need the slightly stronger result that I mention in the first remark; see Tag 00NX for statement and proof). Since the question is local, we may assume $B$ is free of rank $r$ .

Now let $\bar{s} \ra S$ be a geometric point; that is, let $A \ra \bar{K}$ be a map to an algebraically closed field. Then the tensor product $B \tens_A \bar{K}$ is étale of dimension $r$ over $\bar{K}$ . Hence, $X_{\bar s} = X \times_{S} \bar{s}$ is a union of $r$ points. Since $f$ is universally bijective, we have $r = 1$ . Then the result follows from my previous post. $\qedsymbol$

Locally free algebras

Posted on 28 June 2015 by Remy

This is the first in a three-part post about a proof that I contributed to the Stacks project. The result was already there, but I found a slightly easier proof. This post contains a preliminary lemma; the second post contains the result; and the third one contains the application that I was interested in.

Remark. If $M$ is a (locally) free $A$ -module of rank 1, then $\End_A(M) = A$ . Multiplication by $a \in A$ is injective on $M$ if and only if $a$ is not a zero-divisor, and it is surjective if and only if $a \in A\x$ . In particular, if it is surjective, it is also injective.

Lemma. Suppose $f \colon A \ra B$ is a ring homomorphism, such that $B$ is locally free of rank 1 over $A$ . Then $f$ is an isomorphism.

Proof. The question is local on $A$ , so (after replacing $A$ with a suitable localisation) we may assume that $B$ is free of rank 1. Let $b$ be a basis element.

Then we can write $1 = f(a)b$ for some $a \in A$ , hence $b \in B\x$ . Also, we can write $b^2 = f(c)b$ for some $c \in A$ , hence $b = f(c)$ . Therefore, $f$ is surjective, so by the remark above, it is an isomorphism. $\qedsymbol$

Using a different argument, we can also prove:

Lemma. Suppose $f \colon A \ra B$ is a ring homomorphism, such that $B$ is locally free of rank $r > 0$ over $A$ . Then $f$ is injective.

Proof. Since $B$ is locally free of rank $r$ , it is faithfully flat over $A$ . Thus it suffices to prove that $f \tens 1 \colon B \ra B \tens_A B$ is injective. But this map admits a contraction $B \tens_A B \ra B$ given by $b_1 \tens b_2 \rm b_1b_2$ . $\qedsymbol$

Proper + Affine = Finite

Posted on 27 November 2014 by Remy

Another argument I really like. We first present two auxiliary lemmata.

Lemma 1. Let $f \colon A \ra B$ be an injective ring homomorphism such that the associated map $\phi \colon \Spec B \ra \Spec A$ is closed. Then $f^{-1}(B\x) = A\x$ .

Proof. Since $f$ is injective, the map $\phi$ is dominant. Hence, it is a closed surjective map. For such a map, a closed set $Z \sbq \Spec A$ is empty if and only if $\phi^{-1}(Z)$ is empty. Applying this to $Z = V(a)$ for $a \in A$ , we get $V(a) = \varnothing$ if and only if $V(f(a)) = \varnothing$ , i.e. $a \in A\x$ if and only if $f(a) \in B\x$ . $\qedsymbol$

Lemma 2. Let $f \colon A \ra B$ be an injective ring homomorphism such that the associated map $\phi \colon \A^1_B \ra \A^1_A$ (induced by $A[x] \ra B[x]$ ) is closed. Then $f$ is integral.

Proof. Let $b \in B$ . Consider the ideal $I = (xb - 1) \sbq B[x]$ , and let $J = I \cap A[x]$ . Note that $B[x]/I = B[\tfrac{1}{b}]$ , and $A[x]/J$ is the image $C$ of the composite map $A[x] \ra B[x] \ra B[\tfrac{1}{b}]$ (first isomorphism theorem). Write $g$ for the inclusion $C \ra B[\tfrac{1}{b}]$ . Note that the map

$\psi \colon \Spec B\left[\tfrac{1}{b}\right] \ra \Spec C$

induced by $g$ is just the restriction of $\phi$ to the closed subsets $V(I) \ra V(J)$ , hence it is a closed map. Since $g$ is injective, Lemma 1 asserts that

$g^{-1}\left(B\left[\tfrac{1}{b}\right]\x\right) = C\x.$

But $\tfrac{1}{b}$ is invertible in $B[\tfrac{1}{b}]$ (its inverse is $b$ ), and it lies in $C$ since it is the image of $x$ under $A[x] \ra B[\tfrac{1}{b}]$ . Hence, it is invertible in $C$ , i.e. $b \in C$ . That is, we can write

$b = a_0 + a_1 \left(\frac{1}{b}\right) + \ldots + a_n \left(\frac{1}{b}\right)^n$

for certain $a_0, \ldots, a_n \in A$ . Hence, $b^{n+1} - a_0 b^n - \ldots - a_n = 0$ in $B\left[\tfrac{1}{b}\right]$ . Hence, some multiple $b^m \left(b^{n+1} - a_0 b^n - \ldots - a_n\right)$ is $0$ in $B$ , proving that $b$ is integral over $A$ . $\qedsymbol$

Corollary. Let $f \colon A \ra B$ be a ring homomorphism such that the associated map $\Spec B \ra \Spec A$ is proper. Then $f$ is finite.

Proof. Let $I$ be the kernel of $f$ ; then $\Spec B \ra \Spec A$ lands in $\Spec A/I$ . Then $\Spec B \ra \Spec A/I$ is proper (Hartshorne, Corollary II.4.8(e)). Moreover, the map

$\bar{f} \colon A/I \ra B$

is injective, hence integral by Lemma 2. Since $\Spec B \ra \Spec A/I$ is proper, it is of finite type. Hence, $\bar{f}$ is integral and of finite type, hence finite. Hence so is $f$ . $\qedsymbol$

A more geometric version is the following:

Theorem. Let $\phi \colon X \ra Y$ be a morphism of schemes that is both affine and proper. Then $\phi$ is finite.

Proof. Let $U = \Spec A$ be an affine open in $Y$ . Then $V = \phi^{-1}(U)$ is affine (since $\phi$ is an affine morphism); say $V = \Spec B$ . Then the restriction $\phi|_V \colon V \ra U$ is proper (properness is local on the target), hence $B$ is finite over $A$ by the theorem above. This proves that $\phi$ is finite. $\qedsymbol$

Remark. The converse is also true, and in fact much easier: a finite morphism is affine by definition, and proper since in the affine case it is in fact projective. The question whether finite morphisms are always projective turns out to be somewhat subtle: it is true in the EGA sense of projective (it is given by the relative Proj of some sheaf of graded algebras, i.e. it embeds into some projective bundle), but not in the sense of Hartshorne (it embeds into $\P^n_Y$ for some $n$ , i.e. it embeds into a trivial projective bundle).

I believe that the following is an example of the last statement (i.e. a finite morphism which is not H-projective). Let $K$ be a field of infinite degree of imperfection, i.e. $\char K = p$ and $K$ is an infinite extension of $K^p$ . Then Theorem 2 in [B-McL] says that for each $n \in \N$ there exists a finite field extension $L_n$ of $K$ which cannot be generated by fewer than $n$ elements¹. Correspondingly, $\Spec L_n \ra \Spec K$ cannot be embedded into $\P^m_K$ for $m < n$ . Then consider $Y$ an infinite disjoint union of $\Spec K$ (labelled by $\N$ ), and $X$ the disjoint union of $\Spec L_n$ for all $n$ . Then $Y$ is certainly projective in EGA’s sense (although any projective bundle it embeds into cannot have constant dimension). Yet it cannot be embedded into any $\P^n_X$ .

References.
[B-McL] M.F. Becker, S. MacLane, The minimum number of generators for inseparable algebraic extensions. Bull. Amer. Math. Soc. Volume 46-2 (1940), p. 182-186.

¹It is probably not very hard to actually come up with an example of such a field $K$ with extensions $L_n$ . I think that $K = \F_p(x_1, x_2, \ldots)$ and $L_n = K(\sqrt[\uproot{3}p]{x_1}, \ldots, \sqrt[\uproot{3}p]{x_n})$ should do the trick.

Lovely little lemmas

Or maybe ‘amusing observations’

Category Archives: Commutative algebra