Number theory is heavy metal

Posted on 1 April 2019 by Remy

As some of you may be aware, I am a musician as well as a mathematician. I often like to compare my experiences between the two. For example, I found that my approach to the creative process is not dissimilar (in both, I work on the more technical side, with a particular interest in the larger structure), and I face the same problems of excessive perfectionism (never able to finish anything).

This post is about some observations about the material itself, as opposed to my interaction with it.

Universal donor.

A look at the arXiv listing for algebraic geometry shows the breadth of the subject. Ideas from algebraic geometry find their way into many different areas of mathematics; from representation theory and abstract algebra to combinatorics and from number theory to mathematical physics. But when I say algebraic geometry is a universal donor (of ideas, techniques, etc.), I also mean that its applications to other fields far outnumber the applications of other fields to algebraic geometry, and that the field of algebraic geometry is largely self-contained.

Much the same role is played by classical music inside musical composition. Common practise theory is used throughout Western music, whether you’re listening to hip hop, trance, blues, ambient electronic, bluegrass, or hard rock. Conversely, the influence of popular genres music on [contemporary] classical is comparatively little. One could therefore argue that classical music is a universal donor in the field of musical composition.

Universal receptor.

The opposite is the case for number theory. Another vast area, number theory often uses ingenious arguments combining ideas from algebra, combinatorics, analysis, geometry, and many other areas. In practical terms, the amount of material that a number theorist needs to master is immense: whatever solves your particular problem.

So is there any musical genre that plays a similar role? I claim that metal fits the bill. With a vast list of subgenres including thrash metal, black metal, doom metal, progressive metal, death metal, symphonic metal, nu metal, grindcore, hair metal, power metal, and deathcore, metal writing often contains creative combinations of other genres, from classical music to ambient electronic, and from free jazz to hip hop. As Steven Wilson of Porcupine Tree said about his rediscovery of metal:

I said to myself, this is where all the interesting musicians are working! Because for a long time I couldn’t find where all these creative musicians were going… You know in the 70’s they had a lot of creative musicians like Carlos Santana, Jimmy Page, Frank Zappa, Neil Young, I was thinking “where are all these people now?” and I found them, they were working in extreme metal.

I sometimes think of metal as a small microcosmos reflecting the full range of Western (and some non-Western) music, tied together by distorted guitars and fast, technical drumming.

In summary, algebraic geometry is classical music, and number theory is heavy metal.

Not every open immersion is an open immersion

Posted on 13 September 2018 by Remy

An immersion (or locally closed immersion) of schemes is a morphism $f \colon X \to Y$ that can be factored as $X \to U \to Y$ , where $X \to U$ is a closed immersion and $U \to Y$ is an open immersion. If it is moreover an open morphism, it need not be an open immersion:

Example. Let $X$ be a nonreduced scheme, and let $X_{\text{red}} \to X$ be the reduction. This is a closed immersion, whose underlying set is the entire space. Thus, it is a homeomorphism, hence an open morphism. It is not an open immersion, for that would force it to be an isomorphism. $\qedsymbol$

Remark. However, every closed immersion is a closed immersion; see Tag 01IQ.

Finiteness is not a local property

Posted on 27 January 2018 by Remy

In this post, we consider the following question:

Question. Let $A$ be a Noetherian ring, and $M$ and $A$ -module. If $M_\mathfrak p$ is a finite $A_\mathfrak p$ -module for all primes $\mathfrak p \subseteq A$ , is $M$ finite?

That is, is finiteness a local property?

For the statement where local means the property is true on a cover by Zariski opens, see Tag 01XZ. Some properties (e.g. flatness) can also be checked at the level of local rings; however, we show that this is not true for finiteness.

Example 1. Let $A = \mathbb Z$ , and let $M = \bigoplus_{p \text{ prime}} \mathbb Z/p\mathbb Z$ . Then $M_{(p)} = \mathbb Z/p\mathbb Z$ , because localisation commutes with direct sums and $(\mathbb Z/q\mathbb Z)_{(p)} = 0$ if $q \neq p$ is prime. Thus, $M_{(p)}$ is finitely generated for all primes $p$ . Finally, $M_{(0)} = 0$ , because $M$ is torsion. But $M$ is obviously not finitely generated.

Example 2. Again, let $A = \mathbb Z$ , and let $M \subseteq \mathbb Q$ be the subgroup of fractions $\frac{a}{b}$ with $\gcd(a,b) = 1$ such that $b$ is squarefree. This is a subgroup because $\frac{a}{b} + \frac{c}{d}$ can be written with denominator $\lcm(b,d)$ , and that number is squarefree if $b$ and $d$ are. Clearly $M$ is not finitely generated, because the denominators can be arbitrarily large. But $M_{(0)} = \mathbb Q$ , which is finitely generated over $\mathbb Q$ . If $p$ is a prime, then $M_{(p)} \subseteq \mathbb Z_{(p)}$ is the submodule $\frac{1}{p}\mathbb Z_{(p)}$ , which is finitely generated over $\mathbb Z_{(p)}$ .

Another way to write $M$ is $\sum_{p \text{ prime}} \frac{1}{p}\mathbb Z \subseteq \mathbb Q$ .

Remark. The second example shows that over a PID, the property that $M$ is free of rank $1$ can not be checked at the stalks. Of course it can be if $M$ is finitely generated, for then $M$ is finite projective [Tag 00NX] of rank $1$ , hence free since $A$ is a PID.

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 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.

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. $\qedsymbol$

Sites without a terminal object

Posted on 3 April 2017 by Remy

Let $\mathcal C$ be a site with a terminal object $X$ . Then the cohomology on the site is defined as the derived functors of the global sections functor $\Gamma(X,-)$ . But what do we do if the site does not have a terminal object?

The solution is to define $H^i(\mathcal C,-)$ as $\Ext{\mathcal O}{i}(\mathcal O,-)$ , where $\mathcal O$ denotes the structure sheaf if $\mathcal C$ is a ringed site. If $\mathcal C$ is not equipped with a ring structure, we take $\mathcal O$ to be the constant sheaf $\underline{\mathbb Z}$ ; this makes $\mathcal C$ into a ringed site.

Lemma. Let $\mathcal C$ be a site with a terminal object $X$ . Then the above definitions agree, i.e.

$H^i(X,-) = \Ext{\mathcal O}{i}(\mathcal O,-).$

Proof. Note that $\Hom_{\mathcal O}(\mathcal O, \mathscr F) = \Gamma(X, \mathscr F)$ , since any map $\mathcal O(X) \to \mathscr F(X)$ can be uniquely extended to a morphism of (pre)sheaves $\mathcal O \to \mathscr F$ , and conversely every such morphism is determined by its map on global sections. The result now follows since $\Ext{\mathcal O}{i}(\mathcal O, -)$ and $H^i(X,-)$ are defined as the derived functors of $\Hom_{\mathcal O}(\mathcal O,-)$ and $\Gamma(X,-)$ respectively. $\qedsymbol$

Remark. From this perspective, it seems quite magical that for a sheaf $\mathscr F$ of $\mathcal O_X$ -modules on a ringed space $(X,\mathcal O_X)$ , the cohomology groups $\Ext{\mathcal O_X}{i}(\mathcal O_X,\mathscr F)$ and $\Ext{\underline{\Z}}{i}(\underline{\Z},\mathscr F)$ agree. It turns out that this is true in the setting of ringed sites as well; see Tag 03FD.

So why is this useful? Let’s give some examples of sites that do not have a terminal object.

Example. Let $G$ be a group scheme over $k$ . Then we have a stack $BG$ of $G$ -torsors. The objects of $BG$ are pairs $(U,P)$ , where $U$ is a $k$ -scheme and $P$ is a $G$ -torsor over $U$ . Morphisms $(U,P) \to (U',P')$ are pairs $(f,g) \colon (U,P) \to (U',P')$ making the diagram

Rendered by QuickLaTeX.com

commutative. This forces the diagram to be a pullback, since all maps between $G$ -torsors are isomorphisms.

The (large) Zariski site on $BG$ is defined by declaring coverings $\{(U_i, P_i) \to (U,P)\}$ to be families such that $\{U_i \to U\}$ is a Zariski covering (and similarly for the étale and fppf sites).

Now does the category $BG$ have a terminal object? This would be a $G$ -torsor $P_0 \to U_0$ such that every other $G$ -torsor $P \to U$ admits a unique map to it, realising $P$ as the pullback of $P_0$ along $U \to U_0$ . But this object would exactly be the classifying stack $U_0 = BG$ , which does not exist as a scheme (or algebraic space). The fact that a terminal object does not exist is the whole reason we need to define it as a stack in the first place!

Example. Let $X/k$ be a variety in characteristic $p > 0$ ; for simplicity, let’s say $k = \mathbb F_p$ . Then consider the crystalline site of $X/\Spec(\Z/p^n\Z)$ . Roughly speaking, its objects are triples $(U,T,\delta)$ , where $U \to X$ is an open immersion, $U \to T$ is a thickening with a map to $\Spec{\F_p} \to \Spec{\Z/p^n\Z}$ , and $\delta$ is a divided power structure on the ideal sheaf $\mathcal I_U \subseteq \mathcal O_T$ (with a compatibility condition w.r.t. $\Spec{\F_p} \to \Spec{\Z/p^n\Z}$ ). There is a suitable notion of morphisms.

This site does not have a terminal object, basically because there are many thickenings on $U = X$ with the respective compatibilities. (I am admittedly no expert, and it could very well be true that this is not 100% correct. However, I am certain that the crystalline site in general does not have a terminal object.)

Odd degree Betti numbers are even

Posted on 19 May 2016 by Remy

In characteristic 0, it follows from the Hodge decomposition and Hodge symmetry that the Betti numbers $h^i(X) = \dim H^i(X^{\operatorname{an}},\mathbb C)$ of a smooth proper complex variety $X/\mathbb C$ are even when $i$ is odd. In characteristic $p$ however, both Hodge-de Rham degeneration and Hodge symmetry fail (and de Rham cohomology is not a Weil cohomology theory), so we cannot use this method to obtain the result.

On the other hand, in the projective case, we can use hard Lefschetz plus the explicit description of the Poincaré pairing to conclude (we get a perfect alternating pairing, so the dimension has to be even). This leaves open the proper (non-projective) case in positive characteristic. This was settled by Junecue Suh [1]. I will explain the case for finite fields; one can easily reduce the general case to this case.

Notation. Throughout, $K$ will be a $p$ -adic field with ring of integers $W = \O_K$ , residue field $k$ of size $q$ , and (normalised) valuation $v$ such that $v(q) = 1$ (this is the $q$ -valuation on $K$ ).

Throughout, $X$ will be a smooth proper variety over $k$ . We will write $h^i(X)$ for the Betti numbers of $X$ . It can be computed either as the dimension of $H^i\et(\bar X, \Q_\ell)$ , or that of $H^i_{\operatorname{crys}}(X/W)[\frac{1}{p}]$ .

Remark. Recall that if $f$ is the characteristic polynomial of Frobenius acting on $H^i\et(\bar X, \mathbb Q_\ell)$ for $\ell \neq p$ , and $\alpha \in \bar{\mathbb Q}$ is the reciprocal of a root of $f$ , then for every complex embedding $\sigma \colon \bar \Q \to \C$ we have

(1) $\begin{equation*} |\sigma(\alpha)| = q^{\frac{i}{2}}. \end{equation*}$

The same holds for the eigenvalues of Frobenius on crystalline cohomology (in fact, the characteristic polynomials agree). All reciprocal roots are algebraic integers, and $f \in \mathbb Z[t]$ .

Defintion. An algebraic integer $\alpha \in \bar \Q$ is a $q^i$ -Weil integer if it satisfies (1) (for every embedding $\sigma \colon \bar \Q \to \C$ ).

Lemma. Let $f \in \mathbb Q[t]$ be a polynomial, and let $S$ be the multiset of reciprocal roots of $f$ . Assume all $\alpha \in S$ are $q^i$ -Weil integers. Then $v(S) = i - v(S)$ (counted with multiplicity).

Proof. If $\alpha \in S$ , then $\frac{q^i}{\alpha}$ is the complex conjugate with respect to every embedding $\sigma \colon \bar \Q \to \C$ . Thus, it is conjugate to $\alpha$ , hence a root of $f$ as well (with the same multiplicity). Taking valuations gives the result. $\qedsymbol$

Theorem. Let $X$ be smooth proper over $k$ , and let $i$ be odd. Then $h^i(X)$ is even.

Proof. The Frobenius-eigenvalues whose valuation is not $\frac{i}{2}$ come naturally in pairs $(\alpha, \frac{q^i}{\alpha})$ . Now consider valuation $\frac{i}{2}$ . Note that the $p$ -valuation of the semilinear Frobenius $F$ equals the $q$ -valuation of the $K$ -linear Frobenius $F^r$ (which is the one used in computing the characteristic polynomial $f$ ). The sum of the $p$ -valuations of the roots should be an integer, because $f$ has rational coefficients. Thus, there needs to be an even number of valuation $\frac{i}{2}$ eigenvalues, for otherwise their product would not be a rational number. $\qedsymbol$

References.

[1] Suh, Junecue, Symmetry and parity in Frobenius action on cohomology. Compos. Math. 148 (2012), no. 1, 295–303. MR2881317.

Lovely little lemmas

Or maybe ‘amusing observations’

Category Archives: Algebraic geometry