# Local cohomology and cohomology of line bundles on projective spaces

Posted on December 4, 2017 by Dima

tags: likbez, local rings, cohomology, modules, injective hull

all this is terribly unpolished, and arguments are ridden with holes, as is customary on this blog

### Flatness

Frist things first: Nakayama’s lemma.

Statement 1: Let $$R$$ be a ring, $$I$$ an ideal Let $$M$$ be a finitely generated module, and assume $$IM=M$$. Then there exists $$x \in 1 + I$$ such that $$xM=0$$.

Proof: let $$x_1, \ldots, x_n$$ be the generators of $$M$$. Then there exists a matrix $$A=(a_{ij}) \in I^{n \times n}$$ such that $$x_i = \sum a_{ij} x_j$$, or, $$(I-a)x=0$$. Multiply by the matrix formed of minors on the left and get $$\det (Id-A) \cdot I \cdot x = 0$$ where $$x=(x_i)$$. Note that $$\det (Id-A) \in 1 + I$$

Statement 2: Let $$R$$ be a ring, $$I$$ an ideal that annihilates all simple $$R$$-modules (equivalently: if it lies in the intersection of all maximal ideals, i.e. in a local ring, in the maximal ideal), then $$M=0$$.

Proof. Follows from Statement 1, since $$x$$ is invertible. Indeed, otherwise $$(x)$$ would be contained in a maxmial ideal ideal that intersects trivially with $$I$$ or otherwise $$1 \in I$$, but $$I \cap {\mathfrak{m}}\neq 0$$ for any maxmial $${\mathfrak{m}}$$.

But now there is also a more direct proof. Let $$x_1, \ldots, x_n$$ be the minimal set of generators of $$M$$ and assume $${\mathfrak{m}}M = M$$. Then $(1-a_n)x_n = \sum_{i=1}^{n-1} a_i x_i,$ and since $$1-a_n$$ is invertible, $$x_1, \ldots, x_{n-1}$$ generate $$M$$, which contradicts the minimality of the set of generators.

A module $$M$$ is flat iff $$-- \otimes M$$ is an exact functor.

Lemma. $$M$$ is flat if for all ideals $${\mathfrak{p}}\subset R$$ $0 \to {\mathfrak{p}}\otimes M \to M \to M/{\mathfrak{p}}M \to 0$ is exact (and for that it suffices to check that the first arrow is inclusion).

Proposition. Assume $$R$$ is a Dedekind domain (i.e. every ideal is a product of prime ideals; equivalently, 1-dimensional integrally closed). Then an (integral) $$R$$-algebra $$S$$ is flat iff $$R \hookrightarrow S$$ is injective.

Proposition. Let $$f: X={\mathrm{Spec}}S \to Y={\mathrm{Spec}}R$$ be a morphism. Then $$S$$ is a flat $$R$$-algebraic iff $${\mathcal{O}}_{X,x}$$ is flat over $${\mathcal{O}}_{Y,y}$$ for all $$x \in X$$, all $$y = f(x)$$.

Lemma. Let $$R$$ be a local ring withe the maximal ideal $${\mathfrak{m}}$$. Let $$M$$ be a finitely generated $$R$$-module. Then $$M$$ is free iff $${\mathrm{Tor}}_1(M,k) = 0$$ (and moreover both conditions are equivalent to being projective and flat).

Proof. Assume $${\mathrm{Tor}}_1(M,k)$$ vanishes. Since $$M$$ is finitely generated, there is a surjective morphism $$f: R^n \to M$$ where $$n = \dim k$$. Tensoring the short exact sequence $${\mathrm{Ker}}f \to R^n \to M$$ with $$k$$ we get that $${\mathrm{Ker}}f \otimes k = 0$$ which by Nakayama lemma implies $${\mathrm{Ker}}f = 0$$.

Lemma. Setting as above. $$M$$ has a projective resolution of length $$n$$ if and only if $$Tor^n(M,k) = 0$$.

Proof. The non-trivial direction is right-to-left. Assume $\ldots \to P_{n+1} \to P_n \to P_{n-1} \to \ldots \to P_0 \to M \to 0$ is a resolution. Let $$Z_i = {\mathrm{Ker}}d_i \hookrightarrow P_i$$. Then we have short exact sequences $0 Z_i \to P_i \to Z_{i-1} \to 0$ Let $$i=0$$ and tensor with $$k$$, then $\ldots{\mathrm{Tor}}_n(M,k)=0 \to {\mathrm{Tor}}_{n-1}(Z_0,k) \to {\mathrm{Tor}}_{n-1}(P_0, k)=0 \to \ldots$ So $${\mathrm{Tor}}_{n-1}(Z_0,k)=0$$. Arguing inductively, we get $${\mathrm{Tor}}_{n-i-1}(Z_i,k) = 0$$. In particular, $${\mathrm{Tor}}_{1}(Z_{n-2}, k)=0$$ and by the previous lemma it’s free and hence projective. Then $Z_{n-2} \to P_{n_2}\to P_{n-1} \to \ldots$ is a length $$n$$ projective resolution.

### Koszul complex

Let $$R$$ be a local ring with the maximal ideal $${\mathfrak{m}}$$ and residue field $$k=R/{\mathfrak{m}}$$. Then there exists a particularly interesting projective (in fact, free) resolution of $$k$$. Let $$x_1, \ldots, x_n$$ be a set of generators of $${\mathfrak{m}}$$. Let $$K{\langle}\xi_1, \ldots, xi_n {\rangle}$$ be the ring of grassamanian polynomials in $$\xi_1, \ldots, x_n$$ graded by degree, where $$n$$-th graded piece is isomorphic to $$\wedge^n R$$. Define the differential map to be $d(\xi_i) = x_i$ on the 1-homogeneaus piece, and extend it linearly and applying Leibniz’s rule $d(a \wedge b) = da \wedge b + (-1)^{|a||b|}a \wedge db$ The graded pieces of $$K{\langle}\xi_1, \ldots, \xi_n {\rangle}$$ with this differential are element of the Koszul complex.

# From abelianized inertia to divisors

Posted on December 1, 2017 by Dima

tags: curves, Jacobians, finite fields

The goal of this note is to clarify some points from a paper of Bogomolov, Korotiaev and Tschinkel.

Bogomolov’s approach to birational anabelian geometry consists in looking at abelianisations of absolute Galois groups of function fields of varieties over a fixed algebraically closed field $$k$$ and discerning some extra structures that can be computed from the full absolute Galois group in order to reconstruct the function field.

If $$X$$ is a curve over $$k$$, then by results of Florian Pop $${\operatorname{Gal}}(k(X)^{alg}/k(X))$$ is profinite free on $$|X(k)|$$ generators (result independently established by David Harbater: MR1352282 (97b:14035) Harbater, David. Fundamental groups and embedding problems in characteristic p.), and so one cannot distinguish between the curves by the absolute Galois group of their function fields alone.

The anabelian problrem for function fields of curves is reformulated in the BKT paper as follows: let $$(G_1, \mathcal{I}_1)$$ and $$(G_2, \mathcal{I})$$ be a two pairs of abelianizations of absolute Galois groups of functions fields $$k(X_1)$$ and $$k(X_2)$$ of curves of genuse $$\geq 2$$ together with collections of abelianized inertia groups of points.

Let us unpack the terminology.

Let $$L \supset K$$ be a finite extension of fields and $$v: K^\times \to {\mathbb{Z}}$$ a discrete valuation. Assume $$v': L^\times \to {\mathbb{Z}}$$ is a valuation such that $$v'|_K = v|_K$$. The Galois group $${\operatorname{Gal}}(L/K)$$ acts transitively on all such extensions, and the decomposition group $$D_{v'}$$ of $$v'$$ is the subgroup of elements of $${\operatorname{Gal}}(L/K)$$ that fixes $$v'$$: $$v(\sigma\ x) = v(x)$$ for all $$x \in L^\times$$. Decomposition groups of different extensions of $$v$$ to $$L$$ are conjugated. Therefore if the extension $$L/K$$ is Abelian, all decomposition groups coincide. Decomposition group naturally maps to the Galois group of the extension of residue fields: residue field $${\mathcal{O}}_{v'}/{\mathfrak{m}}_{v'}$$ of $$v'$$ over the residue field of $$v$$, the kernel of this homomorphism is called the inertia group. In fact, everything said above applies to non-finitely generated extensions as well. We will denote the abelianized inertia group of any extension of $$v$$ as $$I_{v,a}$$ or simply $$I_v$$.

Here’s how to interpet the decomposition and inertia group in terms of Kummer theory. A cyclic Galois extension $$L/K$$ gives rise to the $$\mu_n$$-torsor $${\operatorname{Spec}}L \otimes_K L$$ defined over $$K$$, and such things are classified by $$H^1$$ of the absolute Galois group: $$H^1(K, \mu_n)$$. Since $$\mu_n$$ carries trivial Galois action, cocycle condition degenerates into the requirement of being a homomorphism from the absolute Galois group of $$K$$ to $$\mu_n$$, and each such homomorphism factors through the maximal abelian factor, let us denote it $${\mathcal{G}}_a$$ after BKT. Thus the $$H^1$$ in question is isomorphic to $${\operatorname{Hom}}({\mathcal{G}}_a, \mu_n)$$. On the other hand, applying group cohomology functor to the Kummer exact sequence gives $$K^\times/(K^\times)^n \cong H^1(K, \mu_n)$$. Now restricting attention to powers of a prime $$l$$, and taking limit, we finally get $\widehat{K^\times} \cong {\operatorname{Hom}}({\mathcal{G}}_{a,l}, {\mathbb{Z}}_l)$ where $$\widehat{K^\times}$$ is the pro-$$l$$ completion, and $${\mathcal{G}}_{a,l}$$ is the maximal Abelian pro-$$l$$ quotient of the absolute Galois group. This identification, when dualized, tells us that $${\operatorname{Hom}}(K^\times, {\mathbb{Z}}_l) \cong {\mathcal{G}}_{a,l}$$.

In terms of this correspondence $D_v = {\operatorname{Hom}}(K^\times/1+m_v, {\mathbb{Z}}_l) \qquad I_v = {\operatorname{Hom}}(K^\times/{\mathcal{O}}_v^\times, {\mathbb{Z}}_l)$ Indeed, there is a natural mapping ${\operatorname{Hom}}(K^\times/1+m_v, {\mathbb{Z}}_l) \to {\operatorname{Hom}}({\mathcal{O}}^\times/1+{\mathfrak{m}}_v, {\mathbb{Z}}_l)$ where $${\mathcal{O}}^\times/1+{\mathfrak{m}}_v$$ is the multiplicative group of the residue field, so the homomorphisms to $${\mathbb{Z}}_l$$ that lie in the inertia must vanish on $${\mathcal{O}}_v^\times$$.

Now let $$X$$ be a smooth projective curve over an algebraically closed field $$k$$, and pick a point $$x \in X(k)$$. In view of the identification above, $$I_x$$ corresponds to the homomorphism obtained as follows. Let $$v_x: K^\times \to {\mathbb{Z}}$$ be the valuation associated with the point $$x$$. Then $$I_{x,a}$$ is the subgroup of $${\operatorname{Hom}}(K^\times, {\mathbb{Z}}_l)$$ topologically generated by $$v_x$$, or, equivalently, the subgroup of maps $$\varphi: K^\times \to {\mathbb{Z}}_l$$ such that $$\varphi({\mathcal{O}}_{v_x}^\times) = 0$$.

Now let $$k=\mathbb{F}_p^{alg}$$, and let $$X$$ be a curve of genuse $$\geq 2$$, $$K=k(X)$$. Then $0 \to K^\times/k^\times \to {\operatorname{Div}}(X) \to {\operatorname{Pic}}(X) \to 0$ Dualize it: $0 \to {\operatorname{Hom}}({\operatorname{Pic}}, {\mathbb{Z}}_l) \to {\operatorname{Hom}}({\operatorname{Div}}, {\mathbb{Z}}_l) \to {\operatorname{Hom}}(K^\times, {\mathbb{Z}}_l) \to {\operatorname{Ext}}^1({\operatorname{Pic}}, {\mathbb{Z}}_l) \to 0$ (and $${\operatorname{Ext}}^1({\operatorname{Div}}, {\mathbb{Z}}_l)$$ vanishes, since $${\operatorname{Div}}$$ is projective as a direct sum of $$X(k)$$ copies of $${\mathbb{Z}}$$). Some identifications:

• $${\operatorname{Hom}}(K^\times, {\mathbb{Z}}_l) \cong {\mathcal{G}}_{a,l}$$ by Kummer theory (above), note that all morphisms here vanish on $$k^\times$$ since it is divisible;
• $${\operatorname{Hom}}({\operatorname{Pic}}(X), {\mathbb{Z}}_l) \cong {\mathbb{Z}}_l$$, since $${\operatorname{Pic}}^0$$ is torsion;
• $${\operatorname{Hom}}({\operatorname{Div}}(X), {\mathbb{Z}}_l)$$ is the $${\mathbb{Z}}_l$$-linear space of maps $$X(k) \to {\mathbb{Z}}_l$$, and $${\operatorname{Hom}}({\operatorname{Pic}}, {\mathbb{Z}}_l)$$ are the constant maps;

Consider the pairing ${\operatorname{Maps}}(X(k), {\mathbb{Q}}_l) \times K^\times/k^\times \to {\mathbb{Q}}_l, \qquad [\mu, f] = \sum_x v_x(f) \mu(x)$ Then $I_x := \{ [a \cdot \delta_x, -]\}_{a \in {\mathbb{Z}}_l}$ where $$\delta_x(x)=1$$, $$\delta_x(y)=0$$ for all $$y \neq x$$.

# On strange efficiency of ultraproducts in mathematics

Posted on November 6, 2017 by Dima