# Temkin's definition of weight function

Posted on January 17, 2018 by Dima

tags: Berkovich spaces, weight, log structures

Let $$k$$ be a non-Archimedean valued field. We want to define a natural metric on the module of Kähler differentials $$\Omega_{Y/X}$$ for a morphism of $$k$$-analytic spaces $$X \to Y$$. This is presented in the utmost generality in this paper, see also the survey at the NATG’15 conference.

A morphism of normed modules $$f: B \to A$$ is called non-expansive if $$||f(b)|| \leq ||b||$$.

Given a morphism of normed rings $$A \to B$$ one can define a norm on the module of differentials $$\Omega_{B/A}$$ so that it is the maxmial norm making the morphism $$d: B \to \Omega_{B/A}$$ non-expansive.

In particular, given a $$k$$-ring $$A$$, the module of Kähler differentials $$\Omega_{A/k}$$ is equipped with the Kähler norm as follows: $||x||_\Omega = \inf_{x = \sum c_i db_i} \max |c_i| |b_i|$ The Kährler semi-norm on $$\Omega_B$$ is characterized by the fact that it is the maximal semi-norm making the differential a non-expansive map.

Let $$A$$ be an Banach algebra over $$k$$, and let $$v \in {\mathcal{M}}(A)$$ be a point of its Banach spectrum. Then an element $$f \in A$$ is a function on $$A$$ with values in residue fields of each point $$v$$, $${\operatorname{\mathcal{H}}}(v)$$, which is naturally normed, and $$|f|$$ is a real-valued function, defined as $$|f|(v) = ||f||_{{\operatorname{\mathcal{H}}}(v)}$$.

If $$v \in X$$ is a point of a Berkovich analytic space, we want to look at the Kähler semi-norm on the complete residue fields $${\operatorname{\mathcal{H}}}(v)$$, so that we can define a real-valued function $$|\omega|$$ by putting it to be $$||\omega||_{{\operatorname{\mathcal{H}}}(v)}$$ at a point $$v$$.

A log structure on a ring $$A$$ is a morphism of multiplictive monoids $$\alpha: M \to A$$ that induces an isomorphism $$M^\times \to A^\times$$.

If $$A$$ is a log $$k$$-ring then the module of log differentials is defined as $$\Omega_{A/k} \oplus (A \otimes M^{gp})$$ (where $$M^{gp}$$ is a groupification of the monoid) module the relations $\begin{array}{c} (0, 1 \otimes c)\\ (da, -a \otimes a) = 0 \\ \end{array}$

Along with the usual derivative $$d$$ we have a log derivative $$\delta$$, $$\delta a = (0, 1 \otimes a)$$, so that elements $$\delta a$$ should be thought of as $$d\log a$$, with the above relation meaning $\begin{array}{c} d\log c = 0, c \in k\\ a d\log a = da, a \in M^{gp}\\ \end{array}$

If $$K$$ is a valued field with a valued ring $$K^\circ$$, then $$K^\circ {\setminus}\{0\} \to K^\circ$$ is a log structure on $$K^\circ$$. We denote $$\Omega^{\log}_{K^\circ/A}$$ the module of log differentials, where the log structure on $$K^\circ$$ is as above, and the log structure on $$A$$ is $$A {\setminus}{\mathrm{Ker}}(A \to K)$$.

Adic seminorm: given a $$K^\circ$$-submodule $$M$$ is defined on $$V=M\otimes K$$ as: $||v|| = \inf {\{ |a| \mid a \in K^\circ, v \in aM \}}$

One checks that $$\Omega^{\log}_{K^\circ/A^\circ} \otimes_{K^\circ} K = \Omega_{K/A}$$, so the image $$\Omega^{\log}_{K^\circ/A^\circ} \to \Omega^{\log}_{K^\circ/A^\circ} \otimes K=\Omega_{K/A}$$ is a lattice.

Theorem. The adic semi-norm on $$\Omega^{log}_{K^\circ/A^\circ}$$ is the maximal norm making the differential non-expansive.

Let $$(V, ||\cdot||)$$ be a finite-dimensional normed vector space. Then a basis $$e_1, \ldots, e_n$$ is called $$r$$-orthogonal if for any $$v=\sum a_i e_i$$ the inequality $$||v|| \geq r\max (|a_i|\cdot||e_i||)$$ holds. If $$r=1$$ and $$||e_i||=1$$ then the basis is called orthogonal.

The valuation $v(\sum a_i t^i) = \max |a_i|$ on $$T_1$$ is called Gauss valuation (there is a more general expression for more variables, and with radii $$\neq 1$$, but let us forfeit it here).

Let $$K/k$$ be an extension of valued fields. Let $$t=(t_1, \ldots, t_n)$$ be a tuple of elements of $$L$$. We call a valuation on $$K$$ $$t$$-monomial if its restriction to $$k[t_1, \ldots, t_n]$$ is a Gauss valuation.

Lemma. Let $$\xi$$ be the Gauss point on $${\mathcal{M}}(T_1)$$, the unit disc. Then $$||dt/t||_{\xi}$$ is 1.

Proof. From the definition one immediately gets $||dt/t||_{\xi} = ||df/(f' t)||_\xi = \inf_f |1/f' t|\cdot|f| = 1$ where $$f=\sum a_i t^i$$, $$f \neq const$$. Indeed, $f't=\sum i a_i t^i$ so its norm coincides with that of $$f$$.

Corollary. The function $$x \mapsto ||dt||_{x}: {\mathcal{M}}(T_1) \to {\mathbb{R}}$$ is the radius function $$r(x)$$, i.e. the infimum of a ball containing the point $$x$$.

Proof. Indeed, any point in the unit disc is the gauss point of a disc of radius $$r(x)$$ and coordinate function of the form $$T-a$$, and $$dt=d(t-a)$$, so from the previous lemma $$1=||dt/t||_x \leq ||dt||_x \cdot |1/t|$$.

# 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.