Posted by Jason Polak on 03. October 2017 · Write a comment · Categories: elementary · Tags: , ,

Let $F$ be a finite field. Did you know that given any function $\varphi:F\to F$, there exists a polynomial $p\in F[x]$ such that $\varphi(a) = p(a)$ for all $a\in F$? It’s not hard to produce such the required polynomial:
$$ p(x) = \sum_{a\in F} \left( \varphi(a)\prod_{b\not= a}(x – b)\prod_{b\not=a}(a-b)^{-1} \prod \right)$$
This works because every nonzero element of $F$ is not a zerodivisor.

The same cannot be said of infinite fields. If $F$ is infinite, then there are functions $\varphi:F\to F$ that cannot be represented as polynomials. That’s because the cardinality of $F[x]$ is the same as that of $F$ when $F$ is infinite. However, the number of functions $F\to F$ is greater than the cardinality of $F$. Therefore, there simply aren’t enough polynomials.

But, one does not have to go to infinite fields. For any prime $q$, there are functions $\Z/q^2\to \Z/q^2$ that cannot be represented as a polynomial. This is true because if $\varphi$ is a polynomial function, then $\varphi(x + q)\equiv \varphi(x)$ modulo $q$. Therefore, any of the $q^{q-2}$ functions $\varphi:\Z/q^2\to\Z/q^2$ satisfying $\varphi(0) = 0$ and $\varphi(q) = 1$ cannot be represented by polynomial.

In Strasbourg Part 1, I promised to deliver a few summaries of the minicourses given at the special week hosted at the Institut de Recherche Mathématique Avancée. In the next few posts, I will highlight a few things that occurred in Bernard le Stum‘s lectures on rigid cohomology. My posts are not meant to be a complete summary, and indeed I would like to be as succint as possible. Furthermore, I intend to add additional material explaining some of of the prerequisites. This first post covering rigid cohomology will try to explain the motivation for considering rigid cohomology in the first place.

Introduction

The lectures on rigid cohomology were given by Bernard le Stum who wrote the only existing textbook on rigid cohomology developed primarily by Pierre Berthelot, so the reader should definitely consult le Stum’s book for more details, as well as consult the slides when they appear, which contain many wonderful examples that I am omitting.

But what is rigid cohomology, and why is it useful?
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