﻿ Algebraic geometry # Welcome to Mathematics

## Commutative Algebra and Algebraic Geometry

You are here : Home Research Algebraic geometry
 Contenus

 *  Algebraic geometry

# Introduction to schemes theory

## Spectrum of ring

### Zariski topology

Let $A$ be a commutative ring with unit. We let $Spec(A)$ denote the set of prime ideals of $A$. We call it the spectrum of the ring $A$. By convention, the unit ideals is not a prime ideal. Thus $Spec(\{0\})=\emptyset$.
We well now endow $Spec(A)$ with a topology structure. For any ideal $I$ of $A$, let $V(I)=\{P\in Spec(A) \ , \ I\subseteq P\}$. If $f\in A$, let $D(f)=Spec(A)\setminus V(fA)$.

#### Proposition:

Let $A$ be a ring. We have the following properties:
1. For any pair of ideals $I,J$ of $A$, we have $V(I)\cup V(J)=V(J\cap J)$,
2. Let $(I_{\alpha})_{\alpha}$ be any family of ideals of $A$. Then $\cap_{\alpha}V(I_{\alpha})=V(\sum_{\alpha}I_{\alpha})$, and
3. $V(A)=\emptyset$ and $V(0)=Spec(A)$.
In particular, there exists a unique topology on $Spec(A)$ whose closed subsets are the sets of the form $V(I)$ for an ideal of $A$. moreover, the sets of the form $D(f)$ , $f\in A$, constitute a base of open subsets on $Spec(A)$.
Let $A$ be a ring. We call the topology defined on the obvious proposition the Zariski topology on $Spec(A)$. An open set of the form $D(f)$ is called a principal open subset , while its complement $V(f)=V(fA)$ is called a principal closed subset.

#### Lemma:

Let $A$ be a ring. Let $I,J$ be two ideals of $A$. The following propositions are true.
1. $\sqrt I=\cap_{P\in V(I)}P$, (i.e the radical of $I$ equals the intersection of the ideals $P\in V(I)$ )
2. We have $V(I)\subseteq V(J)$ if and only if $J\subseteq \sqrt I$.
Let $\varphi$ be a rings homomorphism . Then we have a maps of sets $Spec\varphi:Spec(B)\to Spec(A)$ defined by $P\mapsto \varphi^{-1}(P)$ for very $P\in Spec(B)$.

#### Lemma:

Let $\varphi$ be a rings homomorphism. Let $f=Spec\varphi$ be the map associated to $\varphi$ as above. Then we have:
1. The map $f$ is continuous,
2. If $\varphi$ is surjective, then $f$ induces a homeomorphism from $Spec(B)$ onto the closed subset $V(\ker\varphi)$ of $Spec(A)$,
3. If $\varphi$ is a localization morphism $A\to S^{-1}S$, then $f$ is a homeomorphism from $Spec(S^{-1}A)$ onto the subspace $\{P\in Spec(A) \ , \ P\cap S=\emptyset\}$ of $Spec(A)$.

## Ringed topological spaces

### Sheaves

#### Definition:

Let $X$ be a topological space. A presheave $\mathcal F$ (of abelian groups) on $X$ consists of the following data:
• an Abelian group $\mathcal F(U)$ for every open subset $U$of $X$, and
• a groups homomorphism (restriction map) $\rho_{UV}:\mathcal F(U)\to \mathcal F(V)$ for every pair of open subsets $V\subseteq U$.
Which verify the following conditions:
1. $\mathcal F(\emptyset)=0$;
2. $\rho_{UU}=Id$;
3. If we have three open subsets $W\subseteq V\subseteq U$, then $\rho_{UW}=\rho_{VW}\circ\rho_{UV}$.
An element $s\in \mathcal F(U)$ is called a section of $\mathcal F$ over $U$. We let $s|_V$ denote the element $\rho_{UV}(s)\in \mathcal F(V)$, and we call it the restriction of $s$ to $V$.
 Articles
 Articles
 Maths
 Exercice
 Mohamed AQALMOUN Professeur de l'enseignement supérieur ENS-Fès