Sheaves and Schemes: Properties of Schemes

This is a small note for different properties of schemes.

Definition of Scheme

First of all, quite obviously, we give the definition of isomorphisms of ringed spaces:

Definition
An isomorphism of ringed spaces $(X,\mathcal O_X)$ and $(Y,\mathcal O_Y)$ consists of following data:

1. a homeomorphism $\pi:X\rightarrow Y$
2. an isomorphism of sheaves $\mathcal O_Y\rightarrow \pi_*\mathcal O_X$

Now we can give the definition of schemes

Definition
A scheme $(X,\mathcal O_X)$ is a ringed space such that for any point $p\in X$ there is an open neighbourhood $U$ such that $(U,\mathcal O_X|_U)$ is isomorphic to an affine scheme $(A,\operatorname{spec}(A))$ for some ring $A$

A direct observation is that

Proposition
If $U$ is an open subset of $X$, then $(U,\mathcal O_X|_U)$ is a scheme. It’s called an open subscheme of $X$.

Affine-local Properties

Some properties are defined for any affine open subset, we want to know whether it is enough to check that certain property over a particular affine open cover. First we have a proposition

Proposition
Suppose $\operatorname{Spec}(A)$ and $\operatorname{Spec}(B)$ are affine open subschemes of a scheme $X$. Then $\operatorname{Spec}(A)\cap\operatorname{Spec}(B)$ is the union of open subsets that are simultaneously distinguished open subschemes of $\operatorname{Spec}(A)$ and $\operatorname{Spec}(B)$.

Then if a property $P$ could be checked over a distinguished open cover, we could therefore pass to a different affine cover using this lemma. Such property $P$ is called Affine-local Property. We summaries this in the following lemma:
Affine Communication Lemma
Let $P$ be a property enjoyed by some affine open subsets of a scheme $X$, such that

1. If an affine open subset $\operatorname{Spec}(A)$ has property $P$, then for any $f\in A$, $\operatorname{Spec}(A_f)$ does too
2. If $A$ is covered by $A_{f_i}$ for $1\leq i\leq n$ and all $\operatorname{Spec}(A_{f_i})$ has property $P$, then so does $\operatorname{Spec}(A)$.

If one affine cover of $X$ has property $P$, then all affine cover has property $P$.

This is quite obvious since once $P$ is ture for some affine cover $\{A_i\}$, for any other affine cover $\{B_j\}$, pick $B_j$, apply the proposition above to any $A_i$ where $A_i\cap B_j\neq\emptyset$. Then we obatin an affine cover for $B_j$ where all subsets have property $P$. Since $B_j$ is quaiscompact, we could choose a finite subcover then by lemma, we know that $B_j$ also has property $P$.

Properties

Topological Properties

Quasicompactness

The word “quasicompact” simply means “compact” without Hausdorff. Combine this with the definition of scheme, we have following propostion

Proposition
A scheme $X$ is quasicompact iff it is a finite union of affine open subschemes.

But we have a more interesting proposition below
Proposition
If $X$ is a quasicompact scheme, then every point has a closed point in its closure.

Proof
Assume that $X$ has been covered by affine open subschemes $U_1,…,U_n$, let $p$ be a point in $X$.
WLOG assume that $p\in U_1$, if $p$ is closed in $X$ then we are done. If not, pick a closed point (closed in $U_1$) $p_1\in \bar{p}\cap U_1$. If it’s not closed, its closure $\bar{p}_1$ has a nonempty intersection with $U_2$, pick another closed point (closed in $U_2$) $p_2\in\bar{p}_1\cap U_2$. Since $X$ is $T_0$, and every open neighbourhood of $\bar{p}_1$ contains $p_2$, there must be some open neighbourhood of $p_2$ doesn’t contain $p_1$. Therefore $p_1\not\in\bar{p}_2$, so $\bar{p}_2\cap U_1=\emptyset$. Now if $p_2$ is not closed, assume WLOG that its closure has nonempty intersection with $U_3$, repeat this process until we reached a point $p_n$ closed in $U_n$, and $\bar{p}_n\cap U_i=\emptyset$ for $1\leq i < n$. Therefore $p_n$ is closed in $X$.

Quasiseperateness

A scheme $X$ is called quasiseperated if the intersection of any two quasicompact open sets is quasicompact.

Algebraical Properties

Reduceness

A scheme $X$ is called reduced if $\mathcal O_X(U)$ is reduced for any open subset $U\subset X$.

Proposition
1. $X$ is reduced iff all stalks are reduced
2. If $X$ is a quasicompact scheme, it sufficies to check reduceness at closed points

Proof
If $X$ is reduced, suppose $\mathcal O_{X,p}$ is not reduced, then there is some open neighbourhood $U$ containing $p$ such that $f\in \mathcal O_X(U)$ is nilpotent: there is a smaller open neighbourhood $V$ such that $f^n|_V=0$. But then $f\in\mathcal O_X(V)$ is a nilpotent element. So we get a contradiction!

If all stalks are reduced, since $\mathcal O_X(U)\rightarrow \prod_p \mathcal O_{X,p}$ is an injection, $O_X(U)$ is reduced.

Now if $X$ is quasicompact, and one point $p$ (not necessarily closed) has a nonreduced stalk, we want to show that there is a closed point $q$ whose stalk is also nonreduced: Choose $q$ to be the closed point in the closure of $p$, let $(f,U)$ be the nilpotent element in $\mathcal O_{X,p}$, then there is some $V\subset U$ such that $f^n|_V=0$. Consider the support of $f$ (as a section), it’s closed and $p$ is contained in it. Therefore $q$ is contained in it and $(V,f)$ is a nilpotent element in $\mathcal O_{X,q}$.

Reducessness is an affine-local property. (This is obvious from the stalkwise description)

Integrality

A scheme $X$ is integral if it is nonempty and $\mathcal O_X(U)$ is an integral domain for every nonempty open subset $U$ of $X$.

Proposition
A scheme $X$ is integral iff it is irreducible and reduced.

Proof
If $X$ is integral, then by definition it is reduced. We only have to show that it is irreducible. Suppose that it is reducible, then there are two nonempty open subsets $U_1,U_2$ such that $U_1\cap U_2=\emptyset$. Consider $U=U_1\cup U_2$, the two nonzero elements $(1,0)$ and $(0,1)$ have zero product, which is a contradiction!

Suppose that $X$ is reduced and irreducible. If $X$ is not integral: For some open subset $U\subset X$, there exist $f,g\in\mathcal O_X(U)$ whose product is zero, then let $C_1$ be the zero locus of $f$ as a function and let $C_2$ be the zero locus of $g$ as a function. THen both are closed and $C_1\cup C_2=U$. Since both are nonzero, we have that $C_1$ and $C_2$ are proper subsets of $U$. This contradicts with irreducibility.

Noetherian

A scheme $X$ is called locally Noetherian if there is an affine cover $\{\operatorname{Spec}(A)\}$ where each $A$ is Noetherian. $X$ is called Noetherian if $X$ is also quasicompact.

Proposition
Locally Noetherian schemes are quasiseperated

Scheme Over $A$

A scheme $X$ with a scheme morphism $X\rightarrow \operatorname{Spec}(A)$ is called a scheme over $A$, or an $A$-scheme. If $X$ is an $A$-scheme, then the rings of sections over any open subset are $A$-algebras and the restriction maps are $A$-algebra morphisms.

Finite Type

An $A$-scheme $X$ is called locally of finite type if $X$ could be covered by affine open subsets $\operatorname{Spec}(B_i)$ where all $B_i$ are finitely generated $A$-algebra. $X$ is called Of finite type if $X$ is quasicompact.

Normality

A scheme $X$ is normal if all stalks are normal.

Factoriality

A scheme $X$ is factorial if all stalks are UFD.

Varieties

An affine variety over $k$ is a reduced scheme locally of finite type.


Referrences

  1. THE RISING SEA - Foundations of Algebraic Geometry – Ravi Vakil
  2. Can we always find $q\in\bar{\{p\}}$ with $p\not\in\bar{\{q\}}$?