Disclaimer: Nothing in this note is original.

So far we have talked a lot about algebraic varieties, they have the following properties.

  • algebraic varieties are defined over some algebraically closed fields,
  • they are finitely generated and without nilpotent,
  • they have only closed points (with Zariski topology).

Now we introduce a related but more modern concept, called schemes. In comparison to varieties, schemes have the following properties,

  • They don’t need to be defined over any algebraically closed fields,
  • they provide a broad, robust frame work,
  • they have better algebraic and functorial properties, in a sense we will elaborate later.

A scheme is roughly speaking made of the following three pieces of information.

  • A scheme consists of a set with topological structure and a structure sheaf (means that you can define functions on them),
  • is glued together from affine schemes,
  • is defined in analogy to algebraic varieties.

Affine scheme as a set

Let $R$ be a unital, commutative ring.

Definition. The affine scheme associated to $R$, or the spectrum of $R$ is the set $\text{Spec}(R)$ of all prime ideals of $R$.

Example 1. Let $R = K[X]=\mathcal{O}(X)$, the coordinate ring of an affine variety over some algebraically closed field $K$. Then $\text{Spec}(R)$ is the spectrum, i.e., the prime ideals of $\mathcal{O}(X)$. It is the affine scheme associated to $X$. Note that

\[X = \text{mSpec}(R) \subset \text{Spec}(R)\]

since a point $(a_ {1},\dots,a_ {n})$ is given by the ideal $(x_ {1}-a_ {1},\dots,x_ {n}-a_ {n})$, and $X$ is made of all its points. However the prime ideals are more than just points, it also includes irreducible varieties.

If we consider the elements of a spectrum as generic points, then since

\[\text{Spec}R := \left\{ \text{Points} \right\} \cup \left\{\text{irred. subvarieties} \right\}\]

we should also consider the irreducible subvarieties as points in the scheme. We have generalized the concept of points in varieties.

Example 2. The polynomial ring in one variable $K[T]$ where $K \subset \overline{K}$ (meaning $K$ is algebraically closed). The maximal ideals are of form $(T-t)$ for some $t \in K$, but there is also a prime ideal which is not maximal, i.e. the ideal generated by zero $(0)$. It is trivial indeed yet a prime ideal nevertheless. So when speaking of the scheme, the points include both the usual points on $K$ together with the entire $K$ itself.

Example 3. The ring of integers $\mathbb{Z}$. The spectrum of $\mathbb{Z}$ is

\[\text{Spec}\mathbb{Z} = \left\{ (p) \,\middle\vert\, p \text{ is a prime number } \right\} \cup\left\{ 0 \right\} .\]

Those are the points in the scheme.

Having defined the underling set of a scheme, we would like to continue and somehow define the “functions” and topology on the scheme.

Recall the case with varieties and structure sheaves. Functions on a variety are nothing but maps from the variety to the field over which the variety is defined on, for example for a variety as as subspace of complex manifold, the functions on it are just complex functions. However, regarding scheme, we don’t have a algebraically closed field to begin with! We just have an underlying set, namely the spectrum of some ring $R$. Fortunately, we can construct the field, called the residual field, from the ring. Recall that for a maximal ideal ${\frak m}$ the quotient $R / {\frak m}$ is a field, and for a prime ideal ${\frak p}$ the quotient $R / {\frak p}$ is a integral domain. Given a point ${\frak p}$ in the scheme, which is a prime ideal in $R$, we can consider $R / {\frak p}$, if ${\frak p}$ is also a maximal ideal then we automatically have a field, but if ${\frak p}$ is just prime then we would have an integral domain. Again, lucky for us, we already know how to turn an integral domain into a field: just take the fraction of it! For example, $\mathbb{Z}$ is a integral domain but the rational numbers $\mathbb{Q}$ form a field. So, given a point in the scheme, there is always a way to construct a field based on the point.

Definition. Let $X = \text{Spec}R$ and ${\frak p}\in X$,

  1. the residual field $\kappa({\frak p})$ of $X$ at ${\frak p}$ is defined to be the field of fractions of $R / {\frak p}$.
  2. if $f\in R$, then the value $f({\frak p})$ of $f$ at ${\frak p}$ is the image of $f$ under
\[R \to R / {\frak p} \to \kappa({\frak p}).\]

Writing $f({\frak p})$ seems to suggest that $f$ is a function at ${\frak p}$ thus a function over $\text{Spec}R$, however the residual field varies from point to point in the spectrum so the range of the function is not well defined. It is better reverse the viewpoint and regard ${\frak p}$ a function over $R$, writing ${\frak p}(f)$. However we will not bother with this and use these notation casually, it should not raise any confusion.

Example 4. Let $X$ be an affine variety over field $K$ and $R=\mathcal{O}(X)$ be the coordinate ring of it, then the residual field $\kappa({\frak p})$ is just $K$ itself if ${\frak p}$ is maximal; if ${\frak p}$ is prime but not maximal, then ${\frak p}$ is the ideal of some irreducible subvariety $Y\subset X$, the residual field $\kappa({\frak p})$ is $R / {\frak p}=R / I(Y)$ is nothing but the coordinate ring on $Y$! Amazingly, it reproduces the usual evaluation of polynomial function on points if ${\frak p}$ is a maximal idea. The reader can try to come up with the simplest example and convince yourself of it.

Example 5. Let $R = \mathbb{Z}$ and every element of $\mathbb{Z}$ automatically become a function on $\mathbb{Z}$. Consider the function $7$, $7$ evaluated at $5$ is $7$ in the residual field $\mathbb{Z} / (5)=\mathbb{Z}_ {5}$, which is $[2]$.

Equipped with this, we can define something similar to the zero locus in algebraic varieties.

Let $X=\text{Spec}R$ be an affine scheme, take any subset $S\subset R$. Each element in the scheme has two roles, they are both points and function defined on those points. Of course any element of the ring is a function on the scheme, but only the elements in the scheme are points. First, consider $S$ as functions, then we can talk about there common zero points.

Definition. The zero locus $V(S)$ of set $S$ is

\[V(S) := \left\{ {\frak p}\in X \,\middle\vert\, s({\frak p})=0 \;\forall\; s\in S \right\}\]

where $s$ are regarded as functions. However, $s({\frak p})=0$ means $s = 0 \text{ mod } {\frak p}$, which means $s\in{\frak p}$ where $s$ has become again an element of $R$, thus the zero locus of some subset $S\subset R$ is actually the set that contains all the elements of $S$,

\[V(S) = \left\{ P \,\middle\vert\, S\subset P \right\} .\]

Next, we can consider $S$ as a set of points, then it makes sense to talk about the corresponding ideals whose zero locus reproduces the set $S$.

Definition. The ideals of a set $S$ is

\[I(S) := \left\{ f\in R \,\middle\vert\, f(p)=0 \;\forall\; p\in S \right\}.\]

Given a ring, we can define a scheme on it by picking up all its prime ideals, then these ideals themselves also form a space, with their own subspace and varieties. The construction of a scheme from a ring is like a matryoshka, a ring structure is included in another.

It is relatively straightforward to introduce a topological structure in a scheme, we just endow it with the by now familiar Zariski topology, where the vanishing sets are defined to be closed set. To be specific,

Definition. The Zariski topology on $X=\text{Spec }R$ is the topology whose closed sets are the sets $V(S)$ where $S\subset \text{Spec} R$.

One important consequence of this definition is that, not all points in a scheme are closed! We state without proof that a point ${\frak p}\in X$ is closed iff ${\frak p}$ is a maximal ideal. Non-closed points are called generic points.