# Basic Open Sets

For $f\in A$, let $D(f) := \mathrm{Spec} A - V(f)$, an open subset of Spec A. Note that

$D(f) = \{\mathfrak p \subset A \text{ prime}: \mathfrak p \not\ni f\}$.

Proposition 1.

The collection of $D(f)$ over all $f\in A$ forms a basis for the topology of $\mathrm{Spec} A$.

Proof

Let $U = \mathrm{Spec} A - V(\mathfrak a)$ be an open subset of Spec A. Suppose $\mathfrak p \in U$ so that $\mathfrak p\not\supseteq \mathfrak a$. Pick an $f \in \mathfrak a - \mathfrak p$. Since $f\not\in \mathfrak p$ we have $\mathfrak p \in D(f)$. It remains to show $D(f) \subseteq U$. Indeed if $\mathfrak q\not\ni f$ then certainly $\mathfrak q \not\supseteq \mathfrak a$ so we have $\mathfrak q \in U$. ♦

With this we have:

Proposition 2.

$\mathrm{Spec} A$ is quasi-compact, in the sense that every open cover has a finite subcover.

Note

The modern topological term for this is just compact. However, the term quasi-compact is firmly entrenched in the literature for commutative algebra and algebraic geometry so we will respect the tradition here.

Proof

To prove quasi-compactness, it suffices to check covering via basic open sets. Thus suppose $\mathrm{Spec} A = \cup_i D(f_i)$ is a basic open covering; we need to show there is a finite subcover. Now $\emptyset = \cap_i V(f_i) = V(\mathfrak a)$ where $\mathfrak a = (f_i)$, i.e. the ideal generated by the $f_i$. This can only happen if $\mathfrak a = (1)$ so we can write

$1 = g_{i_1} f_{i_1} + \ldots + g_{i_k} f_{i_k}, \quad g_{i_j} \in A$

for finitely many choices of $f_i$. Hence $(1) = (f_{i_1}, \ldots, f_{i_k})$ so we have

$\cap_{j=1}^k V(f_{i_j}) = \emptyset \implies \cup_{j=1}^k D(f_{i_j}) = \mathrm{Spec} A.$

# More on Zariski Topology

Let $X = \mathrm{Spec} A$ for a ring A. Immediately, we see some differences between X and affine varieties. Firstly, points of X are prime ideals $\mathfrak p \subset A$ and often $\{\mathfrak p\}$ is not closed in X. More generally, we have the following.

Proposition 3.

For $\mathfrak p \in \mathrm{Spec} A$, the closure of $\{\mathfrak p\}$ is $V(\mathfrak p)$. In particular, $\{\mathfrak p\}$ closed if and only if $\mathfrak p$ is a maximal ideal, in which case we call it a closed point.

Proof

The closure C of $\{\mathfrak p\}$ is the intersection of all $V(\mathfrak a)$ containing $\mathfrak p$. This includes $V(\mathfrak p)$ so we have $C \subseteq V(\mathfrak p)$. On the other hand, any closed subset $V(\mathfrak a)$ containing $\mathfrak p$ gives $\mathfrak p \supseteq \mathfrak a$ so $V(\mathfrak p) \subseteq V(\mathfrak a)$. Thus $C = V(\mathfrak p)$. ♦

For example, if A is a domain, then (0) is a prime ideal so $\mathrm{Spec} A = V(0)$ is the closure of the point (0). In a topological space X, a generic point of X is a $p\in X$ such that the closure of $\{p\}$ is X.

Note

In topological lingo, this says that Spec A is not a T1 space. It is, however, a T0 space. Compare this with the case of affine varieties which are T1 (i.e. all points are closed), but not T2. [For example, $V = \mathbb A^1$ has the cofinite topology.]

### Elements of Spec ℤ

Definition.

The intersection of all prime ideals of A is called the nilradical of A, denoted $n(A)$. From the previous article,

$n(A) = \{x \in A: x^n = 0 \text{ for some } n > 0\}.$

Correspondingly, the intersection of all maximal ideals of A is called the Jacobson radical of A, denoted by $J(A)$.

We have the following classification for Jacobson radical.

Proposition 4.

We have $J(A) = \{ x \in A: 1 + xA \subseteq U(A)\}$.

Proof

(⊇) If $x\not\in\mathfrak m$ for maximal ideal $\mathfrak m$ then since $\mathfrak m + xA \supsetneq \mathfrak m$ we have $\mathfrak m + xA = (1)$ so there exists $y\in A, z\in \mathfrak m$ such that $z + xy = 1$ so $1 + x(-y)$ is not a unit, i.e. $1 + xA \not\subseteq U(A)$.

(⊆) Suppose $x\in J(A)$; let $y\in A$. If (1 + xy) is not a unit, it is contained in some maximal ideal $\mathfrak m$. Then $x\not \in \mathfrak m$, or else we would have $1 \in \mathfrak m$. ♦

### Philosophy

Following affine varieties, we now look at rings from a new point of view.

• “Points” correspond to prime ideals $\mathfrak p \subset A$.
• “Functions on points” correspond to elements $f\in A$.
• “Function vanishes on point” corresponds to $f\in \mathfrak p$.

Thus the nilradical (resp. Jacobson radical) is the set of all functions vanishing on all points (resp. closed points). In a way, non-closed points act as “irreducible closed sets” of closed points, so one naturally asks when the nilradical is equal to the Jacobson radical. When this holds, the ring in question is called a Jacobson ring.

# Connected Components

Proposition 5.

• Let $A = B\times C$ be a product of rings. Then $\mathrm{Spec} A$ is the topological disjoint union of $\mathrm{Spec} B$ and $\mathrm{Spec} C$.
• Conversely suppose $\mathrm{Spec A} = V \cup W$ where V, W are disjoint closed subsets, then

$A \cong B \times C, \quad V \cong \mathrm{Spec} B, W \cong \mathrm{Spec} C$.

Proof

The projection maps $A\to B$ gives continuous map $\mathrm{Spec} B \to \mathrm{Spec} A$ which takes $\mathfrak q \mapsto \mathfrak q \times C$. This map is injective, continuous and takes closed set $V(\mathfrak b)$ to closed set $V(\mathfrak b \times C)$. Hence it is a subspace embedding.

Similarly, we have a subspace embedding $\mathrm{Spec} C\to \mathrm{Spec} A$; by exercise A here, the two images form a disjoint union so the first claim is done.

For the second claim, write $V = V(\mathfrak a)$ and $W = V(\mathfrak b)$ so

$\mathrm{Spec} A = V(\mathfrak a) \cup V(\mathfrak b) = V(\mathfrak a \cap \mathfrak b), \quad \emptyset = V(\mathfrak a) \cap V(\mathfrak b) = V(\mathfrak a + \mathfrak b).$

Thus $\mathfrak a + \mathfrak b = (1)$; pick $x\in \mathfrak a$, $y\in \mathfrak b$ such that $x+y=1$ so (x) and (y) are coprime ideals. Now $xy\in \mathfrak a\cap \mathfrak b$ which is contained in the nilradical since it is contained in all prime ideals. Thus $(xy)^n = 0$ for some n > 0 so $(x^n), (y^n)$ are coprime ideals with product zero. By Chinese Remainder Theorem this gives $A \cong A/(x^n) \times A/(y^n)$.

Since $x^n \in \mathfrak a$ we have $V \subseteq V(x^n)$; similarly $W \subseteq V(y^n)$. Since $V(x^n) \cong \mathrm{Spec} A/(x^n)$ and $V(y^n) \cong \mathrm{Spec} A/(y^n)$ also form a disjoint union of Spec A by the first part, we are done. ♦

Pictorially we have

# Irreducible Spaces

Proposition 6.

The space $\mathrm{Spec} A$ is irreducible if and only if the nilradical is a prime ideal.

Proof

(⇐) If the nilradical is prime $\mathfrak p$, and Spec A is a union of two closed subsets, one of them must contain $\mathfrak p \in \mathrm{Spec} A$; hence it must contain all primes. Thus Spec A is irreducible.

(⇒) Conversely, suppose Spec A is irreducible. Pick $x,y\in A$ with $xy\in \mathfrak n(A)$. Then

\begin{aligned} (x)(y) \subseteq n(A) &\implies V(x) \cup V(y) \supseteq V(n(A)) = \mathrm{Spec} A\\ &\implies V(x) \cup V(y) = \mathrm{Spec} A.\end{aligned}

By irreducibility, either V(x) or V(y) is the whole Spec A. Assume the former; then x lies in all prime ideals of A so it is nilpotent, i.e. $x\in n(A)$. ♦

In particular, we have:

Corollary 1.

If $\mathrm{Spec A}$ is irreducible then it has a generic point.

# Prime Chains

Finally, another important aspect of Spec A are the prime chains.

Definition.

A prime chain of A is a strictly increasing sequence of prime ideals

$\mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \ldots \subsetneq \mathfrak p_d$

of A; the length of the chain is d.

For example, in the ring $A = k[X, Y]$ for a field Xwe have $(0) \subset (X) \subset (X, Y)$ which is a prime chain of length 2.

Definition.

The Krull dimension (or just dimension) of a ring A is the supremum of all lengths of prime chains in A.

## Examples

1. The dimension of a field is 0.

2. The dimension of ℤ is 1.

3. We have $\dim A[X] \ge \dim A + 1$, since for any prime chain $(\mathfrak p_i)_{i=0}^d$ of A, we have the prime chain of A[X]:

$\mathfrak p_0[X] \subsetneq \mathfrak p_1[X] \subsetneq \ldots \subsetneq \mathfrak p_d[X] \subsetneq \mathfrak p_d[X] + (X).$

Note that these ideals are all prime since $A[X]/\mathfrak p_i[X] \cong (A/\mathfrak p_i)[X]$ and $A[X]/(\mathfrak p_d[X] +(X)) \cong A/\mathfrak p_d$.

4. From the above, we have $\dim k[X_1, \ldots, X_n] \ge n$ and $\dim \mathbb Z[X_1, \ldots, X_n] \ge n+1$. Eventually, we will see that equality holds!

5. We have $\dim (A\times B) = \sup (\dim A, \dim B)$.

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