# Zariski Topology for Rings

In this article, we generalize earlier results in algebraic geometry to apply to general rings. Recall that points on an affine variety V correspond to maximal ideals $\mathfrak m\subset k[V]$. For general rings, we have to switch to taking prime ideals because of the following.

Lemma 1.

If $f : A \to B$ is a ring homomorphism and $\mathfrak q$ is a prime ideal of B, then $\mathfrak p := f^{-1}(\mathfrak q)$ is a prime ideal of A.

Proof

The composition $A \stackrel f\to B \to B/\mathfrak q$ has kernel $f^{-1}(\mathfrak q) = \mathfrak p$. Hence this induces an injective ring homomorphism $A/\mathfrak p \hookrightarrow B/\mathfrak q$. Since $B/\mathfrak q$ is a domain, so is $A/\mathfrak p$ so $\mathfrak p$ is prime. ♦

However it is not true that if $\mathfrak q$ is maximal in B, $\mathfrak p$ must be maximal in A. For instance consider the inclusion $i:\mathbb Z \hookrightarrow \mathbb Q$. The zero ideal is maximal in $\mathbb Q$ but $i^{-1}((0)) = (0)$ is not maximal in $\mathbb Z$.

Now we define our main object of interest.

Definition.

Given a ring A, the spectrum of A is its set of prime ideals $\mathrm{Spec} A$. The Zariski topology on $\mathrm{Spec A}$ is defined as follows: $V\subseteq \mathrm{Spec A}$ is closed if and only if it is of the following form

$V(S) = \{ \mathfrak p \in \mathrm{Spec A} : \mathfrak p \supset S\},$

for some subset $S\subseteq A$.

Note that since $V(S) = V(\mathfrak a)$ where $\mathfrak a$ is the ideal generated by S, we lose no generality by taking S to be ideals of A. If S = {f} we will write V(f) instead of V({f}) or V((f)) to reduce the clutter.

Immediately, we prove that the above gives a bona fide topology.

Proposition 1.

For any collection of ideals $(\mathfrak a_i)$ of A and ideals $\mathfrak a, \mathfrak b$ of A, we have

• $V(1) = \emptyset, V(0) = \mathrm{Spec} A$,
• $\cap_i V(\mathfrak a_i) = V(\sum_i \mathfrak a_i)$,
• $V(\mathfrak a) \cup V(\mathfrak b) = V(\mathfrak a\cap \mathfrak b) = V(\mathfrak {ab})$.

Proof

The first claim is obvious. For the second, $\mathfrak p$ contains $\mathfrak a_i$ for each i if and only if it contains $\sum_i \mathfrak a_i$.

Finally, for the third claim, we have $\mathfrak {ab} \subseteq \mathfrak a \cap \mathfrak b \subseteq \mathfrak a$ and $\mathfrak a\cap \mathfrak b \subseteq \mathfrak b$ so we have

$V(\mathfrak {ab}) \supseteq V(\mathfrak a\cap \mathfrak b) \supseteq V(\mathfrak a) \cup V(\mathfrak b).$

Finally suppose $\mathfrak p \not\in V(\mathfrak a)\cup V(\mathfrak b)$. Then $\mathfrak p \not\supseteq \mathfrak a$ and $\mathfrak p \not\supseteq \mathfrak b$ so there are $x \in \mathfrak a - \mathfrak p$ and $y \in \mathfrak b - \mathfrak p$. Then $xy \in \mathfrak{ab} - \mathfrak p$ since $\mathfrak p$ is prime so $\mathfrak p \not\in V(\mathfrak{ab})$. Hence if  $\mathfrak p \in V(\mathfrak{ab})$ then $\mathfrak p\in V(\mathfrak a) \cup V(\mathfrak b)$. ♦

# Structure of Spec A

Thanks to Zorn’s lemma, we have the following results on the structure of Spec A.

Proposition 2.

If A is a ring, every proper ideal $\mathfrak a\subsetneq A$ is contained in a maximal ideal.

Proof

Let $\Sigma$ be the collection of all proper ideals $\mathfrak b\subsetneq A$ containing $\mathfrak a$, ordered partially by inclusion. We claim that every chain $\Sigma' \subseteq \Sigma$ has an upper bound. Since $\mathfrak a \in \Sigma$ we may assume without loss of generality $\Sigma' \ne\emptyset$.

Take $\mathfrak c = \cup_{\mathfrak b \in \Sigma'} \mathfrak b$. Let us show that $\mathfrak c$ is an ideal of A.

Clearly $0 \in \mathfrak c$. Let $x,y\in \mathfrak c$ and $a\in A$. Then $x\in \mathfrak b$ and $y\in \mathfrak b'$ for some $\mathfrak b, \mathfrak b' \in \Sigma'$. Since $\Sigma'$ is totally ordered, either $\mathfrak b'\subseteq \mathfrak b$ or $\mathfrak b' \subseteq \mathfrak b'$. Assuming the former, this gives $x, y\in \mathfrak b$ so $ax - y\in \mathfrak b \subseteq \mathfrak c$. Thus $\mathfrak c$ is an ideal of A.

Since none of the $\mathfrak b\in \Sigma'$ contains 1, we have $1\not\in \mathfrak c$. Hence $\mathfrak c \in \Sigma$ is an upper bound of $\Sigma'$.

By Zorn’s lemma $\Sigma$ has a maximal element, which is precisely a maximal ideal of A containing $\mathfrak a$

We say that $\mathfrak p \in \mathrm{Spec} A$ minimal if it is a minimal element with respect to inclusion.

Proposition 3.

Any $\mathfrak p\in \mathrm{Spec} A$ contains a minimal prime $\mathfrak q$.

Proof

Let $\Sigma$ be the collection of all prime ideals of A contained in $\mathfrak p$, ordered partially by reverse inclusion, i.e. $\mathfrak q \le \mathfrak q'$ if $\mathfrak q \supseteq \mathfrak q'$. To apply Zorn’s lemma we need to show that every chain $\Sigma'\subseteq \Sigma$ has a maximal element (i.e. minimal with respect to inclusion). Without loss of generality we may assume $\Sigma' \ne \emptyset$.

Let $\mathfrak p' = \cap_{\mathfrak q \in \Sigma'} \mathfrak q$, an ideal of A. We need to show that it is prime. Suppose $x,y\in A - \mathfrak p'$ so there exist $\mathfrak q_1, \mathfrak q_2\in \Sigma'$ such that $x\not\in \mathfrak q_1$ and $y\not\in \mathfrak q_2$. Since $\Sigma'$ is a chain either $\mathfrak q_1 \subseteq \mathfrak q_2$ or $\mathfrak q_2 \subseteq \mathfrak q_1$; assuming the former we have $x,y\not\in \mathfrak q_1$. Since $\mathfrak q_1$ is prime we have $xy\not\in \mathfrak q_1$ so $xy\not\in \mathfrak p'$.

Now apply Zorn’s lemma to obtain a maximal element, which is a minimal prime contained in $\mathfrak p$. ♦

Corollary.

If A is non-trivial, $\mathrm{Spec} A$ must contain a maximal ideal and a minimal prime ideal.

# Homomorphisms

Suppose $f:A\to B$ is a ring homomorphism. We saw that this induces a map

$f^* : \mathrm{Spec} B \to \mathrm{Spec} A,\quad \mathfrak q \mapsto f^{-1}(\mathfrak q)$.

Proposition 4.

The map $f^*$ is continuous with respect to the Zariski topology; in fact

$(f^*)^{-1}(V(\mathfrak a)) = V(f(\mathfrak a)).$

Note that $f(\mathfrak a)$ is not an ideal of B in general, but that is okay since we defined V on arbitrary subsets.

Proof

Let $\mathfrak q \in \mathrm{Spec} B$. This lies in the LHS if and only if $f^*(\mathfrak q) = f^{-1}(\mathfrak q)$ contains $\mathfrak a$, if and only if $\mathfrak q$ contains $f(\mathfrak a)$. ♦

Clearly the identity ring homomorphism on A induces the identity map on $\mathrm{Spec A}$, also for ring homomorphisms $f:A\to B$ and $g:B\to C$ we have

$(g\circ f)^* = f^* \circ g^* : \mathrm{Spec} C \to \mathrm{Spec} A.$

Exercise A

Prove that for an ideal $\mathfrak a\subseteq A$, the canonical map $\pi : A \to A/\mathfrak a$ induces

$\pi^* : \mathrm{Spec}(A / \mathfrak a) \to \mathrm{Spec} A$

which is a subspace embedding onto the closed subset $V(\mathfrak a)$.

# Comparison with Algebraic Geometry

Proposition 5.

Let $V = V(\mathfrak a)$ be a closed subset of $\mathrm{Spec} A$. Then

$\bigcap \{ \mathfrak p : \mathfrak p \in V\} = r(\mathfrak a)$,

the radical of $\mathfrak a$.

Proof

(⊇) Suppose $x\in A$, $x^n \in \mathfrak a$ for some n > 0. Now each $\mathfrak p \in V$ contains $\mathfrak a$ and hence contains $x^n$. Since $\mathfrak p$ is prime we have $x\in \mathfrak p$.

(⊆) Suppose $x\in A$, $\mathfrak a$ does not contain any power of x. We will prove the existence of $\mathfrak p \supseteq \mathfrak a$ not containing x.

Let $\Sigma$ be the set of all ideals $\mathfrak b\supseteq \mathfrak a$ not containing any power of x; thus $\mathfrak a \in \Sigma$. Order $\Sigma$ by inclusion. We will show that any chain $\Sigma' \subseteq \Sigma$ has an upper bound. Indeed, let $\mathfrak c = \cup_{\mathfrak b \in \Sigma'} \mathfrak b$. As before, since $\Sigma'$ is a chain of ideals ordered by inclusion, $\mathfrak c$ is an ideal of A. It does not include any power of x so $\mathfrak c\in \Sigma$.

Hence by Zorn’s lemma, $\Sigma$ has a maximal element $\mathfrak p$; by construction $\mathfrak p$ contains $\mathfrak a$ but not any power of x. It remains to show $\mathfrak p$ is prime: suppose not so there are $y, z\in A - \mathfrak p$ with $yz \in \mathfrak p$. Now $\mathfrak p + (y)$ and $\mathfrak p + (z)$ strictly contain $\mathfrak p$, so by maximality of $\mathfrak p$, we have $\mathfrak p + (y), \mathfrak p + (z) \not\in\Sigma$. Hence

\left. \begin{aligned} \exists m > 0,\ x^m \in \mathfrak p+(y) \\ \exists n>0, \ x^n \in \mathfrak p + (z)\end{aligned} \right\} \implies x^{m+n} \in (\mathfrak p + (y))(\mathfrak p + (z)) \subseteq \mathfrak p + (yz) \subseteq \mathfrak p.

a contradiction. Thus $\mathfrak p$ is prime. ♦

Exercise B

From proposition 5, prove that we have a bijection

The correspondence reverses inclusion. For $\cap V_i$ and $V_1 \cup V_2$ in Spec A, write down the corresponding operations for radical ideals of A.

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