## 10.26 Irreducible components of spectra

We show that irreducible components of the spectrum of a ring correspond to the minimal primes in the ring.

Lemma 10.26.1. Let $R$ be a ring.

For a prime $\mathfrak p \subset R$ the closure of $\{ \mathfrak p\} $ in the Zariski topology is $V(\mathfrak p)$. In a formula $\overline{\{ \mathfrak p\} } = V(\mathfrak p)$.

The irreducible closed subsets of $\mathop{\mathrm{Spec}}(R)$ are exactly the subsets $V(\mathfrak p)$, with $\mathfrak p \subset R$ a prime.

The irreducible components (see Topology, Definition 5.8.1) of $\mathop{\mathrm{Spec}}(R)$ are exactly the subsets $V(\mathfrak p)$, with $\mathfrak p \subset R$ a minimal prime.

**Proof.**
Note that if $ \mathfrak p \in V(I)$, then $I \subset \mathfrak p$. Hence, clearly $\overline{\{ \mathfrak p\} } = V(\mathfrak p)$. In particular $V(\mathfrak p)$ is the closure of a singleton and hence irreducible. The second assertion implies the third. To show the second, let $V(I) \subset \mathop{\mathrm{Spec}}(R)$ with $I$ a radical ideal. If $I$ is not prime, then choose $a, b\in R$, $a, b\not\in I$ with $ab\in I$. In this case $V(I, a) \cup V(I, b) = V(I)$, but neither $V(I, b) = V(I)$ nor $V(I, a) = V(I)$, by Lemma 10.17.2. Hence $V(I)$ is not irreducible.
$\square$

In other words, this lemma shows that every irreducible closed subset of $\mathop{\mathrm{Spec}}(R)$ is of the form $V(\mathfrak p)$ for some prime $\mathfrak p$. Since $V(\mathfrak p) = \overline{\{ \mathfrak p\} }$ we see that each irreducible closed subset has a unique generic point, see Topology, Definition 5.8.6. In particular, $\mathop{\mathrm{Spec}}(R)$ is a sober topological space. We record this fact in the following lemma.

Lemma 10.26.2. The spectrum of a ring is a spectral space, see Topology, Definition 5.23.1.

**Proof.**
Formally this follows from Lemma 10.26.1 and Lemma 10.17.9. See also discussion above.
$\square$

Lemma 10.26.3. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime.

the set of irreducible closed subsets of $\mathop{\mathrm{Spec}}(R)$ passing through $\mathfrak p$ is in one-to-one correspondence with primes $\mathfrak q \subset R_{\mathfrak p}$.

The set of irreducible components of $\mathop{\mathrm{Spec}}(R)$ passing through $\mathfrak p$ is in one-to-one correspondence with minimal primes $\mathfrak q \subset R_{\mathfrak p}$.

**Proof.**
Follows from Lemma 10.26.1 and the description of $\mathop{\mathrm{Spec}}(R_\mathfrak p)$ in Lemma 10.17.5 which shows that $\mathop{\mathrm{Spec}}(R_\mathfrak p)$ corresponds to primes $\mathfrak q$ in $R$ with $\mathfrak q \subset \mathfrak p$.
$\square$

Lemma 10.26.4. Let $R$ be a ring. Let $\mathfrak p$ be a minimal prime of $R$. Let $W \subset \mathop{\mathrm{Spec}}(R)$ be a quasi-compact open not containing the point $\mathfrak p$. Then there exists an $f \in R$, $f \not\in \mathfrak p$ such that $D(f) \cap W = \emptyset $.

**Proof.**
Since $W$ is quasi-compact we may write it as a finite union of standard affine opens $D(g_ i)$, $i = 1, \ldots , n$. Since $\mathfrak p \not\in W$ we have $g_ i \in \mathfrak p$ for all $i$. By Lemma 10.25.1 each $g_ i$ is nilpotent in $R_{\mathfrak p}$. Hence we can find an $f \in R$, $f \not\in \mathfrak p$ such that for all $i$ we have $f g_ i^{n_ i} = 0$ for some $n_ i > 0$. Then $D(f)$ works.
$\square$

Lemma 10.26.5. Let $R$ be a ring. Let $X = \mathop{\mathrm{Spec}}(R)$ as a topological space. The following are equivalent

$X$ is profinite,

$X$ is Hausdorff,

$X$ is totally disconnected.

every quasi-compact open of $X$ is closed,

there are no nontrivial inclusions between its prime ideals,

every prime ideal is a maximal ideal,

every prime ideal is minimal,

every standard open $D(f) \subset X$ is closed, and

add more here.

**Proof.**
First proof. It is clear that (5), (6), and (7) are equivalent. It is clear that (4) and (8) are equivalent as every quasi-compact open is a finite union of standard opens. The implication (7) $\Rightarrow $ (4) follows from Lemma 10.26.4. Assume (4) holds. Let $\mathfrak p, \mathfrak p'$ be distinct primes of $R$. Choose an $f \in \mathfrak p'$, $f \not\in \mathfrak p$ (if needed switch $\mathfrak p$ with $\mathfrak p'$). Then $\mathfrak p' \not\in D(f)$ and $\mathfrak p \in D(f)$. By (4) the open $D(f)$ is also closed. Hence $\mathfrak p$ and $\mathfrak p'$ are in disjoint open neighbourhoods whose union is $X$. Thus $X$ is Hausdorff and totally disconnected. Thus (4) $\Rightarrow $ (2) and (3). If (3) holds then there cannot be any specializations between points of $\mathop{\mathrm{Spec}}(R)$ and we see that (5) holds. If $X$ is Hausdorff then every point is closed, so (2) implies (6). Thus (2), (3), (4), (5), (6), (7) and (8) are equivalent. Any profinite space is Hausdorff, so (1) implies (2). If $X$ satisfies (2) and (3), then $X$ (being quasi-compact by Lemma 10.17.8) is profinite by Topology, Lemma 5.22.2.

Second proof. Besides the equivalence of (4) and (8) this follows from Lemma 10.26.2 and purely topological facts, see Topology, Lemma 5.23.8.
$\square$

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