How to show $X=Spec(A)$ has maximal components $V(p)$?

Question

This is from Atiyah Macdonald, Chapter 1, Execrise 20.

If $A$ is a ring and $X=Spec(A)$, then the irreducible components of $X$ are the closed sets $V(p)$, where $p$ is a minimal prime ideal of $A$.

It is not difficult to prove the closed sets $V(p)$ are irreducible. However it isn't clear to me how to prove the second part. It is clear that when closed sets of the form $V(q),q\in X$ then $V(p)$ did provide a maximal irreducible closed set. Neverthelss it is not clear to me why every closed set in $X$ must be of this form(they only provide a basis of closed sets). So theoretically there is the possiblity of some closed set $U$ that contains $V(p)$, and in particular contains some prime ideal $k$ such that $p\not\subset k$. Then what can we do to prove this cannot hold and $k$ must contain $p$?

One way to address this is to take the closure of $U\cap X_{p}=K$. Since the $V(q),q\in X$ form a basis there is some $V(l)\subset \overline K$. Then $V_{l}\cup V_{p}\subset U$, while $V_{l}\cup V_{p}=V_{l\cap p}$. But by definition $l\cap p=p$; so such $V_{l}$ could not exist, and $K$ must be empty.

I am wondering if this approach is long-winded, as I found almost every other proof used the fact every irreducible component of $X$ must be of the form $V_{r(p)}$, where $r(p)$ denotes the radical of $p$. Why this is true, and how it might help to simplify the above proof?

Answer 1

Let $I$ be an ideal of $A$. We denote by $rad(I)$ the set $= \{x \in A\colon x^n \in I$ for some integer $n > 0$, where $n$ depends on $x\}$. It is easy to see that $rad(I)$ is an ideal and $V(I) = V(rad(I))$.

Let $E$ be a subset of $X = Spec(A)$. We denote by $\mathfrak{I}(E)$ the ideal $\bigcap_{p \in E} p$.

Lemma 0 Let $I$ be an ideal of $A$. Then $\mathfrak{I}(V(I)) = rad(I)$.

Proof: This is an immediate consequence of proposition 1.8 of Atiyah-MacDonald(it states that $\mathfrak{I}(X) = rad(0)$.

Lemma 1 Let $E$ be a subset of $X$. Then $V(\mathfrak{I}(E))$ is the closure of $E$.

Proof: Let $\bar E$ be the closure of $E$. Since $E \subset V(\mathfrak{I}(E))$, $\bar E \subset V(\mathfrak{I}(E))$.

Conversely suppose $E \subset V(I)$ for some ideal $I$. Then $I \subset \mathfrak{I}(E)$. Hence $V(\mathfrak{I}(E)) \subset V(I)$. Hence $V(\mathfrak{I}(E)) \subset \bar E$. QED

Lemma 2 $X = Spec(A)$ is irreducible if and only if $A/rad(0)$ is an integral domain.

Proof: Since $X = Spec(A/rad(0))$, we can assume $rad(0) = 0$.

Suppose $X$ is reducible. There exist proper closed subsets $F_1, F_2$ of $X$ such that $X = F_1 \cup F_2$. Hence $0 = \mathfrak{I}(X) = \mathfrak{I}(F_1) \cap \mathfrak{I}(F_2)$. By Lemma 1, $V(\mathfrak{I}(F_1)) = F_1$. Since $F_1 \neq X$, $\mathfrak{I}(F_1) \neq 0$. Similarly $\mathfrak{I}(F_2) \neq 0$. Hence $A$ is not an integral domain.

Conversely suppose $A$ is not an integral domain. There exist elements $f, g$ of $A$ such that $f \neq 0, g \neq 0$, $fg = 0$. Then, by Lemma 0, $V(f) \neq X$, $V(g) \neq X$ and $X = V(f) \cup V(g)$. Hence $X$ is reducible. QED

Lemma 3 Let $I$ be an ideal of $A$. Then $V(I)$ is irreducible if and only if $rad(I)$ is a prime ideal.

Proof: This follows immediately from Lemma 2.

By Lemma 3, an irreducible closed subset of $X$ is of the form $V(q)$, where $q$ is a prime ideal of $A$. Hence $V(p)$ is a maximal irreducible closed subset of $X$, if $p$ is a minimal prime ideal of $A$.