Proposition

Let $A$ be a ring and let $X$ be the set of all prime ideals of $A$. For each subset $E$ of $A$, let $V(E)$ denote the set of all prime ideals of $A$ which contain $E$. Prove that

if $a$ is the ideal generated by $E$, then $V(E) = V(a) = V(r(a))$.
$V(0) = X, V(1) = \emptyset$.
If $(E_i)_{i \in I}$ is any family of subsets of $A$, then $V(\cup_{i \in I} E_i) = \cap_{i \in I} V(E_i)$.
$V(a \cap b) = V(ab) = V(a) \cup V(b)$ for any ideals $a, b$ of $A$.

Solution

1

Since $E \subset a$, $V(a) \subset V(E)$.
Let $p \in V(E)$. Then $p$ is an ideal containing $E$. Since $a$ is the smallest ideal containing $E$, $a \subset p$. Therefore, $p \in V(a)$.

Therefore, $V(a) = V(E)$.

Since $a \subset r(a)$, $V(r(a)) \subset V(a)$.

Let $p \in V(a)$. Then $p$ is a prime ideal containing $a$. Let $x \in r(a)$. If $x \in p$, we are done.

Suppose $x \notin p$. Then $x^m \in a \subset p$ for some $m \geq 1$. Then $x \cdot x^{m - 1} \in p$, so either $x$ or $x^{m - 1}$ is in $p$. Since we assumed $x \notin p$, $x^{m - 1} \in p$. However, by repeating this process, it will show that $x \in p$ after $m - 1$ steps. Therefore, $x$ must be in $p$.

This means $r(a) \subset p$, so $p \in V(r(a))$. Therefore, $V(r(a)) = V(a)$.

2

Since every prime ideal contains $0$, $V(0) = X$. Since every prime ideal is a proper ideal, no prime ideal contains $1$. Thus $V(1) = \emptyset$.

3

\[\begin{align*} p \in V(\cup_{i \in I} E_i) &\iff \forall i \in I, E_i \subset p \\ &\iff \forall i \in I, p \in V(E_i) \\ &\iff p \in \cap_{i \in I} V(E_i). \end{align*}\]

4

Since $ab \subset a$ and $ab \subset b$, $ab \subset a \cap b$. Therefore, $V(a \cap b) \subset V(ab)$.

Let $p \in V(ab)$. Then $ab \subset p$. Let $x \in a \cap b$. Then $x^2 \in ab \subset p$, so $x \in p$. Therefore, $a \cap b \subset p$, so $p \in V(a \cap b)$.

This shows $V(ab) = V(a \cap b)$.

Since $a \cap b \subset a$ and $a \cap b \subset b$, $V(a) \subset V(a \cap b)$ and $V(b) \subset V(a \cap b)$. Thus, $V(a) \cup V(b) \subset V(a \cap b)$.

Let $p \in V(a \cap b)$. If $a \subset p$, then $p \in V(a)$. Suppose otherwise. Let $x \in b$ and $y \in a \setminus p$. Then $xy \in a \cap b \subset p$. Since $p$ is prime and $y \notin p$, $x \in p$. Therefore, $b \subset p$, so $p \in V(b)$.

Thus we have $p \in V(a) \cup V(b)$, so we have $V(a) \cup V(b) = V(a \cap b)$.