Proposition

Let $A$ be a ring, $\mathfrak{R}$ its nilradical. Show that the following are equivalent:

$A$ has exactly one prime ideal;
every element of $A$ is either a unit or nilpotent;
$A/\mathfrak{R}$ is a field.

Solution

By Proposition 1.8[Atiyah], $\mathfrak{R}$ is the intersection of all the prime ideals of $A$. We will use this property in the following solution without explicitly mentioning it. We will assume that $A \ne (0)$.

$1 \implies 2$

Suppose $A$ has only one prime ideal, $p$. Then $\mathfrak{R} = p$.

Since every maximal ideal is prime, $A$ is a local ring. Then $p$ is its maximal ideal. Let $x \in A$ be given. Suppose that $x$ is not a unit. By Corollary 1.5[Atiyah], $x$ is in some maximal ideal. Since the only maximal ideal in $A$ is $p$, $x \in p = \mathfrak{R}$. By definition, $\mathfrak{R}$ is the set of all nilpotent elements. Thus $x$ must be a nilpotent element.

$2 \implies 3$

Let $x + \mathfrak{R} \in A / \mathfrak{R}$. Suppose $x + \mathfrak{R} \ne 0$. In other words, $x \notin \mathfrak{R}$. Since $x$ is the set of all nilpotent elements, $x$ must be a unit. Then $x^{-1} \in A$, so $(x^{-1} + \mathfrak{R})(x + \mathfrak{R}) = 1 + \mathfrak{R} = 1 \in A / \mathfrak{R}$. Therefore, $A / \mathfrak{R}$ is a field.

$3 \implies 1$

Let $p \subset A$ be a prime ideal. Then $\mathfrak{R} \subset p$. If we can show that $\mathfrak{R} = p$, then it implies that $\mathfrak{R}$ is the only prime ideal in $A$. Let $I = \{ x + \mathfrak{R} \mid x \in p \}$ be an ideal in $A / \mathfrak{R}$. By Proposition 1.2[Atiyah], $I = (0)$ or $(1)$. Let $y \in A \ p$ and $x \in p$. Then $y - x \notin p$. Therefore, $y - x \notin \mathfrak{R}$. This implies $y + \mathfrak{R} \ne x + \mathfrak{R}$, so $y + \mathfrak{R} \notin I$. Thus $I \ne (1)$.

Since $I = (0)$, $p \subset \mathfrak{R}$, so $p = \mathfrak{R}$.