Lemma 10.18.2. Let $R$ be a ring. The following are equivalent:
$R$ is a local ring,
$\mathop{\mathrm{Spec}}(R)$ has exactly one closed point,
$R$ has a maximal ideal $\mathfrak m$ and every element of $R \setminus \mathfrak m$ is a unit, and
$R$ is not the zero ring and for every $x \in R$ either $x$ or $1 - x$ is invertible or both.
Proof. Let $R$ be a ring, and $\mathfrak m$ a maximal ideal. If $x \in R \setminus \mathfrak m$, and $x$ is not a unit then there is a maximal ideal $\mathfrak m'$ containing $x$. Hence $R$ has at least two maximal ideals. Conversely, if $\mathfrak m'$ is another maximal ideal, then choose $x \in \mathfrak m'$, $x \not\in \mathfrak m$. Clearly $x$ is not a unit. This proves the equivalence of (1) and (3). The equivalence (1) and (2) is tautological. If $R$ is local then (4) holds since $x$ is either in $\mathfrak m$ or not. If (4) holds, and $\mathfrak m$, $\mathfrak m'$ are distinct maximal ideals then we may choose $x \in R$ such that $x \bmod \mathfrak m' = 0$ and $x \bmod \mathfrak m = 1$ by the Chinese remainder theorem (Lemma 10.15.4). This element $x$ is not invertible and neither is $1 - x$ which is a contradiction. Thus (4) and (1) are equivalent. $\square$
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