The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.17.2. Let $R$ be a ring. The following are equivalent:

  1. $R$ is a local ring,

  2. $\mathop{\mathrm{Spec}}(R)$ has exactly one closed point,

  3. $R$ has a maximal ideal $\mathfrak m$ and every element of $R \setminus \mathfrak m$ is a unit, and

  4. $R$ 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 10.14.4). 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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