---- > [!definition] Definition. ([[local ring]]) > A **local ring** $R$ is one with a unique [[maximal ideal]]: a single proper ideal $\mathfrak{m}$ containing every other proper ideal. ^definition > [!basicproperties] > - Any nonunit $a \in R$ generates an ideal $(a)=aR \not \ni 1$, and is hence proper so that $(a) \subset \mathfrak{m}$. Contrapositively, the elements of $R-\mathfrak{m}$ are precisely the [[unit|units]] of $R$: $R-\mathfrak{m}=R^{*}.$ > - If $A$ is already local with unique maximal ideal $\mathfrak{m}$, then $A_{\mathfrak{m}} \cong A$, because any element in $A-\mathfrak{m}$ is a [[local ring#^properties|already invertible]]: since $A-\mathfrak{m}=A^{*}$, we have (being super explicit)$\begin{align} A_{\mathfrak{m}}= \left\{ \frac{a}{s} : a \in A , s \in A - \mathfrak{m} \right\}&=\left\{ \frac{a}{s} \ \frac{s ^{-1}}{s ^{-1}}:a \in A, s \in A^{*} \right\} \\ &= \{ a s ^{-1} : a \in A, s \in A^{*} \} \\ &= As ^{-1}, \text{ where } s \in A^{*} \\ & = A \end{align}$ ^properties > [!basicexample] Example. (Localizing at $\mathfrak{p}$ produces a local ring) > The [[localization]] of a [[commutative ring|commutative]] [[ring]] $A$ at a [[prime ideal]] $\mathfrak{p}$, $A_{\mathfrak{p}}$, is local. The [[ideal]] $\mathfrak{m}$ given by the [[extension of an ideal|extension]] of $\mathfrak{p}$ under the [[localization|localization map]], $\mathfrak{m}=\left\{ \frac{a}{s}: a \in \mathfrak{p}, s \in A - \mathfrak{p} \right\},$ is the unique [[maximal ideal|maximal]] [[ideal]]. > Indeed, suppose $\mathfrak{J} \subset A_{\mathfrak{p}}$ is an [[ideal]] not contained in $\mathfrak{m}$, allowing us to fix some $\frac{r}{t} \in \mathfrak{J}$ with $r \notin \mathfrak{p}, t \notin \mathfrak{p}$. But then $\frac{r}{t}$ is invertible, and multiplying through by its inverse produces $1 \in \mathfrak{J}$, and therefore $\mathfrak{J}=(1)=A_{\mathfrak{p}}$. ^basic-example > [!basicexample] Example. ($\frac{A}{\mathfrak{m}^{n}}$ is a local ring) > Let $\mathfrak{m}$ be a [[maximal ideal]] of a [[commutative ring]] $A$. Then for $n \geq 1$, $\frac{A}{\mathfrak{m}^{n}}$ is [[local ring]]. Indeed, the [[prime ideal|prime ideals]] of $\frac{A}{\mathfrak{m}^{n}}$ [[The correspondence theorem for rings|correspond to]] [[prime ideal|prime ideals]] of $A$ containing $\mathfrak{m}^{n}$. But these all contain $\mathfrak{m}$, hence all equal $\mathfrak{m}$. So $\frac{A}{\mathfrak{m}^{n}}$ has a unique prime ideal $\mathfrak{m} / \mathfrak{m}^{n}$; by [[existence of maximal ideals in commutative rings|any proper ideal has a maximal ideal containing it]] this is a (unique) [[maximal ideal]]. ^basic-example ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```