Definition 10.18.1. A *local ring* is a ring with exactly one maximal ideal. The maximal ideal is often denoted $\mathfrak m_ R$ in this case. We often say “let $(R, \mathfrak m, \kappa )$ be a local ring” to indicate that $R$ is local, $\mathfrak m$ is its unique maximal ideal and $\kappa = R/\mathfrak m$ is its residue field. A *local homomorphism of local rings* is a ring map $\varphi : R \to S$ such that $R$ and $S$ are local rings and such that $\varphi (\mathfrak m_ R) \subset \mathfrak m_ S$. If it is given that $R$ and $S$ are local rings, then the phrase “*local ring map $\varphi : R \to S$*” means that $\varphi $ is a local homomorphism of local rings.

## 10.18 Local rings

Local rings are the bread and butter of algebraic geometry.

A field is a local ring. Any ring map between fields is a local homomorphism of local rings.

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.
$\square$

The localization $R_\mathfrak p$ of a ring $R$ at a prime $\mathfrak p$ is a local ring with maximal ideal $\mathfrak p R_\mathfrak p$. Namely, the quotient $R_\mathfrak p/\mathfrak pR_\mathfrak p$ is the fraction field of the domain $R/\mathfrak p$ and every element of $R_\mathfrak p$ which is not contained in $\mathfrak pR_\mathfrak p$ is invertible.

Lemma 10.18.3. Let $\varphi : R \to S$ be a ring map. Assume $R$ and $S$ are local rings. The following are equivalent:

$\varphi $ is a local ring map,

$\varphi (\mathfrak m_ R) \subset \mathfrak m_ S$, and

$\varphi ^{-1}(\mathfrak m_ S) = \mathfrak m_ R$.

For any $x \in R$, if $\varphi (x)$ is invertible in $S$, then $x$ is invertible in $R$.

**Proof.**
Conditions (1) and (2) are equivalent by definition. If (3) holds then (2) holds. Conversely, if (2) holds, then $\varphi ^{-1}(\mathfrak m_ S)$ is a prime ideal containing the maximal ideal $\mathfrak m_ R$, hence $\varphi ^{-1}(\mathfrak m_ S) = \mathfrak m_ R$. Finally, (4) is the contrapositive of (2) by Lemma 10.18.2.
$\square$

Let $\varphi : R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime and set $\mathfrak p = \varphi ^{-1}(\mathfrak q)$. Then the induced ring map $R_\mathfrak p \to S_\mathfrak q$ is a local ring map.

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