## 10.18 Local rings

Local rings are the bread and butter of algebraic geometry.

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.

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:

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 10.15.4). 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:

1. $\varphi$ is a local ring map,

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

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

4. 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.

Comment #865 by on

I think there is missing a definition of the terminology "fraction field". This is the first tag to use the terminology, but I can't find an actual definition of it. There should be one, right? Or am I missing it?

Comment #872 by on

Actually, it is in Example 9.3.4. To find it you have to look for "field of fractions" in the search feature. I have added the term "fraction field" to the example, see this commit. Yes, maybe we should make a formal definition... but somebody else can do this sometime.

Comment #6033 by Lars Hesselholt on

Maybe you already had this discussion, but "fraction field" is really not good terminology; "field of fractions" or "quotient field" are both good.

Comment #8512 by Daniel Apsley on

Example suggestion: Let $R$ be a local ring and $\mathfrak{p}$ a prime ideal which is not maximal. Then, the localization map $R \to R_{\mathfrak{p}}$ is not a local ring map.

Comment #8554 by Elizabeth Henning on

+1 for the example suggestion #8512 above

Comment #8709 by Cesar Massri on

Another item in Lema 10.18.3 (or 07BJ) that can be useful is: (2 bis) $\varphi(\mathfrak{m}_R^k)\subseteq \mathfrak{m}_S$ for some $k>0$. This item gives the equivalence between being local and being continuos (in the adic topology). The proof of (2bis) => (2) is straightforward: Let $r\in\mathfrak{m}_R$ be such that $\varphi(r)$ is invertible, then $\varphi(r^k)$ is invertible, but $\varphi(r^k)\in \mathfrak{m}_S$

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