## 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. If $R$ is a local ring, then the maximal ideal is often denoted $\mathfrak m_ R$ and the field $R/\mathfrak m_ R$ is called the residue field of the local ring $R$. We often say “let $(R, \mathfrak m)$ be a local ring” or “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.

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, by Lemma 10.17.5 every prime ideal of $R_\mathfrak p$ is contained in the prime ideal $\mathfrak p R_\mathfrak p$ (hence this is a maximal ideal and the only maximal ideal of $R_\mathfrak p$). The residue field of $R_\mathfrak p$ is denoted $\kappa (\mathfrak p)$; we call it the residue field of $\mathfrak p$; by Proposition 10.9.14 we may identify $\kappa (\mathfrak p)$ with the field of fractions of the domain $R/\mathfrak p$. Via the composition

$\mathop{\mathrm{Spec}}(\kappa (\mathfrak p)) \to \mathop{\mathrm{Spec}}(R_\mathfrak p) \to \mathop{\mathrm{Spec}}(R)$

the unique point of the source maps to the point $\mathfrak p$ of the target.

Let $\varphi : R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime and consider the prime $\mathfrak p = \varphi ^{-1}(\mathfrak q)$ of $R$. Since $\varphi (\mathfrak p) \subset \mathfrak q$ the induced ring map

$R_\mathfrak p \to S_\mathfrak q,\quad r/g \mapsto \varphi (r)/\varphi (g)$

is a local ring map and we obtain an induced map of residue fields $\kappa (\mathfrak p) \to \kappa (\mathfrak q)$.

Example 10.18.2. If $R$ is a local ring and $\mathfrak p \subset R$ is a non-maximal prime ideal, then $R \to R_\mathfrak p$ is not a local homomorphism.

Lemma 10.18.3. 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. $\square$

Lemma 10.18.4. 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 10.18.3. $\square$

Remark 10.18.5. A fundamental commutative diagram associated to a ring map $\varphi : R \to S$ and a prime $\mathfrak p \subset R$ is the following

$\xymatrix{ \kappa (\mathfrak p) \otimes _ R S = S_{\mathfrak p}/{\mathfrak p}S_{\mathfrak p} & S_{\mathfrak p} \ar[l] & S \ar[r] \ar[l] & S/\mathfrak pS \ar[r] & (R \setminus \mathfrak p)^{-1}S/\mathfrak pS \\ \kappa (\mathfrak p) = R_{\mathfrak p}/{\mathfrak p}R_{\mathfrak p} \ar[u] & R_{\mathfrak p} \ar[u] \ar[l] & R \ar[u] \ar[r] \ar[l] & R/\mathfrak p \ar[u] \ar[r] & \kappa (\mathfrak p) \ar[u] }$

In this diagram the outer left and outer right columns are identical. On spectra the horizontal maps induce homeomorphisms onto their images and the squares induce fibre squares of topological spaces (see Lemmas 10.17.5 and 10.17.7). This shows that $\mathfrak p$ is in the image of the map on Spec if and only if $S \otimes _ R \kappa (\mathfrak p)$ is not the zero ring. If there does exist a prime $\mathfrak q \subset S$ lying over $\mathfrak p$, i.e., with $\mathfrak p = \varphi ^{-1}(\mathfrak q)$ then we can extend the diagram to the following diagram

$\xymatrix{ \kappa (\mathfrak q) = S_{\mathfrak q}/{\mathfrak q}S_{\mathfrak q} & S_{\mathfrak q} \ar[l] & S \ar[r] \ar[l] & S/\mathfrak q \ar[r] & \kappa (\mathfrak q) \\ \kappa (\mathfrak p) \otimes _ R S = S_{\mathfrak p}/{\mathfrak p}S_{\mathfrak p} \ar[u] & S_{\mathfrak p} \ar[u] \ar[l] & S \ar[u] \ar[r] \ar[l] & S/\mathfrak pS \ar[u] \ar[r] & (R \setminus \mathfrak p)^{-1}S/\mathfrak pS \ar[u] \\ \kappa (\mathfrak p) = R_{\mathfrak p}/{\mathfrak p}R_{\mathfrak p} \ar[u] & R_{\mathfrak p} \ar[u] \ar[l] & R \ar[u] \ar[r] \ar[l] & R/\mathfrak p \ar[u] \ar[r] & \kappa (\mathfrak p) \ar[u] }$

In this diagram it is still the case that the outer left and outer right columns are identical and that on spectra the horizontal maps induce homeomorphisms onto their image.

Lemma 10.18.6. Let $\varphi : R \to S$ be a ring map. Let $\mathfrak p$ be a prime of $R$. The following are equivalent

1. $\mathfrak p$ is in the image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$,

2. $S \otimes _ R \kappa (\mathfrak p) \not= 0$,

3. $S_{\mathfrak p}/\mathfrak p S_{\mathfrak p} \not= 0$,

4. $(S/\mathfrak pS)_{\mathfrak p} \not= 0$, and

5. $\mathfrak p = \varphi ^{-1}(\mathfrak pS)$.

Proof. We have already seen the equivalence of the first two in Remark 10.18.5. The others are just reformulations of this. $\square$

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 $\varphi(r^k)\in \mathfrak{m}_S$

Comment #9117 by on

OK, I tried to improve this section a bit. I didn't add the equivalent condition suggested by Cesar Massri but I will in the future if it turns out to help. See changes

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