Localization

Local Rings

In this post, we define localization. Briefly speaking, localization can be thought of as the process of making a ring \(A\) into a local ring.

Definition 1 A ring \(A\) is a local ring if \(A\) has a unique maximal ideal.

Then we can show the following equivalence.

Proposition 2 For a ring \(A\), the following are equivalent.

1. \(A\) is a local ring.
2. There exists an ideal \(\mathfrak{m}\subsetneq A\) such that all non-units of \(A\) belong to \(\mathfrak{m}\).
3. The collection of all non-units of \(A\) forms an ideal.

Proof

First assume 1, and suppose an arbitrary non-unit of \(A\), \(a\in A\), is given. Then, since \(a\) is not a unit, \((a)\) satisfies \((a)\subsetneq A\), so as an ideal of \(A\), by [Algebraic Structures] §Definition of a Ring, ⁋Theorem 10 (Krull) it is contained in some maximal ideal. But since \(A\) has a unique maximal ideal \(\mathfrak{m}\), we must have \((a)\subseteq \mathfrak{m}\), and therefore \(a\in \mathfrak{m}\).

Now assume 2 and show 3. For this, it suffices to show that the collection of non-units of \(A\) is closed under addition. First, from \(\mathfrak{m}\neq A\) we know that \(\mathfrak{m}\) contains no units of \(A\). From this we see that the collection of all non-units of \(A\) must equal \(\mathfrak{m}\).

Finally, we must assume condition 3 and show condition 1. For any ideal \(\mathfrak{a}\subsetneq A\), by the preceding observation we know that \(\mathfrak{a}\) consists only of non-units, and therefore \(\mathfrak{a}\) is contained in the ideal of all non-units of \(A\), \(\mathfrak{m}\). On the other hand, \(\mathfrak{m}\) is a maximal ideal, because any element of \(A\setminus \mathfrak{m}\) is a unit of \(A\), so the only ideal containing \(\mathfrak{m}\) is \(A\).

Localization of Modules

As explained in the previous post, to see the localization of a ring we define the localization of a module more generally.

Definition 3 For a ring \(A\), a subset \(S\) is multiplicatively closed if any product of elements of \(S\) again belongs to \(S\).

In particular, since \(1\) can be regarded as the product of a family indexed by the empty set, by definition \(1\in S\).

Definition 4 For a ring \(A\), an \(A\)-module \(M\), and a multiplicative subset of \(A\), \(S\), the localization at \(S\) of \(M\) is the \(A\)-module \(S^{-1}M\) defined as follows.

1. As a set, \(S^{-1}M\) is the quotient set obtained by defining on \(M\times S\) the following equivalence relation:

   \[(x,s)\sim (x',s')\iff \text{there exists $t\in S$ such that $t(s'x-sx')=0$}\]

   Here, the equivalence class containing \((x,s)\) is denoted by \(x/s\).

2. On \(S^{-1}M\), the \(A\)-module structure is defined as follows:

\[\frac{x}{s}+\frac{x'}{s'}=\frac{s'x+sx'}{ss'},\qquad a\cdot \frac{x}{s}=\frac{ax}{s}.\]

We should verify that the operations defined in condition 2 actually give an \(A\)-module structure, but this is not difficult. Instead, we add a few observations. First, for any \(t\in S\) and \(x/s\in S^{-1}M\),

\[\frac{tx}{ts}=\frac{x}{s}\]

holds. This is simply because \(1(txs-tsx)=0\). Also, we can examine the identity and inverse with respect to addition in \(S^{-1}M\): first, for any \(s,s'\in S\), we can verify that

\[\frac{0}{s}=\frac{0}{s'}\]

holds, and from the equation

\[\frac{0}{s'}+\frac{x}{s}=\frac{x}{s}+\frac{0}{s'}=\frac{0s+s'x}{ss'}=\frac{s'x}{s's}=\frac{x}{s}\]

we see that this is the identity with respect to addition in \(S^{-1}M\). By a similar calculation, one can verify that the inverse of any \(x/s\) is \((-x)/s\).

The above calculations are no different from the addition and multiplication of fractions we have done since middle school. Taking this as intuition, we can define the canonical map from \(M\) to \(S^{-1}M\), \(\epsilon: M \rightarrow S^{-1}M\), by \(x\mapsto x/1\). Unfortunately, in general \(\epsilon\) may not be an inclusion; the reason is obvious upon reflection and is captured in the following proposition.

Proposition 5 In the above situation, \(\epsilon(x)=0\) if and only if there exists \(s\in S\) such that \(sx=0\). In particular, if \(M\) is finitely generated, then \(S^{-1}M=0\) if and only if some \(s\in S\) annihilates \(M\).

Proof

If

\[\epsilon(x)=x/1=0=0/1\]

then there exists \(s\in S\) such that

\[s(1x-0\cdot1)=sx=0\]

holds. The above argument also holds in the reverse direction.

Localization of Rings

The simplest example of localization is the ring of fractions examined in [Algebraic Structures] §Field of Fractions, ⁋Definition 2. Here we took \(M=A\). In particular, we also saw that if \(A\) is an integral domain, then its ring of fractions \(\Frac(A)\) is a field. ([Algebraic Structures] §Field of Fractions, ⁋Proposition 6)

As another example, similarly taking \(M=A\), for \(A\) with a prime ideal \(\mathfrak{p}\), setting \(S=A\setminus \mathfrak{p}\) allowed us to consider \(A_\mathfrak{p}=S^{-1}A\). Using Definition 4, this can also be applied to any \(A\)-module \(M\), and the resulting \(A\)-module is denoted by \(M_\mathfrak{p}\).

Both of the above examples have a multiplication structure in addition to the addition structure and scalar multiplication by \(A\) defined in Definition 4. Explicitly, this structure is given by

\[\frac{x}{s}\frac{x'}{s'}=\frac{xx'}{ss'}\]

so we can regard \(S^{-1}A\) as an \(A\)-algebra.

The localization of a ring has the following universal property.

Proposition 6 Fix rings \(A,B\) and, for \(A\), a multiplicative subset \(S\). If under a ring homomorphism \(f:A \rightarrow B\), the image of \(S\), \(f(S)\subseteq B\), satisfies \(f(S)\subseteq B^\times\), then there exists a unique ring homomorphism \(\overline{f}: S^{-1}A \rightarrow B\) such that \(\overline{f}\circ\epsilon=f\) holds.

Proof

Suppose that \(f\) satisfying the given condition is given. If there exists \(\overline{f}: S^{-1}A \rightarrow B\) satisfying the given condition, then for any \(a/s\in S^{-1}A\),

\[\overline{f}\left(\frac{a}{s}\right)=\overline{f}\left(\frac{a}{1}\frac{1}{s}\right)=\overline{f}(\epsilon(a)\epsilon(s)^{-1})=\overline{f}(\epsilon(a))\overline{f}(\epsilon(s)^{-1})=f(a)f(s)^{-1}\]

must hold, so if \(\overline{f}\) exists, it is uniquely determined by the formula above. Now it suffices to show that, defined by the above formula \(\overline{f}(a/s)=f(a)f(s)^{-1}\), \(\overline{f}: S^{-1}A \rightarrow B\) is a ring homomorphism, and this is merely a simple calculation.

From this, the functoriality of localization can also be shown.

Localization and Ideals

Meanwhile, there is a particular relationship between localization and ideals. First, let us define the following.

Definition 7 For a ring homomorphism \(f:A \rightarrow B\), an ideal of \(A\), \(\mathfrak{a}\), and an ideal of \(B\), \(\mathfrak{b}\), we define the following.

1. Under \(f\), the contraction of \(\mathfrak{b}\) is defined as the ideal of \(A\), \(f^{-1}(\mathfrak{b})\), and is denoted by \(\mathfrak{b}^c\).
2. Under \(f\), the extension of \(\mathfrak{a}\) is defined as the ideal generated by the image \(f(\mathfrak{a})\) in \(B\), and is denoted by \(\mathfrak{a}^e\).

To make the first definition, one must prove that \(f^{-1}(\mathfrak{b})\) is an ideal, but this proof is easy. While the above notations are useful, they are relatively less intuitive, so after this post we will write \(f^{-1}(\mathfrak{b})\) and \(f(\mathfrak{a})B\) instead of the above notations.

Proposition 8 For any ring \(A\), multiplicative subset \(S\), localization \(S^{-1}A\), and canonical map \(\epsilon:A \rightarrow S^{-1}A\), the following hold.

1. For any ideal \(\mathfrak{b}\subseteq S^{-1}A\), we have \(\mathfrak{b}=\mathfrak{b}^{ce}\).

2. For any ideal \(\mathfrak{a}\subseteq A\),

   \[\mathfrak{a}^{ec}=\{a\in A\mid\text{there exists $s\in S$ satisfying $sa\in \mathfrak{a}$}\}\]

   holds. In particular, \(\mathfrak{a}^e=S^{-1}A\) if and only if \(\mathfrak{a}\cap S\neq\emptyset\).

Therefore, there exists an inclusion-preserving bijection between the prime ideals of \(S^{-1}A\) and the prime ideals of \(A\) that do not meet \(S\).

Proof

1. First, \(\mathfrak{b}^{ce}\subseteq \mathfrak{b}\) always holds in general. For the reverse direction, let \(a/s\in \mathfrak{b}\). Then \(s(a/s)=a/1\) must belong to \(\mathfrak{b}\), so \(a\in \mathfrak{b}^c\) holds. Hence \(a/1\in \mathfrak{b}^{ce}\), and from this we see that \(a/s=(1/s)(a/1)\in \mathfrak{b}^{ce}\).
2. Let us denote the right-hand side of the given equation by \(\mathfrak{a}'\) for convenience. Then first, for any \(a'\in \mathfrak{a}'\), there exists \(s\) such that \(sa'\in \mathfrak{a}\). Now from \(a'/1=sa'/s\in \mathfrak{a}^e\), we see that \(a'\in \mathfrak{a}^{ec}\). Conversely, for any \(a\in \mathfrak{a}^{ec}\), we can find \(a'\in \mathfrak{a}\) and \(s\in S\) satisfying \(a/1=a'/s\). Then there exists a suitable \(t\in S\) such that \(tsa=ta'\in \mathfrak{a}\), and since \(ts\in S\), we have \(a\in \mathfrak{a}'\) by definition. Also,

\[\mathfrak{a}^e=S^{-1}A\iff 1/1\in \mathfrak{a}^e\iff 1\in \mathfrak{a}^{ec}\iff \text{there exists $s\in S$ s.t. $s1\in \mathfrak{a}$}\iff \mathfrak{a}\cap S\neq \emptyset\]

Now from the result of 2, given any prime ideal \(\mathfrak{b}\subseteq S^{-1}A\), we know that \(\mathfrak{b}^c\) is a prime ideal of \(A\) not meeting \(S\). ([Algebraic Structures] §Field of Fractions, ⁋Proposition 10) Conversely, let \(\mathfrak{a}\subseteq A\) be a prime ideal of \(A\) not meeting \(S\). Then \(\mathfrak{a}^e\) is a prime ideal of \(S^{-1}A\). Suppose \((b/t)(b'/t')\in \mathfrak{a}^e\) for arbitrary \(b/t,b'/t'\). Then there exist \(a\in \mathfrak{a}\) and \(s\in S\) such that \((bb')/(tt')=a/s\), and hence there exists \(u\in S\) such that \(utt'a=usbb'\in \mathfrak{a}\). Now from \(\mathfrak{a}\cap S=\emptyset\), we know that \(us\not\in \mathfrak{a}\), and since \(\mathfrak{a}\) is a prime ideal, \(bb'\in \mathfrak{a}\) holds. Therefore \(b\in \mathfrak{a}\) or \(b'\in \mathfrak{a}\), and \(\mathfrak{a}^e\) is a prime ideal. That these correspondences are mutual inverses follows naturally from the results of 1 and 2.

The following is immediate from the above proposition.

Corollary 9 The localization of a Noetherian ring is Noetherian.

Proof

Suppose an ascending chain of ideals of \(S^{-1}A\)

\[\mathfrak{b}_0\subseteq \mathfrak{b}_1\subseteq\cdots\]

is given. Then

\[\mathfrak{b}_0^c\subseteq \mathfrak{b}_1^c\subseteq\cdots\]

is an ascending chain of ideals in the Noetherian ring \(A\), so there exists \(N\) such that whenever \(n>N\), \(\mathfrak{b}_n^c=\mathfrak{b}_{n+1}^c\). Now for such \(n\),

\[\mathfrak{b}_n=\mathfrak{b}_n^{ce}=\mathfrak{b}_{n+1}^{ce}=\mathfrak{b}_{n+1}\]

Meanwhile, from Proposition 8, for any prime ideal \(\mathfrak{p}\) of \(A\) it is immediate that \(\mathfrak{p}^e=\mathfrak{p}A_\mathfrak{p}\) is the unique maximal ideal of \(A_\mathfrak{p}\). That is, \(A_\mathfrak{p}\) is a local ring, and its quotient \(A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\) is a field.

Definition 10 For a ring \(A\) and a prime ideal \(\mathfrak{p}\), we call the field \(A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\) the residue field of \(A\) at \(\mathfrak{p}\) and denote it by \(\kappa(\mathfrak{p})\).

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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