Localization

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

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 called a local ring if it 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. Every non-unit of \(A\) lies in some proper ideal \(\mathfrak{m}\subsetneq A\).
3. The set of all non-units of \(A\) forms an ideal.

Proof

First assume (1), and let \(a\in A\) be an arbitrary non-unit. Then \((a)\) is an ideal of \(A\), so by [Algebraic Structures] §Definition of Rings, ⁋Theorem 9 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 hence \(a\in \mathfrak{m}\).

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

Finally, assume (3) and show (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 \(\mathfrak{m}\) consisting of all non-units of \(A\). 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 \(A\) itself.

Localization of Modules

As explained in the previous post, to study localization of rings we define localization of modules in greater generality.

Definition 3 A subset \(S\) of a ring \(A\) is multiplicatively closed if the product of any 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, we have \(1\in S\) by definition.

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

1. As a set, \(S^{-1}M\) is the quotient of \(M\times S\) by the equivalence relation

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

   We write the equivalence class of \((x,s)\) as \(x/s\).

2. The \(A\)-module structure on \(S^{-1}M\) is defined by

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

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

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

This holds simply because \(1(txs-tsx)=0\). Also, we can examine the additive identity and inverses in \(S^{-1}M\): for any \(s,s'\in S\) we can verify that

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

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 additive identity 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 define the canonical map \(\epsilon: M \rightarrow S^{-1}M\) by \(x\mapsto x/1\). Unfortunately, \(\epsilon\) need not be injective in general; the reason is obvious 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 \(M\) is annihilated by \(S\).

Proof

If

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

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

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

The above argument also works 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 take \(M=A\). 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, again taking \(M=A\) and letting \(S=A\setminus \mathfrak{p}\) for a prime ideal \(\mathfrak{p}\) of \(A\), we can consider \(A_\mathfrak{p}=S^{-1}A\). Using Definition 4 we can apply this to any \(A\)-module \(M\), and the resulting \(A\)-module is denoted \(M_\mathfrak{p}\).

Both of the above examples carry a multiplication structure in addition to the addition 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 a multiplicative subset \(S\) of \(A\). If the image \(f(S)\subseteq B\) of a ring homomorphism \(f:A \rightarrow B\) satisfies \(f(S)\subseteq B^\times\), then there exists a unique ring homomorphism \(\overline{f}: S^{-1}A \rightarrow B\) such that \(\overline{f}\circ\epsilon=f\).

Proof

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

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

so if \(\overline{f}\) exists it is uniquely determined by the above formula. It remains to show that \(\overline{f}: S^{-1}A \rightarrow B\) defined by \(\overline{f}(a/s)=f(a)f(s)^{-1}\) is a ring homomorphism, which is a straightforward calculation.

From this the functoriality of localization can also be shown.

Localization and Ideals

Meanwhile, there is a specific relationship between localization and ideals. First we define the following.

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

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

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

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 \(\mathfrak{b}\subset S^{-1}A\), we have \(\mathfrak{b}=\mathfrak{b}^{ce}\).

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

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

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

Therefore, there is 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 \(a\in \mathfrak{b}^c\). Hence \(a/1\in \mathfrak{b}^{ce}\), and from this we see \(a/s=(1/s)(a/1)\in \mathfrak{b}^{ce}\).
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 \(sa'\in \mathfrak{a}\). 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 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 (2), given any \(\mathfrak{b}\subseteq S^{-1}A\) we know that \(\mathfrak{b}^c\) is a prime ideal of \(A\) not meeting \(S\). ([Algebraic Structures] §Field of Fractions, ⁋Proposition 9) 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 \(utt'a=usbb'\in \mathfrak{a}\). Since \(\mathfrak{a}\cap S=\emptyset\) we know \(us\not\in \mathfrak{a}\), and since \(\mathfrak{a}\) is a prime ideal we have \(bb'\in \mathfrak{a}\). 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 result of (2).

The following is immediate from the above proposition.

Corollary 9 The localization of a Noetherian ring is Noetherian.

Proof

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

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

we have

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

which is an ascending chain of ideals in the Noetherian ring \(A\), so there exists \(N\) such that for all \(n>N\) we have \(\mathfrak{b}_n^c=\mathfrak{b}_{n+1}^c\). 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 field \(A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\) is well-defined.

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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