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localization of a commutative ring

.

DEFINITION 04.2   Let $ f$ be an element of a commutative ring $ A$ . Then we define the localization $ A_f$ of $ A$ with respect to $ f$ as a ring defined by

$\displaystyle A_{f}=A[X]/(X f -1)
$

where $ X$ is a indeterminate.

In the ring $ A_f$ , the residue class of $ X$ plays the role of the inverse of $ f$ . Therefore, we may write $ A[1/f]$ instead of $ A_f$ if there is no confusion.

One may define localization in much more general situation. The reader is advised to read standard books on commutative algebras.

LEMMA 04.3   Let $ f$ be an element of a commutative ring $ A$ . Then there is a canonically defined homeomorphism between $ O_f$ and $ \operatorname{Spec}(A_f)$ . (It is usual to identify these two via this homeomorphism.)

PROOF.. Let $ i_f:A \to A_f$ be the natural homomorphism. We have already seen that we have a continuous map

$\displaystyle \operatorname{Spec}(i_f):\operatorname{Spec}(A_f)\to \operatorname{Spec}(A).
$

We need to show that it is injective, and that it gives a homeomorphism between $ \operatorname{Spec}(A_f)$ and $ O_f$ .

Let us do this by considering representations.

  1. $ \mathfrak{p}\in \operatorname{Spec}(A)$ corresponds to a representation $ \rho_\mathfrak{p}$ .

  2. % latex2html id marker 962
$ \mathfrak{q}\in \operatorname{Spec}(A_f)$ corresponds to a representation % latex2html id marker 964
$ \rho_\mathfrak{q}$ .

  3. $ \operatorname{Spec}(i_f)$ corresponds to a restriction map $ \rho\mapsto \rho \circ i_f$ .

Now, for any $ \mathfrak{p}\in \operatorname{Spec}(A)$ , $ \rho_\mathfrak{p}$ extends to $ A_f$ if and only if the image $ \rho_\mathfrak{p}(f)$ of $ f$ is invertible, that means, % latex2html id marker 980
$ \rho_\mathfrak{p}(f)\neq 0$ . In such a case, the extension is unique. (We recall the fact that the inverse of an element of a field is unique.)

It is easy to prove that $ \operatorname{Spec}(i_f)$ is a homeomorphism. % latex2html id marker 947
$ \qedsymbol$

Let $ A$ be a ring. Let $ f\in A$ . It is important to note that each element of $ A_f$ is written as a ``fraction''

% latex2html id marker 990
$\displaystyle \frac{x}{f^k} \qquad (x\in A; k\in \mathbb{N}).
$

One may introduce $ A_f$ as a set of such formal fractions which satisfy ordinary computation rules. In precise, we have the following Lemma.

LEMMA 04.4   Let $ A$ be a ring, $ f$ be its element. Let us consider the following set

$\displaystyle S=\{ (x ,f^k); x\in A; k\in \mathbb{N}\}.
$

We introduce on $ S$ the following equivalence law.

% latex2html id marker 1007
$\displaystyle (x ,f^k) \sim (y,f^l) \iff (y f^k-x f^l)f^N=0 \qquad (\exists N\in \mathbb{N})
$

Then we may obtain a ring structure on $ S/\sim$ by introducing the following sum and product.

$\displaystyle (x /f^k) + (y/f^l)=(x f^l +y f^k/ f^{k+l})
$

$\displaystyle (x /f^k) (y/f^l)=(x y / f^{k+l})
$

where we have denoted by $ (x/f^k)$ the equivalence class of $ (x,f^k)\in S$ .

COROLLARY 04.5   Let $ A$ be a ring, $ f$ be its element. Then we have $ A_f=0$ if and only if $ f$ is nilpotent.

Likewise, for any $ A$ -module $ M$ , we may define $ M_f$ as a set of formal fractions

% latex2html id marker 1038
$\displaystyle \frac{m}{f^k} \qquad (m\in M; k\in \mathbb{N}).
$

which satisfy certain computation rules.

Subsections
next up previous
Next: Existence of a point Up: Algebraic geometry and Ring Previous: ring homomorphism and spectrum
2017-07-21