Localization of a commutative ring

Let $𝑅$ be a commutative ring and $𝑆$ be a multiplicative submonoid. The localization of $𝑅$ at $𝑆$ is a certain universal solution to the problem of adding inverses for each element in $𝑆$ . Specifically, it consists of a ring $𝑆 - 1 𝑅$ equipped with a ring homomorphism $𝑗 𝑅, 𝑆 : 𝑅 \to 𝑆 - 1 𝑅$ such that

we have $𝑗 (𝑆) \subseteq (𝑆 - 1 𝑅) \times$ , the group of units of $𝑆 - 1 𝑅$ ;
if $𝜑 : 𝑅 \to 𝑇$ is a ring homomorphism with $𝜑 (𝑅) \in 𝑇 \times$ , then there exists a unique factorization $¯ 𝜑 : 𝑆 - 1 𝑅 \to 𝑇$ so that $𝜑 = ¯ 𝜑 𝑗$ .

If we instead have set of elements $𝑆$ , we denote the localization at the submonoid generated by $𝑆$ by $𝑅 [𝑆 - 1]$ .

Caveat emptor

If the set $𝑆$ is not regular, then $𝑗 : 𝑅 \to 𝑅 [𝑆 - 1]$ may not be injective, in which case $𝑅 [𝑆 - 1]$ is not a true adjunction of a ring. See Regulars and zero-divisors.

Construction

The localization is typically constructed as follows. Elements of $𝑆 - 1 𝑅$ are formal quotients $𝑥 / 𝑠$ for some $𝑥 \in 𝑅$ and $𝑠 \in 𝑆$ . Addition and multiplication are defined in the usual manner, i.e.

𝑥 𝑠 1 + 𝑦 𝑠 2 = 𝑠 2 𝑥 + 𝑠 1 𝑦 𝑠 1 𝑠 2, 𝑥 𝑠 1 \cdot 𝑦 𝑠 2 = 𝑥 𝑦 𝑠 1 𝑠 2,

and equality is defined by

𝑥 1 𝑠 1 = 𝑥 2 𝑠 2 ⟺ \exists 𝑠 \in 𝑆 \to 𝑠 𝑠 2 𝑥 1 = 𝑠 𝑠 1 𝑥 2 .

The morphism $𝑗$ is defined by $𝑥 \mapsto 𝑥 / 1$ .

Proof of universal property

To see ^L1, note that the inverse of $𝑗 (𝑠) = 𝑠 / 1$ is just $1 / 𝑠$ . For ^L2, suppose $𝜑$ is such a homomorphism. It is clear that any putative factorization must satisfy $¯ 𝜑 : 𝑥 / 𝑠 \mapsto 𝜑 (𝑥) 𝜑 (𝑠) - 1$ , so it suffices to show this map is well-defined in general. Suppose $𝑥 1, 𝑥 2 \in 𝑅$ and $𝑠, 𝑠 1, 𝑠 2 \in 𝑆$ so that $𝑠 𝑠 2 𝑥 1 = 𝑠 𝑠 1 𝑥 2$ . Then

𝜑 (𝑠) 𝜑 (𝑠 2) 𝜑 (𝑥 1) = 𝜑 (𝑠) 𝜑 (𝑠 1) 𝜑 (𝑥 2) .

Multiplying both sides by $𝜑 (𝑠 𝑠 1 𝑠 2) - 1$ gives

𝜑 (𝑥 1) 𝜑 (𝑠 1) - 1 = 𝜑 (𝑥 2) 𝜑 (𝑠 2) - 1,

so indeed $¯ 𝜑 : 𝑆 - 1 𝑅 \to 𝑇$ is well defined.

Explanation of terminology

Suppose our ring is that of smooth scalar fields on a $𝐶 \infty$ -manifold $𝑀$ , $𝑓 \in 𝐶 \infty (𝑀)$ and $𝑈 = s u p p 𝑓$ is its support. Then the localization $𝐶 \infty (𝑀) [1 / 𝑓]$ corresponds to the subalgebra of $𝐶 \infty (s u p p 𝑓)$ consisting of scalar fields of the form $(𝑔 ↾ 𝑈) / (𝑓 ↾ 𝑈) 𝑚$ for some $𝑔 \in 𝐶 \infty (𝑀)$ and $𝑚 \in ℕ$ .

Regulars and zero-divisors

Let $𝑅$ be a ring, $𝑎 \in 𝑅$ be an element, $𝑆 = ⟨ 𝑎 ⟩$ be the multiplicative monoid generated thereby. Then

If $𝑎$ is regular, then $𝑗 : 𝑅 ↪ 𝑆 - 1 𝑅$ is injective.

We can consider two special cases where $𝑎$ is a zero-divisor:

If $𝑎$ is nilpotent, then $𝑆 - 1 𝑅 ≅ 0$ .
If $𝑎$ is an idempotent then $𝑗 : 𝑅 ↠ 𝑆 - 1 𝑅$ is surjective and $𝑆 - 1 𝑅 = 𝑅 / ⟨ 1 - 𝑎 ⟩$ .

Proof

For ^j1, suppose we have $𝑥, 𝑦 \in 𝑅$ such that $𝑗 (𝑥) = 𝑗 (𝑦)$ , i.e. $𝑥 / 1 = 𝑦 / 1$ . By the definition of equality in $𝑆 - 1 𝑅$ , this means there exists some $ℓ \in ℕ$ such that $𝑎 ℓ 𝑥 = 𝑎 ℓ 𝑦$ , which by regularity of $𝑎$ can only occur if $𝑥 = 𝑦$ . Therefore $𝑗$ is injective.

For ^j2, note that $𝑎$ being nilpotent implies $0 \in 𝑆$ . Given any $𝑥 / 𝑎 ℓ, 𝑦 / 𝑎 𝑘 \in 𝑅 - 1 𝑆$ , we then have $0 𝑎 𝑘 𝑥 = 0 = 0 𝑎 ℓ 𝑦$ so $𝑥 / 𝑎 ℓ = 𝑦 / 𝑎 𝑘$ . Therefore $𝑆 - 1 𝑅$ is trivial.

For ^j3, first note that $𝑗 (𝑎) = 𝑗 (1) = 1$ , since $𝑎 \cdot 1 \cdot 𝑎 = 𝑎 \cdot 𝑎 \cdot 1$ . It follows that $𝑗$ is surjective, since $𝑥 / 𝑎 ℓ = 𝑥 / 1 = 𝑗 (𝑥)$ for $ℓ \in ℕ$ . By the First isomorphism theorem, we just need to show that $k e r 𝑗 = ⟨ 1 - 𝑎 ⟩$ . It is clear that $⟨ 1 - 𝑎 ⟩ \subseteq k e r 𝑗$ . To show that $k e r 𝑗 \subseteq ⟨ 1 - 𝑎 ⟩$ , suppose $𝑥 \in k e r 𝑗$ , i.e. $𝑗 (𝑥) = 𝑥 / 1 = 0 / 1 = 0$ . Then either $𝑥 = 0$ or $𝑎 𝑥 = 0$ . In the first case we are done, and in the second case we have $𝑎 𝑥 = 0$ whence $𝑥 = 𝑥 (𝑎 - 1) \in ⟨ 𝑎 - 1 ⟩$ . Therefore $k e r 𝑗 = ⟨ 1 - 𝑎 ⟩$ as required.