TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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Universal property of localisation: maps that invert $S$ factor uniquely through $S^{- 1} R$

Statement

Let $f : R \to A$ be a unital homomorphism of commutative rings such that $f (s)$ is a unit for every $s \in S$ . There is a unique unital ring homomorphism $\tilde{f} : S^{- 1} R \to A$ satisfying $\tilde{f} \circ λ_{S} = f$ , namely $\tilde{f} (r / s) = f (r) f (s)^{- 1} .$

Facts & Assumptions

Given: A multiplicative subset $S$ of a commutative ring $R$ and a unital ring homomorphism $f : R \to A$ taking every element of $S$ to a unit.

[F1]

Fraction equality means that $u (r s^{'} - r^{'} s) = 0$ for some $u \in S$ (Equality, vanishing, and the kernel of the localisation map).

[F2]

Units form a group under multiplication, so products and inverses of units are units and inverses are unique (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring).

[F3]

Localisation arithmetic is $r / s + r^{'} / s^{'} = (r s^{'} + r^{'} s) / (s s^{'})$ and $(r / s) (r^{'} / s^{'}) = r r^{'} / (s s^{'})$ (The localisation relation is an equivalence relation and fraction arithmetic is well defined).

Proof

technique · constructive

1.1

Define $\tilde{f} (r / s) = f (r) f (s)^{- 1}$ . If $r / s = r^{'} / s^{'}$ , [F1] gives $u (r s^{'} - r^{'} s) = 0$ for some $u \in S$ . Applying $f$ and cancelling the unit $f (u)$ yields $f (r) f (s^{'}) = f (r^{'}) f (s)$ ; multiplying by $f (s)^{- 1} f (s^{'})^{- 1}$ proves the definition is independent of representatives.

F1F2construct

1.2

Using [F3] and the homomorphism laws for $f$ , direct calculation shows that $\tilde{f}$ preserves addition, multiplication, zero, and one. Also $\tilde{f} (r / 1) = f (r) f (1)^{- 1} = f (r)$ , so $\tilde{f} \circ λ_{S} = f$ .

F2F3algebra

2.1

If $g : S^{- 1} R \to A$ is another such homomorphism, then $g (r / 1) = f (r)$ and $g (s / 1) = f (s)$ . Since $(s / 1) (1 / s) = 1$ , uniqueness of inverses gives $g (1 / s) = f (s)^{- 1}$ ; hence $g (r / s) = g (r / 1) g (1 / s) = \tilde{f} (r / s)$ for every fraction.

F2step 1.2

3.1

If $0 \in S$ , the hypothesis says that $f (0) = 0$ is a unit of $A$ , so $A$ is the zero ring. The construction and uniqueness above still apply, with the unique map between zero rings.

F2step 1.1step 1.2step 2.1discharge-construct∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 20 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

The Stacks Project, Proposition 10.9.3 (standard reference, not scraped)

Universal property of localisation: maps that invert SS factor uniquely through S−1RS−1R