TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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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

1.1F1F2construct

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.

1.2F2F3algebra

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 $\tilde{f} \circ λ_{S} = f$ .

2.1F2step 1.2

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.

3.1F2step 1.1step 1.2step 2.1discharge-construct∎

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.

Depends on

Dependency tree · 2 levels

Sources

The Stacks Project, Proposition 10.9.3