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Every faithfully flat ring map is injective

Statement

Assume the Axiom of Choice.

Every faithfully flat homomorphism of commutative rings is injective.

Facts & Assumptions

Given: The Axiom of Choice and a faithfully flat homomorphism $f : R \to S$ .

[L1]

Faithful flatness of the ring map means that $S$ is a faithfully flat $R$ -module, hence in particular a flat $R$ -module (Flat and faithfully flat modules and ring homomorphisms).

[L2]

For a faithfully flat module $T$ , a nonzero module cannot tensor to zero: if $N \otimes_{R} T = 0$ , then $N = 0$ (For a flat module, faithful flatness is equivalent to detecting nonzero modules and residue fields).

Proof

technique · direct

1.1L1given

Let $K = \ker f$ . Since $S$ is flat over $R$ by [L1], tensoring the exact sequence $0 \to K \to R \overset{f}{\to} S$ with $S$ remains exact: $0 \to K \otimes_{R} S \to R \otimes_{R} S \to S \otimes_{R} S .$

2.1step 1.1algebra

Under the canonical identification $R \otimes_{R} S ≅ S$ , the middle map in step 1.1 is $s \mapsto 1 \otimes s$ . Multiplication $μ : S \otimes_{R} S \to S$ , $μ (s \otimes t) = s t$ , is a left inverse to this map, so the middle map is injective. Therefore $K \otimes_{R} S = 0$ .

3.1L1L2step 2.1

Because $S$ is faithfully flat over $R$ by [L1], [L2] applied to $N = K$ shows that $K = 0$ .

4.1algebra∎

Hence $f$ is injective.

Depends on

Used by

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, A Primer of Commutative Algebra, Proposition 11.12 (standard reference, not scraped)