Let f : X —> S be a morphism of finite type. The *relative assassin Ass(X/S) of X/S* is the set of points x of X which are embedded points of their fibres. So if f has reduced fibres or if f has fibres which are S_1, then these are just the generic points of the fibres, but in general there may be more. If T —> S is a morphism of schemes then it isn’t quite true that Ass(X_T/T) is the inverse image of Ass(X/S), but it is almost true, see Remark Tag 05KL.

Definition: We say X is *S-pure* if for any x ∈ Ass(X/S) the image of the closure {x} is closed in S, and if the same thing remains true after any etale base change.

Clearly if f is proper then X is pure over S. If f is quasi-finite and separated then X is S-pure if and only if X is finite over S (see Lemma Tag 05K4). It turns out that if S is Noetherian, then purity is preserved under arbitrary base change (see Lemma Tag 05J8), but in general this is not true (see Lemma Tag 05JK). If f is flat with geometrically irreducible (nonempty) fibres, then X is S-pure (see Lemma Tag 05K5).

A key algebraic result is the following statement: Let A —> B be a flat ring map of finite presentation. Then B is projective as an A-module if and only if Spec(B) is pure over Spec(A), see Proposition Tag 05MD. The current proof involves several bootstraps and starts with proving the result in case A —> B is a smooth ring map with geometrically irreducible fibres.

I challenge any commutative algebraist to prove this statement without using the language of schemes. You will find another challenge in the next post.