Behavior of canonical divisors under purely inseparable base changes

Lemma A.1.

Let k be a field. Let

be k-morphisms of k-varieties which satisfy the following properties:

1. X,Y, and T are normal k-varieties.

2. T is an affine scheme.

3. π is a proper surjective morphism such that π*⁢𝒪Y=𝒪T and f is a finite surjective morphism.

Then S:=Spec⁡H0⁢(X,OX) completes the commutative diagram

such that S is a normal k-variety, where ρ is a proper surjective k-morphism such that ρ*⁢OX=OS and g is a finite surjective k-morphism.

Proof.

Fix an affine open cover X=⋃i∈IXi and set Yi:=f-1⁢(Xi). Clearly, S satisfies the commutative diagram in the lemma. Note that

and

Thus S is an affine integral k-scheme. We show that S satisfies the required properties in the lemma. For this, it suffices to show that g is a finite morphism by the Eakin–Nagata theorem. Taking the separable closure of K⁢(Y)/K⁢(X), we may assume that either g is separable or purely inseparable.

Suppose that K⁢(Y)/K⁢(X) is a separable. Let L⊃K⁢(Y)⊃K⁢(X) be the Galois closure and set G:=Gal⁢(L/K⁢(X)). Let g:Z→Y be the normalization of Y in L and set Zi:=g-1⁢(Yi). Since the composite morphism Z→Y→T is proper, the ring Γ⁢(Z,𝒪Z) is a finitely generated k-algebra. We obtain

Therefore, Γ⁢(S,𝒪S) is a finitely generated k-algebra and Γ⁢(Z,𝒪Z) is a finitely generated Γ⁢(S,𝒪S)-module, hence so is Γ⁢(Y,𝒪Y)=Γ⁢(T,𝒪T). Thus, g is finite.

Therefore, we may assume that K⁢(Y)/K⁢(X) is purely inseparable. We can find e∈ℤ>0 such that K⁢(X)⊃K⁢(Y)pe, in particular, Γ⁢(Xi,𝒪X)⊃Γ⁢(Yi,𝒪Y)pe. We have

where k⁢[A] means the minimum k-algebra containing A. Here Γ⁢(T,𝒪T) is a finitely generated k⁢[Γ⁢(T,𝒪T)pe]-module because it is an integral extension. Therefore, Γ⁢(T,𝒪T) is a finitely generated Γ⁢(S,𝒪S)-module. We are done. ∎

Remark A.2.

Lemma A.1 fails when T is not affine. Actually, there is a finite surjective morphism

On the other hand, Y has a proper morphism to a curve but X does not.

Lemma A.3.

Let k be a field. Let Y be a proper normal k-variety. Let Y0⊂Y be a non-empty open subset and let

be a proper surjective k-morphism to an affine k-variety Z such that π*⁢OY0=OZ. We fix an embedding K⁢(Z)⊂K⁢(Y) induced by π. Then

Proof.

Set Y1:={y∈Y:R⊂𝒪Y,y}. We show Y0⊂Y1. Take y∈Y0. Then, we obtain

This implies y∈Y1.

We prove the inverse inclusion Y0⊃Y1. We take y∈Y1. Then, we obtain the following commutative diagram of inclusions:

Take a normal projective compactification Z⊂Z¯ and the normalization of a blowup of the indeterminacy of Y⇢Z¯, denoted by π′:Y′→Z¯. We obtain

Fix a point y′ over y. Then, we obtain R⊂𝒪Y′,y′. Hence, π′⁢(y′) is a point z in Z=Spec⁡R. It is sufficient to show that π-1⁢(z)=π′⁣-1⁢(z). Note that π-1⁢(z) is proper over k and that π-1⁢(z)=π′⁣-1⁢(z)∩Y0 is an open subset of π′⁣-1⁢(z). Since π*⁢𝒪Y0=𝒪Z, the field K⁢(Z) is algebraically closed in K⁢(Y). Therefore, π*′⁢𝒪Y′=𝒪Z¯. In particular, π′⁣-1⁢(z) is connected, which implies π-1⁢(z)=π′⁣-1⁢(z). Indeed, otherwise π-1⁢(z)=π′⁣-1⁢(z)∩Y0 is not proper over k. ∎

Proposition A.4.

Let k⊂k′ be a field extension. Let X be a proper geometrically normal and geometrically connected variety over k. Let X⊃X0→V and X⊗kk′⊃Y0→W be MRCC fibrations of X and X⊗kk′, respectively. Then dim⁡V=dim⁡W.

Proof.

For a proper variety V, set r⁢(V):=dim⁡V-dim⁡W where V⊃V0→W is an MRCC fibration. Note that r⁢(V) is well-defined because W and W1 are birational for another MRCC fibration V⊃V10→W1. It is enough to show r⁢(X)=r⁢(X⊗kk′).

We show r⁢(X)≤r⁢(X⊗kk′). Take an RCC fibration X⊃X0→Z such that the dimension of a general fiber is r⁢(X). Then, taking the base change to k′, we obtain an RCC fibration X⊗kk′⊃X0⊗kk′→Z⊗kk′. Therefore r⁢(X)≤r⁢(X⊗kk′).

It suffices to prove r⁢(X)≥r⁢(X×kk′). We obtain an RCC fibration X⊗kk′⊃Y0→Z such that dim⁡X-dim⁡Z=r⁢(X⊗kk′).

We prove that we may assume that [k′:k]<∞. We obtain a family of RCC fibrations X⊗kR⊃YR0→ZR over some intermediate ring k⊂R⊂k′ which is of finite type over k. Taking a general closed point of Spec⁡R, we may assume [k′:k]<∞.

If k′/k is purely inseparable, the assertion can be easily proved. Thus, we may assume that k′/k is a finite separable extension. Moreover, by taking the Galois closure of k′/k, we may assume that k′/k is a finite Galois extension.

Assume that k′/k is a finite Galois extension. In this case, we have an MRCC fibration Y:=X⊗kk′⊃Y0→Z such that dim⁡Y-dim⁡Z=r⁢(X⊗kk′). Since k′/k is a Galois extension, so is K⁢(Y)=K⁢(X⊗kk′)/K⁢(X). Set

By shrinking Z, we may assume Z=Spec⁡RZ. We fix an embedding K⁢(Z)⊂K⁢(Y). Let σi*:Y→Y be the induced automorphism.

We show K⁢(Z)=σi⁢(K⁢(Z)) for every integer 1≤i≤N. Fix 1≤i≤N. We obtain another MRCC fibration

Then, we see that Spec⁡RZ and Spec⁡σi⁢(RZ) are birational. This implies K⁢(Z)=σi⁢(K⁢(Z)).

Thus, each σi induces an birational automorphism σi:Z⇢Z. As G={σ1,…,σN} is a finite group, we can find a non-empty open subset Z′⊂Z such that the induced rational map Z′⇢Z′ is an isomorphism, i.e. automorphism. By replacing Z with Z′, we may assume that σi:Z→Z is an automorphism. In particular, σi⁢(RZ)=RZ.

We show Y0=σi*⁢(Y0) for every i. By Lemma A.3, we obtain

Therefore,

Set X0:=β⁢(Y0), where β:Y→X. Then, by Y0=σi⁢(Y0), we obtain

By Lemma A.1, we obtain the commutative diagram

where S is an affine normal k-variety, T→S is a finite surjective morphism, and X0→S is a proper surjective morphism. Since fibers of Y0→T are RCC, so are the fibers of X0→S. Therefore X⊃X0→S is an RCC fibration, which implies