CANCELLATION THEOREM FOR ALGEBRAIC LINE BUNDLES

Robert Dryło

PREPRINT IMUJ 2005/13

Abstract

Let 𝕂 be an algebraically closed field of characteristic 0. In this paper we consider the following cancellation problem. Let X be an affine variety and E an algebraic line bundle on X. Suppose that a variety Y and an isomorphism Y × 𝕂^{m} ≅ E × 𝕂^{m} are given. Is it true that Y ≅ E? We give an affirmative solution if X is either non-𝕂-uniruled or it is unirational and has a non-uniruled component at infinity.

1 Introduction

We shall work in the category of algebraic varieties over an algebraically closed field 𝕂 of characteristic 0. By a variety we will usually mean an irreducible algebraic variety.

A variety X has the cancellation property if it satisfies the condition: if Y is a variety such that for some m ≥ 0 there is an isomorphism X × 𝕂^{m} ≅ Y × 𝕂^{m} then X ≅ Y. Furthermore, we say that a variety X has the strong cancellation property if every isomorphism f : X × 𝕂^{m} → Y × 𝕂^{m} satisfies the condition: for each x ∈ X there exists y ∈ Y such that f({x} × 𝕂^{m}) = {y} × 𝕂^{m} (clearly, then f induces the isomorphism between X and Y).

It is well-known that affine curves have the cancellation property (a more general result was proved by Abhyankar, Eakin and Heinzer in [1]), but surfaces may not have this property (see Danielewski [3]).

Zariski’s Cancellation Problem asks if 𝕂^{n} has the cancellation property. The affirmative answer for n = 2 is due to Fujita [6] and Miyanishi-Sugie [15]. For n > 2 this problem remains open.

A variety of non-negative logarithmic Kodaira dimension has the strong cancellation property, which was proved by Iitaka and Fujita in [8]. This property has also an affine variety which is either non-𝕂-uniruled or it is unirational of dimension greater than 1 and has a non-uniruled component at infinity (see [4]).

In this paper we extend the last result, namely, we will prove the following

Theorem 1. Suppose that an affine variety X is either non-𝕂-uniruled or it is unirational of dimension greater than 1 and has a non-uniruled component at infinity. Then

(i) X has the strong cancellation property. (ii) any algebraic line bundle on X has the cancellation property.

We use the following terminology.

A variety X of positive dimension n is said to be uniruled (𝕂-uniruled) if there exists a variety W of dimension n − 1 and a dominant rational map W × ℙ^{1}(𝕂) −−→ X (a dominant morphism W × 𝕂 → X). A reducible variety is said to be uniruled (𝕂-uniruled) if all its irreducible components are uniruled (𝕂-uniruled).

A variety X of dimension n is said to be unirational if there exists a dominant rational map ℙ^{n} −−→ X.

A projective variety which contains a variety X as an open subset is called a compactification of X.

An affine variety X has a non-uniruled component at infinity if there exists a compactification \overline{X} of X such that some component of the set \overline{X} \ X is non-uniruled (it is well-known that for any compactification \overline{X} of X the set \overline{X} \ X is of pure dimension dim X − 1).

In the sequel π_{X} denotes the projection X × 𝕂^{m} ∋ (x, t) ↦ x ∈ X.

2 Proof of Theorem 1.

Lemma 1. Let X be an affine variety. Suppose that there exists a variety Y and a dominant morphism f : Y × 𝕂^{m} → X such that f({b} × 𝕂^{m}) has positive dimension for some b ∈ Y. Then X is 𝕂-uniruled. Furthermore, if Y is unirational then X has only uniruled components at infinity.

Proof. It suffices to give the proof for m = 1 (to see this observe that there is a line L in 𝕂^{m} such that f({b} × L) has positive dimension, so changing coordinates we may assume that there exists a dominant morphism (Y × 𝕂^{m−1}) × 𝕂 → X not contracting of {b′} × 𝕂 for some b′ ∈ Y × 𝕂^{m−1}).

The proof is by induction on r := dim Y. Let n := dim X. If r = n − 1 then X is 𝕂-uniruled by definition. Furthermore, if Y is unirational with a dominant rational map g : ℙ^{r} −−→ Y then we have a dominat morphism res g × id_{𝕂} : U × 𝕂 → Y × 𝕂, where U is the domain of g. Consequently we have a dominant morphism U × 𝕂 → X. Since ℙ^{r} × ℙ^{1} is a smooth compactification of U × 𝕂 and (ℙ^{r} × ℙ^{1}) \ (U × 𝕂) is uniruled, for every compactification \overline{X} of X the set \overline{X} \ X is uniruled by [11, th.4].

Suppose r ≥ n. Let Z = {y ∈ Y : dim f({y} × 𝕂) = 0}. This is a closed subset of Y, because if X is contained in 𝕂^{N} and f = (f_{1}, ..., f_{N}) then Z = ⋂_{i=1,...,N} ⋂_{s,t∈𝕂}{y ∈ Y : f_{i}(y, s) − f_{i}(y, t) = 0}. In particular, if V is any non-empty open subset in Y then we may always assume that b ∈ V. Let X_{0} be an open subset of X such that dim f^{−1}(x) = r+1−n for each x ∈ X_{0}. We may assume that f({b} × 𝕂) meets X_{0}. Choose a ∈ f({b} × 𝕂) ∩ X_{0} and a hyperplane section H of X passing through a and not containing f({b} × 𝕂). Clearly, f^{−1}(H) is of pure dimension r. Let S be a component of f^{−1}(H) such that S ∩f^{−1}(a)∩({b} × 𝕂) ≠ ∅. Put f̃ = res f : S → H and π̃ = res π_{Y} : S → Y. These morphisms are dominant, since dim f̃^{−1}(a) = r + 1 − n and dim π̃^{−1}(b) = 0.

First we show that X is 𝕂-uniruled. Choose c ∈ S ∩ f^{−1}(a) ∩ ({b} × 𝕂) and a hypersurface H′ contained in S such that c ∈ H′ and dim(H′ ∩ f̃^{−1}(a)) < dim f̃^{−1}(a). Then res f : H′ → H is dominant. Consequently res f : π̃(H′) × 𝕂 → X is dominant, since H ∪ f({y} × 𝕂) is contained in the closure of f(π̃(H′) × 𝕂). This proves that X is 𝕂-uniruled.

Suppose now that Y is unirational. Let Y_{0} be a non-singular open subset of Y such that dim π̃^{−1}(y) = 0 and {y} × 𝕂 is not contained in f^{−1}(H) for each y ∈ Y_{0}. We may assume that there is an open subset U in ℙ^{r} and a finite morphism U → Y_{0}, and that b ∈ Y_{0}. Let us choose an unirational hypersurface H′′ contained in Y_{0} such that b ∈ H′′ and dim(H′′∩π̃(f̃^{−1}(a))) < dim f̃^{−1}(a). Since H′′ is locally principal in Y_{0} (we have assumed that Y_{0} is smooth), π̃^{−1}(H′′) is of pure dimension r − 1 and dim(π̃^{−1}(H′′) ∩ f̃^{−1}(a)) < dim f̃^{−1}(a). Hence res f : π̃^{−1}(H′′) → H is dominant. So similarly as above, res f : H′′ × 𝕂 → X is dominant. This concludes the proof.

Now we come to the following

Problem 1. Let R be a ring (commutative with identity). Suppose that A is an R-algebra and there is an R-isomorphism of polynomial rings R[T_{1}, ....T_{n+1}] ≅ A[T_{1}, ..T_{n}]. Is it true that A is R-isomorphic to R[T_{1}]?

This problem was considered in [1], [2], [9] and [13]. Generally the answer is negative. Asanuma gave a counterexample in [2] if char R > 0 (see also section 4). In the next section we prove that if R is a coordinate ring of a smooth affine variety then the answer is affirmative. Although this is a consequence of more general results from papers mentioned above, we will give a short geometric proof.

In the proof of Theorem 1 we shall use the following

Theorem 2. (Hamann, [9].) If R is a ℚ-algebra then Problem 1 has an affirmative solution.

We will need also an elementary

Lemma 2. Let X be a variety and p_{i} : E_{i} → X be an algebraic line bundle on X for i = 1, 2. Then E_{1} ≅ E_{2} as line bundles on X provided that there exists m ≥ 0 and an isomorphism f : E_{1} × 𝕂^{m} → E_{2} × 𝕂^{m} such that the following diagram is commutative

\xymatrix{E_{1} × 𝕂^{m} \ar[d]_{π_{E_{1}}} \ar[rr]^{f} &  & E_{2} × 𝕂^{m} \ar[d]^{π_{E_{2}}} \\ E_{1} \ar[dr]_{p_{1}} &  & E_{2} \ar[dl]^{p_{2}} \\  & X}

Proof. Suppose that E_{i} is given by an open cover {U_{α}} of X and by transition functions g_{α,β}^{i} : U_{α} ∩ U_{β} → 𝕂^{∗} for i = 1, 2. Clearly, we can identify E_{i} × 𝕂^{m} with the direct sum of E_{i} and the trivial bundle X × 𝕂^{m}. Hence

G_{α,β}^{i} = ( \begin{matrix}g_{α,β}^{i} & 0 \\ 0 & I_{m} \end{matrix} )

are transition functions for E_{i} × 𝕂^{m} on U_{α} ∩ U_{β}, where I_{m} is the identity in GL(𝕂^{m}). Obviously, f induces the family of polynomial isomorphisms of trivial bundles f_{α} : U_{α} × 𝕂^{m+1} → U_{α} × 𝕂^{m+1} such that for each u ∈ U_{α} ∩ U_{β},

f_{α}(u,^{.})G_{α,β}^{1}(u) = G_{α,β}^{2}(u)f_{β}(u,^{.}).

Denote by Jf_{α}(u) the Jacobi matrix of f_{α}(u,^{.}) for u ∈ U_{α}, and by h_{α} the Jacobian of f_{α}, i.e., h_{α} : U_{α} ∋ u ↦ det Jf_{α}(u,^{.}) ∈ 𝕂^{∗}. Then for u ∈ U_{α} ∩ U_{β},

h_{α}(u)g_{α,β}^{1}(u) = g_{α,β}^{2}(u)h_{β}(u).

Thus the family {h_{α}} determines the isomorphism between E_{1} and E_{2}. This concludes the proof.

Proof of theorem 1. The following easy observation will be needed: if V_{1} and V_{2} are affine varieties and V_{1} × 𝕂^{m} ≅ V_{2} × 𝕂^{m} then V_{1} dominates V_{2} and conversely. In particular, if V_{1} is unirational then so is V_{2}.

(i) Let f : Y × 𝕂^{m} → X × 𝕂^{m} be an isomorphism. Applying Lemma 1 to π_{X} ∘ f, we find a morphism g : Y → X such that π_{X} ∘ f = g ∘ π_{Y}. This implies that X has the strong cancellation property.

(ii) Let p : E → X be an algebraic line bundle on X and f : Y × 𝕂^{m} → E × 𝕂^{m} an isomorphism. Again, by Lemma 1 there exists a morphism q : Y → X such that the following diagram is commutative

\xymatrix{Y × 𝕂^{m} \ar[dd]_{π_{Y}} \ar[r]^{f} & E × 𝕂^{m} \ar[d]^{π_{X}} \\  & E \ar[d]^{p} \\ Y \ar[r]^{q} & X}

Therefore if E is trivial on an affine open subset U of X then res q : q^{−1}(U) → U is a trivial bundle by the geometric version of Theorem 2. Thus q : Y → X is a line bundle and Lemma 2 concludes the proof.

Remark 1. Theorem 1 remains true if we assume that X \ Sing X is either non-𝕂-uniruled or it is unirational of dimension greater than 1 and has a non-uniruled hypersurface at infinity.

For any variety X we denote, here and in the sequel, the singular locus by Sing X and the set of non-singular points by Reg X = X \ Sing X.

A variety X of dimension n has a non-uniruled hypersurface at infinity if there exists a compactification \overline{X} of X such that the set \overline{X} \ X has a non-uniruled component of dimension n − 1.

Proof. The above proof works also in this situation, we only need to modify Lemma 1 slightly. Furthermore, the following observation will be needed: if f : Y × 𝕂^{m} → X × 𝕂^{m} (f : Y × 𝕂^{m} → E × 𝕂^{m}, where p : E → X is a line bundle on X) is an isomorphism then it induces the isomorphism res f : Reg Y × 𝕂^{m} → Reg X × 𝕂^{m} (res f : Reg Y × 𝕂^{m} → Reg E × 𝕂^{m}).

The details are left to the reader.

3 More about Problem 1

We give now a solution of Problem 1 when R is the coordinate ring of a smooth affine variety over an algebraically closed field 𝕂 of arbitrary characteristic.

Proposition 1. Let X be an affine smooth variety. Let Y be a variety with a morphism q : Y → X and let f : X × 𝕂^{m+1} → Y × 𝕂^{m} be an isomorphism such that π_{X} = q ∘ π_{Y} ∘ f. Then there exists an isomorphism g : X × 𝕂 → Y such that q ∘ g = π_{X}.

\xymatrix{X × 𝕂^{m+1} \ar[ddr]_{π_{X}} \ar[r]^{f}_{≅} & Y × 𝕂^{m} \ar[d]^{π_{Y}} \\  & Y \ar[d]^{q} & X × 𝕂 \ar[l]_{g}^{≅} \ar[dl]^{π_{X}} \\  & X}

Proof. First observe that all fibers of q are isomorphic to 𝕂. Indeed, f takes π_{X}^{−1}(x) ≅ 𝕂^{m+1} onto q^{−1}(x) × 𝕂^{m}, so q^{−1}(x) ≅ 𝕂, because affine curves have the cancellation property. Take now the zero section s_{0} : X ∋ x ↦ (x, 0) ∈ X × 𝕂^{m+1}. Then s : X ∋ x ↦ π_{Y} (f(s_{0}(x))) ∈ Y is a section of q, i.e., q ∘ s = id_{X}. In particular, s is a closed immersion and Γ := s(X) meets transversally all fibers of q. We shall show that q : Y → X is a line bundle.

Since X and Y are smooth, we have isomorphisms π_{X}^{∗} : Pic(X) → Pic(X × 𝕂^{m+1}) and π_{Y}^{∗} : Pic(Y) → Pic(Y × 𝕂^{m}) (see [10, p.134,141]). Hence q^{∗} : Pic(X) → Pic(Y) is an isomorphism too. For a prime divisor Γ on Y there exists a divisor D on X such that Γ and q^{∗}(D) are linearly equivalent. Let {U_{i}} be an open affine cover of X such that D ∩U_{i} is principal in U_{i}. Clearly, q^{∗}(D)∩q^{−1}(U_{i}) is principal in q^{−1}(U_{i}), and so is Γ∩q^{−1}(U_{i}). This implies that the ideal of Γ∩q^{−1}(U_{i}) is generated in the coordinate ring 𝕂[q^{−1}(U_{i})] by some function F_{i} ∈ 𝕂[q^{−1}(U_{i})]. Now F_{i} restricts to a coordinate function on π^{−1}(x) for each x ∈ U_{i}, because q^{−1}(x) ≅ 𝕂 and Γ meets q^{−1}(x) at only one point transversally. Therefore the morphism q^{−1}(U_{i}) ∋ y ↦ (q(y), F_{i}(y)) ∈ U_{i} × 𝕂 is bijective, so it is an isomorphism by Zariski’s Main Theorem. Thus we have showed that q : Y → X is a line bundle. Now Lemma 2 concludes the proof.

4 Final remarks

We shall begin with

Question 1. Is it true that an affine variety with a non-uniruled component at infinity has the cancellation property?

By Theorem 1 the answer is affirmative if we add the unirationality assumption, as well as for any line bundle on a non-uniruled variety.

It is a good place to mention our result from [5] which is connected with the stable equivalence problem (see also [12])

Let X be a smooth affine variety and let H be a non-uniruled hypersurface in X. Suppose that f : X × 𝕂^{m} → Y × 𝕂^{m} is an isomorphism such that f(H × 𝕂^{m}) = H′ × 𝕂^{m}, where H′ is a hypersurface in a variety Y. Then for each x ∈ X there exists y ∈ Y such that f({x} × 𝕂^{m}) = {y} × 𝕂^{m}.

Question 2. Suppose that an affine variety X has the strong cancellation property. Does it follow that X × 𝕂 has the cancellation property?

Again, Theorem 1 gives the affirmative answer for some varieties. This question was already considered by Asanuma in [2], where he showed that if char 𝕂 > 0 then the answer is no. His example was a rational curve with the coordinate ring 𝕂[T^{n}, T^{n+1}], where n > 1 (this is also a counterexample to Problem 1).

On the other hand if char 𝕂 = 0 we have the following

Corollary 1. If X and Y are affine curves then the surface X × Y has the cancellation property.

Proof. The hardest case, when X ≅ Y ≅ 𝕂 follows from [6] and [15].

If X is non-isomorphic to 𝕂 then X \ Sing X is non-𝕂-uniruled. (This is a consequence of the following arguments: (1) every smooth affine and 𝕂-uniruled curve is isomorphic to 𝕂; (2) if C is an affine curve then every non-constant morphism 𝕂 → C is finite and hence surjective.) Thus, by Remark 1, X × 𝕂 has the cancellation property.

The case when neither X ≅ 𝕂 nor Y ≅ 𝕂 again follows from Remark 1, since the set X × Y \ Sing(X × Y) = [(Reg X) × Y] ∩ [X × Reg Y)] is non-K-uniruled. This concludes the proof.

We give one more result a proof of which is left as an exercise.

Proposition 2. Theorem 1 remains true for any affine variety X such that every dominant morphism X → X is birational (for example, X satisfies this condition if it is of hyperbolic type, i.e., \overline{κ}(X) = dim X; see [7, p. 335]).

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Institute of Mathematics Jagiellonian University Reymonta 4 30-059 Kraków, Poland E-mail: Robert.Drylo@im.uj.edu.pl