Fields Institute Communications Volume 00, 0000

V-rational fields

Jack Ohm

Department of Mathematics, Louisiana State University, Baton Rouge, Louisiana 70803

                                  jwho@earthlink.net

Abstract. Given a field extension K/k and a collection V of valuations of k, the field K is said to be V-rational if it is the quotient field of a UFD D such that every unit of D is also a unit of every element of V and every essential valuation of D contracts to an element of V. A criterion is given for K to not be V-rational when K/k is an algebraic extension of degree 2.

1 Introduction

Suppose we are given a rational function ξ in a variable x with coefficients in a field k_{0}, in factored form; say k_{0} = Q and

ξ = 5(x^{3} − x − 1)(x^{5} − 2)^{7}/(x^{3} + 3).

Question: Is the field extension k_{0}(x, \sqrt{ξ})/k_{0} rational or not? That is, does there exist t ∈ K = k_{0}(x, \sqrt{ξ}) such that K = k_{0}(t)? Claim: K/k_{0} is not rational.

To begin with, we are allowed to throw away repeated factors of ξ in forming K = k_{0}(x, \sqrt{ξ}), so we may replace ξ by the polynomial

f := 5(x^{3} − x − 1)(x^{5} − 2)(x^{3} + 3).

Next, break f into two polynomials, not both of odd degree, say

\begin{matrix}f_{1} = & 5(x^{3} − x − 1)(x^{5} − 2), \\ f_{2} = & x^{3} + 3. \end{matrix}  (1)

Consider then the valuations of k_{0}(x)/k_{0} having odd value at f_{i}:

\begin{matrix}Odd_{k_{0}(x)}(f_{1}) := & \{V_{x^{3}−x−1}, V_{x^{5}−2}\} , \\ Odd_{k_{0}(x)}(f_{2}) := & \{V_{x^{3}+3}, V_{1/x}\} . \end{matrix}  (2)

Here V_{p} denotes the valuation whose ring is k_{0}[x]_{(p)}, and V_{1/x} the valuation whose ring is k_{0}[1/x]_{(1/x)}.

Note that Odd_{k_{0}(x)}(f_{1})∩Odd_{k_{0}(x)}(f_{2}) = ∅. Moreover, f_{1} = (f_{1}/\sqrt{f})^{2}f_{2} implies Odd_{K}(f_{1}) = Odd_{K}(f_{2}), where Odd_{K}(f_{i}) denotes the set of valuations of K/k_{0}

\begin{matrix}c \\ ◯ \end{matrix}0000 American Mathematical Society

1991 Mathematics Subject Classification. Primary 12J10, 14G04, 14H05; Secondary 14H45.

which have odd value at f_{i}. But every element of Odd_{K}(f_{i}) contracts to an element of Odd_{k_{0}(x)}(f_{i}), so we conclude that Odd_{K}(f_{i}) = ∅.

Let us detour for a moment to look at the valuations of the rational function field k_{0}(t)/k_{0}: they are just the W_{p}, p an irreducible polynomial in k_{0}[t], and W_{1/t}. Also, every element ξ ∈ k_{0}(t) \ k_{0} has a unique decomposition of the form

ξ = cp_{1}^{n_{1}} ⋯ p_{r}^{n_{r}},

where 0 ≠ c ∈ k_{0}, p_{i} ≠ p_{j} are monic irreducible polynomials in k_{0}[t], and n_{i} is a nonzero integer. The valuations of k_{0}(t)/k_{0} having nonzero value at ξ are just the W_{p_{i}}, and also W_{1/t} if deg ξ (:= n_{1} deg p_{1} + ⋯ + n_{r} deg p_{r}) ≠ 0. In addition, note that ξ ∈ k_{0}[t] iff all the n_{i} are > 0, and \sqrt{ξ/c} ∈ k_{0}(t) iff all the n_{i} are even, or equivalently, iff Odd_{k_{0}(t)}(ξ/c) = ∅.

Returning to our argument (with ξ = f_{1}), we see that Odd_{K}(f_{1}) = ∅ implies \sqrt{f_{1}/c} ∈ K if K = k_{0}(t). Since K also equals k_{0}(x, \sqrt{f}), this in turn would imply there exist α and β in k_{0}(x) such that

\sqrt{f_{1}/c} = α + β\sqrt{f},

or equivalently, such that

\{ \begin{matrix}f_{1}/c = & α^{2} + β^{2}f, \\ 0 = & 2αβ. \end{matrix}  (3)

(We now need 2 ≠ 0, i.e., char k_{0} ≠ 2.) If β = 0, then f_{1}/c would be a square in k_{0}(x), and this is not the case since Odd_{k_{0}(x)}(f_{1}) ≠ ∅; while if α = 0, then f_{1}/c = β^{2}f, and this is not the case since Odd_{k_{0}(x)}(f_{1}) ≠ Odd_{k_{0}(x)}(f) because Odd_{k_{0}(x)}(f_{2}) ≠ ∅. Thus, the supposition that K = k_{0}(t) leads to a contradiction, and we conclude that K/k_{0} is not rational.

We propose to make here a detailed analysis of the above argument, and to push it as far as it will go. This results in the valuation theoretic criterion of Theorem 4.3. To illustrate the criterion, we use it to derive a couple of classical results, one of them (Theorem 4.4) involving the non-rationality of function fields k_{0}(x, \sqrt{f})/k_{0} like the above, and the other (Theorem 4.5) involving the analogous non-factorial property of the ring of integers in an imaginary quadratic number field.

This approach to the question of non-rationality is the 1-dimensional case of an argument of Colliot-Thélène and Ojanguren (cf. [3, Appendice] and, especially, [10]), who use the 2-dimensional case of the argument to give a truly elementary example of a unirational, non-rational 3-fold, i.e., of a finitely generated field extension K/k_{0} of tr. deg 3 which is a subextension of a pure transcendental extension but which is not itself pure transcendental. (This is also known as the “Lüroth problem”; see, for example, [8], [9] for some related remarks and references.) The present paper may be regarded as an introduction to the Pisa lecture notes [10] of Ojanguren.

We begin in §2 by introducing the basic valuation-theoretic definitions for a field extension k/k_{0}, with two cases in mind: the function field case, where k = k_{0}(x), and the arithmetic case, where k = Q; and in §3 we interpret these definitions for the case of k = k_{0}(x). Section 4 contains our V-criterion, with applications to the arithmetic case Q(\sqrt{−m}) and the function field case k_{0}(x, \sqrt{f})/k_{0}; and in §5 we discuss the modifications required in characteristic 2. Section 6 places the problem in a quadratic form setting and sketches the function field generalization to the 2-dimensional case, where the field k_{0}(x, \sqrt{f}) is replaced by a field k_{0}(x, y)(ϕ), ϕ a conic with coefficients in k_{0}(x, y).

1.1. Notation We use V for both a valuation of a field k and its ring, and we shall use valuation and valuation ring interchangably. The value group of V will be denoted G_{V} and the residue field k_{V}. Moreover, we shall use ^{∗} to denote image under the residue homomorphism V → k_{V}, i.e., if ξ ∈ V, ξ^{∗} is the image of ξ in k_{V}. Further notation: Z = integers; Q = rationals; \overline{k_{0}} = algebraic closure of k_{0}; DVR abbreviates “discrete, rank 1, valuation ring”; U(D) denotes the units of the domain D.

2 Basic definitions

Let k be a field and 0 ≠ ξ ∈ k, and let V denote a valuation of k and V a collection of valuations of k.

We begin by introducing some valuation-theoretic definitions.

2.1 Odd and even We say that V (ξ) is even, or that V is even at ξ, if V (ξ) ∈ 2G_{V}, i.e., if there exists η ∈ k and a unit u of V such that ξ = uη^{2}. If V is not even at ξ, we say that V is odd at ξ.

If V is odd at ξ_{1} and even at ξ_{2}, then V is odd at ξ_{1}ξ_{2}. If V is a DVR, i.e., if G_{V} = Z, and V is odd at ξ_{1} and ξ_{2}, then V is even at ξ_{1}ξ_{2}, but this is false for an arbitrary valuation V.

Now let

Odd_{V}(ξ) := {valuations V ∈ V | V (ξ) is odd} .

If ξ = ξ_{1}ξ_{2}, then

Odd_{V}(ξ) ⊆ Odd_{V}(ξ_{1}) ∪ Odd_{V}(ξ_{2}),  (4)

and if Odd_{V}(ξ_{1}) ∩ Odd_{V}(ξ_{2}) = ∅, then = holds. Conversely, if V consists of DVRs and = holds in (4), then the intersection is empty.

Thus,

2.2. Proposition Given nonzero elements ξ, ξ_{1} of a field k, let ξ_{2} = ξ/ξ_{1}, and consider the following:

i) Odd_{V}(ξ_{1}) ∩ Odd_{V}(ξ_{2}) = ∅.

ii) Odd_{V}(ξ) = Odd_{V}(ξ_{1}) ∪ Odd_{V}(ξ_{2}).
iii) Odd_{V}(ξ_{1}) ⊆ Odd_{V}(ξ).

Then i) ⇒ ii) ⇒ iii); and if V consists of DVRs, then also iii) ⇒ i).

2.3 V-factors Let ξ_{1} and ξ be nonzero elements of a field k. Then ξ_{1} will be called a V-factor of ξ if

Odd_{V}(ξ_{1}) ∩ Odd_{V}(ξ/ξ_{1}) = ∅,  (5)

and a proper V-factor of ξ if, in addition, ∅ ≠ Odd_{V}(ξ_{1}) ≠ Odd_{V}(ξ).

If V consists of DVRs (as it does in our applications), this definition takes the following simple form:

2.4. Corollary Suppose V is a collection of DVRs of k. Then ξ_{1} is a V-factor (resp., proper V-factor) of ξ iff Odd_{V}(ξ_{1}) is a subset (resp., nonempty proper subset) of Odd_{V}(ξ).

Multiplication of ξ or ξ_{1} by a nonzero square does not affect the V-factor property. Moreover, if ξ_{1} is a V-factor (resp., proper V-factor) of ξ, then ξ_{2} = ξ/ξ_{1} is also a V-factor (resp., proper V- factor), and Odd_{V}(ξ) is the disjoint union of Odd_{V}(ξ_{1}) and Odd_{V}(ξ_{2}).

If K is an overfield of k and V is even at ξ, then any extension of V to a valuation of K is also even at ξ. Said differently, if V′ is a valuation of K which is odd at ξ ∈ k, then the contraction of V′ to k is also odd at ξ.

2.5. Proposition Let V be a collection of valuations of a field k, let ξ be a nonzero element of k, let K = k(\sqrt{ξ}), and let V(K) be the set of extensions of elements of V to K. If ξ_{1} is a proper V-factor of ξ, then Odd_{V(K)}(ξ_{1}) = ∅.

Proof We have ξ_{1} = (ξ_{1}/\sqrt{ξ})^{2}ξ_{2}, where ξ_{2} = ξ/ξ_{1}; therefore
Odd_{V(K)}(ξ_{1}) = Odd_{V(K)}(ξ_{2}).

But an element of Odd_{V(K)}(ξ_{i}) would contract to an element of Odd_{V}(ξ_{i}), hence to an element of Odd_{V}(ξ_{1})∩Odd_{V}(ξ_{2}). Since this intersection is empty, Odd_{V(K)}(ξ_{i}) = ∅.

For the next proposition, note that Odd_{V}(ξ_{1}) ≠ ∅ implies \sqrt{ξ_{1}} ∉ k; and therefore the condition k(\sqrt{ξ_{1}}) ≠ k(\sqrt{ξ}) is equivalent to \sqrt{ξ}_{1} ∉ k(\sqrt{ξ}).

2.6. Proposition Suppose char k ≠ 2 and ξ, ξ_{1} are nonzero elements of k. Then

i) k(\sqrt{ξ_{1}}) = k(\sqrt{ξ}) (if and) only if there exists β ∈ k such that ξ_{1} = β^{2}ξ.

ii) If ∅ ≠ Odd_{V}(ξ_{1}) ≠ Odd_{V}(ξ), then k(\sqrt{ξ_{1}}) ≠ k(\sqrt{ξ}) (equivalently, \sqrt{ξ_{1}} ∉ k(\sqrt{ξ})).

Proof Suppose k(\sqrt{ξ_{1}}) = k(\sqrt{ξ}). If k(\sqrt{ξ_{1}}) = k(\sqrt{ξ}) = k, then β = \sqrt{ξ_{1}/ξ} works; so we may assume \sqrt{ξ} and \sqrt{ξ_{1}} are not in k. Since \sqrt{ξ_{1}} ∈ k(\sqrt{ξ}) ≠ k, there exist α, β ∈ k such that \sqrt{ξ_{1}} = α + β\sqrt{ξ}, or equivalently, such that

\{ \begin{matrix}ξ_{1} & = & α^{2} + β^{2}ξ \\ 0 & = & 2αβ. \end{matrix}  (6)

Since 2 ≠ 0 and ξ_{1} is not a square in k, α = 0; hence ξ_{1} = β^{2}ξ.

Now suppose Odd_{V}(ξ_{1}) ≠ Odd_{V}(ξ). Then ξ_{1} ≠ β^{2}ξ, so by the previous assertion k(\sqrt{ξ_{1}}) ≠ k(\sqrt{ξ}).

2.7. Corollary (to 2.6-(ii)) Suppose char k ≠ 2 and 0 ≠ ξ ∈ k. Then for every proper V-factor ξ_{1} of ξ, \sqrt{ξ_{1}} ∉ k(\sqrt{ξ}).

We have in mind two particular applications of these concepts: one is to the case that V is the set of all valuations of Q, and the other to the case that V is the set of all valuations of k_{0}(x)/k_{0}. The interpretation of these definitions in the case of Q is straight-forward, and in the case of k_{0}(x) a bit less obvious because of the extra valuation V_{1/x} “at infinity”. We take up the latter case next.

3 Example: k = k_{0}(x)

Let x be transcendental over the field k_{0}, let V be the set of all valuations of k_{0}(x)/k_{0}, and let P be the set of all monic irreducible polynomials of k_{0}[x].

As noted in § 1,

V = {V_{p} | p ∈ P } ∪ {V_{1/x}},

where V_{p} := k_{0}[x]_{(p)}, and V_{1/x} := k_{0}[1/x]_{(1/x)}. Moreover, every ξ ∈ k_{0}(x) \ k_{0} has a unique expression

ξ = cp_{1}^{n_{1}} ⋯ p_{r}^{n_{r}},  (7)

with 0 ≠ c ∈ k_{0}; p_{i} monic irreducible in k_{0}[x]; p_{i} ≠ p_{j} for i ≠ j; and 0 ≠ n_{i} ∈ Z. We say that ξ is without repeated factors in k_{0}[x] if n_{i} = ±1 for i = 1, …, r.

Note that ξ ∈ k_{0}[x] \ k_{0} iff the exponents n_{i} are all > 0; and \sqrt{ξ/c} ∈ k_{0}(x) iff n_{1}, …, n_{r} are all even.

We define

deg_{x} ξ = n_{1} deg_{x} p_{1} + ⋯ + n_{r} deg_{x} p_{r}.

The valuations of k_{0}(x)/k_{0} which are nonzero at ξ are V_{p_{1}}, …, V_{p_{r}}, and possibly V_{1/x}, where V_{1/x}(ξ) = − deg_{x} ξ. Therefore,

Odd_{V}(ξ) = \{ \begin{matrix}\{V_{p_{i}} | n_{i} is odd\} & if deg_{x} ξ is even, \\ \{V_{p_{i}} | n_{i} is odd\} ∪ \{V_{1/x}\} & if deg_{x} ξ is odd. \end{matrix}  (8)

Let f ∈ k_{0}[x] \ k_{0} be of degree n and without repeated factors, i.e., such that all the exponents in (7) are 1. We define

deg Odd_{V}(f) = ∑{[k_{V} : k_{0}] | V ∈ Odd_{V}(f)}.

Since deg_{x} p_{i} = [k_{V_{p_{i}}} : k_{0}] and [k_{V_{1/x}} : k_{0}] = 1,

deg Odd_{V}(f) = \{ \begin{matrix}n & if n is even, \\ n + 1 & if n is odd. \end{matrix}  (9)

For elements ξ_{1}, ξ_{2} in k_{0}(x) \ k_{0}, Odd_{V}(ξ_{1}) ∩ Odd_{V}(ξ_{2}) = ∅ iff no monic irreducible p occurs with odd exponent in both of their k_{0}[x]-representations (7) and not both ξ_{1} and ξ_{2} have odd degree in x. Thus,

3.1. Proposition Let ξ, ξ_{1} ∈ k_{0}(x) \ k_{0}, and let ξ_{2} = ξ/ξ_{1}. Then ξ_{1} is a proper V-factor of ξ iff each ξ_{i} has at least one prime factor with odd exponent in its irreducible k_{0}[x]-representation (7), no p occurs with odd exponent in both irreducible representations, and not both ξ_{1} and ξ_{2} have odd degree in x.

3.2. Corollary Suppose f_{1}, f in k_{0}[x] \ k_{0} and f is without repeated factors in k_{0}[x]. If deg f is odd, then f_{1} is a proper V-factor of f iff f_{1} is a proper factor of f in k_{0}[x]. If deg f is even, then f_{1} is a proper V-factor of f iff f_{1} is a proper factor of f in k_{0}[x] and deg f_{1} is even.

In particular, every proper factor f_{1} of f such that deg_{x} f_{1} is even is a proper V-factor.

3.3. Corollary For f ∈ k_{0}[x] \ k_{0}, the following are equivalent:

i) f has a proper V-factor;

ii) f has a proper factor of even degree in k_{0}[x];

iii) either f has exactly 2 irreducible factors in k_{0}[x] and one of these has even degree, or f has at least 3 irreducible factors in k_{0}[x].

4 V-rational

Fix a field extension K/k and a collection V of valuations of k.

Let D be a UFD with quotient field K, and let P = {p_{i}} be a complete set of representatives for the irreducible elements of D. Then every nonzero ξ ∈ K has a unique representation

ξ = c ∏_{p_{i}∈P} p_{i}^{n_{i}},  (10)

where c is a unit of D and the n_{i} are integers, with at most finitely many n_{i} nonzero. Note that \sqrt{ξ/c} ∈ K iff all n_{i} are even iff Odd_{V(K)}(ξ/c) = ∅, where V(K) = {D_{(p)} | p ∈ P }.

4.1 V-rational We shall say that K is V-rational if there exists a UFD D having quotient field K such that i) every D_{(p)}, p an irreducible element of D, contracts to an element of V, and ii) U(D) ⊆ U(V) (:= ∩{U(V) | V ∈ V}).

Note that ii) may be replaced by iii) U(D) = U(V), and ii) implies U(D) ⊆ k.

4.2. Examples i) Let k = k_{0}(x_{1}, …, x_{n}), with the x_{i} algebraically independent over k_{0}, and let K be a finite algebraic extension of k; in other words, K/k_{0} is a function field of dimension n. If V is the set of all valuations (or the set of all DVRs) of k/k_{0}, then K is V-rational iff there exists a UFD having quotient field K and units k_{0}^{×}. In particular, K is not V-rational implies K/k_{0} is not rational (i.e., pure transcendental).

ii) Let V be the set of all valuations of the rationals Q, let m be a square-free integer > 0, and let \overline{Z} denote the integral closure of Z in the imaginary quadratic number field Q(\sqrt{−m}). Since \overline{Z} is a Dedekind domain with torsion class group, every proper overring of \overline{Z} in Q(\sqrt{−m}) is a quotient ring of \overline{Z} w.r.t. a multiplicative system (cf. [4]), hence has strictly larger unit group than ±1 = U(V). Therefore, Q(\sqrt{−m}) is V-rational implies \overline{Z} is itself a UFD. The converse holds if m ≠ 1 or 3; for it is known that the units of \overline{Z} are ±1, except if m = 1 or 3 (cf. [11, p. 62, Proposition 1]). Hence, Q(\sqrt{−m}), with m ≠ 1, 3, is V-rational iff \overline{Z} is a UFD.

We can now state our criterion for not being V-rational.

4.3. Theorem Let V be a collection of valuations of a field k of characteristic ≠ 2, and let 0 ≠ ξ ∈ k. If there exists a proper V-factor ξ_{1} of ξ in k, then k(\sqrt{ξ}) is not V-rational.

Proof Let K = k(\sqrt{ξ}), and suppose ξ_{1} is a proper V-factor of ξ. By 2.5, we have Odd_{V(K)}(ξ_{1}) = ∅, where V(K) denotes the set of all extensions of elements of V to K.

Suppose K is V-rational. Then by definition ξ_{1} has a K-representation (10) with c ∈ U(V). Since Odd_{V(K)}(ξ_{1}) = ∅, all the exponents n_{i} must be even; hence \sqrt{ξ_{1}/c} ∈ K. But c has value 0 at every valuation of V, so Odd_{V}(ξ_{1}) = Odd_{V}(ξ_{1}/c). Therefore ξ_{1}/c is also a proper V-factor of ξ; and then, by 2.7, \sqrt{ξ_{1}/c} ∉ K, a contradiction.

Our definition of V-rational was designed to include the following applications of 4.3:

4.4. Theorem (Application to Q(\sqrt{−m})) Let m ∈ Z be a positive integer without repeated prime factors. If m has at least two distinct prime factors in Z, then the integral closure of Z in Q(\sqrt{−m}) is not a UFD.

Proof Let V be the set of all valuations of Q. Since a composite integer has a proper V-factor in Q, we conclude from 4.3 that Q(\sqrt{−m}) is not V-rational. But then by Example (ii) of 4.2, the integral closure of Z in Q(\sqrt{−m}) is not a UFD.

4.5. Theorem (Application to k_{0}(x, \sqrt{f})/k_{0}) Let f ∈ k_{0}[x] \ k_{0} be without repeated factors in k_{0}[x], and assume char k_{0} ≠ 2. If f has a proper factor f_{1} in k_{0}[x] of even degree, then the extension k_{0}(x, \sqrt{f})/k_{0} is not rational.

Proof Let V be the set of all valuations of k_{0}(x)/k_{0}. By 3.2 every proper factor of f of even degree is a proper V-factor, so by 4.3 K is not V-rational. Thus, there does not exist a UFD having quotient field K and units U(V) = k_{0}^{×}, and in particular, there does not exist t ∈ K such that k_{0}(t) = K.

4.6. Corollary Let \overline{k_{0}} be the algebraic closure of k_{0}, and assume char k_{0} ≠ 2. If f ∈ k_{0}[x] \ k_{0} has at least 3 distinct (linear) factors in the ring \overline{k_{0}}[x], then k_{0}(x, \sqrt{f})/k_{0} is not rational. In particular, if f is of degree ≥ 3 and separable (i.e., f has no repeated linear factors in \overline{k_{0}}[x]), then k_{0}(x, \sqrt{f})/k_{0} is not rational.

Proof If k_{0}(x, \sqrt{f}) = k_{0}(t), then \overline{k_{0}}(x, \sqrt{f}) = \overline{k_{0}}(t), so the corollary follows from 4.5 applied to \overline{k_{0}}.

Remarks (on Corollary 4.6) For f as in 4.6, what can one say if deg f = 1 or 2, or if f is not separable? The case of deg f = 1 or 2 is disposed of as follows. If deg f = 1, then k_{0}(x, \sqrt{f})/k_{0} is rational. If deg f = 2, then k_{0}(x, \sqrt{f})/k_{0} is the function field of an absolutely irreducible conic Y^{2} = aX^{2} + bX + c (assuming char k_{0} ≠ 2), and it is easily seen (e.g., [2, p. 304, Thm. 8] or [9, p. 338, Thm. 1.4]) that k_{0}(x, \sqrt{f})/k_{0} is rational iff the (affine) conic has a point with coordinates in k_{0}.

As for the separability assumption, the corollary holds without it, but I know of no way of treating the inseparable case without appealing to the Riemann-Roch theorem. (The corollary does not follow from Hurwitz’s genus formula, which only applies in case k_{0} is perfect.) The difficulty arises from the fact that an inseparable f picks up repeated factors in passing to \overline{k_{0}} and thus \overline{k_{0}}(x, \sqrt{f})/\overline{k_{0}} may be rational even though k_{0}(x, \sqrt{f})/k_{0} is not. The particular cases that cause a problem are when f is either irreducible and purely inseparable or when f is a product of 2 purely inseparable irreducible polynomials of odd degree.

5 Revisit of §4 in characteristic 2

The characteristic ≠ 2 hypothesis in 4.3 is needed for the last line of the proof, where the property \sqrt{ξ_{1}/c} ∉ k(\sqrt{ξ}) is obtained by applying 2.7, which requires characteristic ≠ 2. We can delete the characteristic 2 hypothesis from 4.3 if we replace it by the hypothesis that \sqrt{ξ_{1}} ∉ k(\sqrt{ξ}) for every proper V-factor ξ_{1} of ξ. Thus, consider the following definition:

5.1 V-minimal Let V be a collection of valuations of a field k. A nonzero element ξ of k will be called V-minimal if for every proper V-factor ξ_{1} of ξ, k(\sqrt{ξ_{1}}) ≠ k(\sqrt{ξ}) (equivalently, \sqrt{ξ_{1}} ∉ k(\sqrt{ξ})).

If char k ≠ 2, then by 2.7 every nonzero element of k is V-minimal. Also, the V-minimal property is unaffected if ξ is multiplied by a nonzero square from k. Now Theorem 4.3 becomes:

5.2. Theorem (4.3, arbitrary characteristic) Let V be a collection of valuations of a field k, and let 0 ≠ ξ ∈ k be V-minimal. If there exists a proper V-factor ξ_{1} of ξ in k, then k(\sqrt{ξ}) is not V-rational.

Similarly, the “char k_{0} ≠ 2” hypothesis in 4.5 and 4.6 may be replaced by “f is V-minimal”, where V is the set of all valuations of k_{0}(x)/k_{0}.

In the next two propositions we give an idea of what the V-minimal elements may look like in the case that k = k_{0}(x).

5.3. Proposition Let V be the set of all valuations of k_{0}(x)/k_{0} and f ∈ k_{0}[x]\ k_{0} be without repeated factors in k_{0}[x]. If deg f is odd (resp., even), then f is Vminimal iff for every proper factor (resp., every proper factor of even degree) f_{1} of f in k_{0}[x], k_{0}(x, \sqrt{f}) ≠ k_{0}(x, \sqrt{f_{1}})).

Proof Apply 3.2.

For example, an irreducible f is V-minimal.

In view of 5.3 the next proposition, with, say, k_{0} a perfect field of characteristic 2, shows that an f ∈ k_{0}[x] \ k_{0} without repeated factors and of the form f = xg with deg g an even integer > 0 cannot be V-minimal.

5.4. Proposition Suppose char k_{0} = 2 and f ∈ k_{0}[x] \ k_{0} is without repeated factors in k_{0}[x]. If every coefficient of f is a square in k_{0}, then k_{0}(x, \sqrt{f}) = k_{0}(\sqrt{x}) (hence k_{0}(x, \sqrt{f})/k_{0} is rational).

Proof We have

f = a + xb,

where a, b ∈ k_{0}[x] have only terms of even degree. Since the coefficients of a and b are squares in k_{0} and char k_{0} = 2, we can write a = a_{0}^{2}, b = b_{0}^{2}, (a_{0}, b_{0} ∈ k_{0}[x]). Moreover b_{0} ≠ 0 since f is assumed to be without repeated factors in k_{0}[x]. Therefore

((\sqrt{f} − a_{0})/b_{0})^{2} = x,

so k_{0}(x, \sqrt{f}) = k_{0}(\sqrt{x}).

5.5. Remark If char k_{0} ≠ 2, and f ∈ k_{0}[x] \ k_{0} is without repeated factors in k_{0}[x], then it follows from 2.6 and unique factorization in k_{0}(x) that k_{0} is algebraically closed in k_{0}(x, \sqrt{f}). This fails in characteristic 2: for example, if f = 1 + ax^{2} with \sqrt{a} ∉ k_{0}, then k_{0}(x, \sqrt{f}) = k_{0}(x, \sqrt{a}). In fact, Lang-Tate prove in [7, p. 622] that if f = ax^{2} + b and k_{0}(x, \sqrt{f})/k_{0} is not rational, then k_{0} is algebraically closed in k_{0}(x, \sqrt{f}) only if [k_{0}(\sqrt{a}, \sqrt{b}) : k_{0}] = 4.

6 Function fields of quadratic forms

Fix a field extension L/k_{0} of characteristic ≠ 2. Our fields will be assumed to be extensions of k_{0}.

In order to explain how our argument is related to an elementary case of the non-rationality criterion of Colliot-Thélène and Ojanguren, we must first give a sketch of their argument (cf. [3], [10]). If D is a field or a subring of a field, by a D-form ψ we shall mean a nonsingular quadratic form with coefficients in D. Recall that nonsingular means that the determinant of the matrix of ψ is a unit of D. We denote isomorphism of D-forms by ≃_{D} and Witt equivalence by ≡_{D}. If D is a field or a polynomial ring in one variable over a field, it is known that every D-form is equivalent over D to a diagonal D-form, denoted ⟨a_{1}, …, a_{n}⟩; and the underlying nonsingular assumption implies that the a_{i} are units of D. (References for quadratic forms: [6], [12].) The function field, denoted k_{0}(ψ), of the form ψ = ⟨a_{1}, …, a_{n}⟩, is by definition the quotient field of k_{0}[X_{1}, …, X_{n}]/(a_{1} + a_{2}X_{2}^{2} + ⋯ + a_{n}X_{n}^{2}). (Note: If n > 2, a_{1} + a_{2}X_{2}^{2} + … + a_{n}X_{n}^{2} is automatically irreducible; if n = 2, one must assume irreducibility. Moreover, k_{0}(ψ) is independent of which indeterminate is used to dehomogenize the form.)

6.1. Definition Let (V, π) be a DVR of L/k_{0}, where π denotes a fixed generator of the maximal ideal of V, and let L_{V} denote the residue field of V. Let ψ = ⟨g_{1}, …, g_{n}⟩ := g_{1}X_{1}^{2} + ⋯ + g_{n}X_{n}^{2} (g_{i} ∈ L) be a diagonal L-form, and suppose g_{i_{1}}, …, g_{i_{s}}, i_{1} < ⋯ < i_{s}, are those g_{i} of odd value. Each element g_{i} has a unique expression of the form g_{i} = u_{i}π^{V (g_{i})}, where u_{i} is a unit of V. Let \overline{u}_{i} denote the image of u_{i} in L_{V}. The reduction ∂_{V} (ψ) of ψ at (V, π) is defined by

∂_{V} (ψ) = 0 if V (g_{i}) is even for all i = 1, …, n,
= ⟨\overline{u}_{i_{1}}, …, \overline{u}_{i_{s}}⟩ otherwise.

(11)

The map ∂_{V} can be shown to preserve Witt equivalence, and therefore defines a homomorphism from the Witt group of L-forms into the Witt group of L_{V} -forms. Moreover, the kernel of this homomorphism is independent of the choice of π.

6.2. Definition Let W be the set of all DVRs of L/k_{0} and ψ be a diagonal L-form. We define

Ram_{L}ψ = {V ∈ W | ∂_{V} (ψ) ≢_{L_{V}} 0}.

Furthermore, we say that ψ is unramified (w.r.t. L/k_{0}) if ψ ≢_{L} 0 and Ram_{L}ψ = ∅.

6.3. Remark Since ∂_{V} preserves Witt equivalence, if ψ and ψ′ are Witt equivalent L-forms, then Ram_{L}ψ = Ram_{L}ψ′. Note also that since 1-dimensional forms are not Witt equivalent to 0, for an L-form of the shape ψ = ⟨1, −g⟩, Ram_{L}ψ = {V ∈ W | V (g) is odd}, i.e., Ram_{L}ψ = Odd_{L}g.

6.4. Theorem Suppose L/k_{0} is a rational field extension. If ψ is an unramified L-form, then there exists a k_{0}-form ψ_{0} such that ψ ≡_{L} ψ_{0}.

6.5. Corollary (Non-rationality criterion) Suppose L/k_{0} is a field extension and k_{0} is algebraically closed. If there exists an even dimensional unramified L-form ψ, then L/k_{0} is not rational.

Proof If L/k_{0} is rational, then by 6.4 there exists a k_{0}-form ψ_{0} such that ψ ≡_{L} ψ_{0}, and such a ψ_{0} is also even dimensional by the definition of ≡_{L}. But k_{0} is algebraically closed implies every even dimensional k_{0}-form is Witt equivalent to 0. Thus, ψ ≡_{L} ψ_{0} ≡_{k_{0}} 0, a contradiction to ψ ≢_{L} 0.

For the proof of 6.4, it suffices to prove the following descent lemma.

6.6. Lemma Suppose there exists a field L_{1} such that L = L_{1}(t), t algebraically independent over L_{1}. If there exists an unramified L-form ψ, then there exists an unramified L_{1}-form ψ_{1} such that ψ ≡_{L} ψ_{1}.

6.7. Definition If ψ is an L-form and D a subring of L, we shall say that ψ is defined over D if there exists a D-form ψ_{0} such that ψ ≡_{L} ψ_{0}.

Outline of proof of 6.6 If V is a DVR of L/k_{0}, then ∂_{V} (ψ) ≡_{L_{V}} 0 iff ψ is defined over V. Thus, Ram_{L}ψ = ∅ implies ψ is defined over every DVR of L/k_{0}. In particular, ψ is defined over every localization of L_{1}[t], which, in turn, implies (the hard part) that ψ is defined over L_{1}[t].

It is easily seen that a (nonsingular) L_{1}[t]-form is equivalent over L_{1}[t] to a diagonal L_{1}[t]-form ψ_{1}, and such a diagonal form has entries which are units of L_{1}[t], hence which are in L_{1}. Thus, ψ_{1} is an L_{1}-form and ψ ≡_{L} ψ_{1}.

Finally, Ram_{L_{1}}ψ_{1} = ∅ because every DVR of L_{1}/k_{0} may be extended to a DVR of L/k_{0} in such a way that the residue extension is simple transcendental, and such an extension does not disturb Ram.

Next, following [10], we want to describe how this criterion may be used to construct a unirational, non-rational function field. Let k be a field and f, g_{1}, g_{2} be three nonzero elements of k, and let g = g_{1}g_{2}. Consider then the three linked Pfister k-forms

\{ \begin{matrix}ϕ & = & ⟨1, −f, −g, fg⟩ & = & ⟨1, −f⟩ ⊗ ⟨1, −g⟩, \\ ϕ_{1} & = & ⟨1, −f, −g_{1}, fg_{1}⟩ & = & ⟨1, −f⟩ ⊗ ⟨1, −g_{1}⟩, \\ ϕ_{2} & = & ⟨1, −f, −g_{2}, fg_{2}⟩ & = & ⟨1, −f⟩ ⊗ ⟨1, −g_{2}⟩. \end{matrix}

Let ϕ^{−} = ⟨1, −f, −g⟩ and define

K = k(ϕ^{−}) := quotient field of k[Z, W]/(1 − fZ^{2} − gW^{2}) = k(z, \sqrt{(1 − fz^{2})/g}).

6.8. Lemma If Ram_{k}ϕ_{1} and Ram_{k}ϕ_{2} are non-empty, disjoint sets, then

ϕ_{1} ≡_{K} ϕ_{2} ≢_{K} 0, and Ram_{K}ϕ_{i} = ∅.

Proof First, ϕ_{1} ≡_{K} ϕ_{2}, for

ϕ_{1} − ϕ_{2} ≡_{k} ⟨−g_{1}, fg_{1}⟩ − ⟨−g_{2}, fg_{2}⟩ ≡_{k} −g_{1}⟨1, −f, −g_{1}g_{2}, fg_{1}g_{2}⟩ ≡_{K} 0.

Since an element of Ram_{K}ϕ_{i} contracts to an element of Ram_{k}ϕ_{i}, it follows from Ram_{k}ϕ_{1} ∩ Ram_{k}ϕ_{2} = ∅ that Ram_{K}ϕ_{i} = ∅.

It remains to see that ϕ_{i} ≢_{K} 0. If, on the contrary, ϕ_{i} ≡_{K} 0, then also ϕ_{i} ≡_{k(ϕ)} 0 since k(ϕ) is a simple transcendental extension of K = k(ϕ^{−}) (cf. [5, pp. 72–73] or [1, pp. 1681–1682]). Therefore by the Cassels-Pfister subform theorem (cf. [6, p. 262, Theorem 2.8]) and the fact that ϕ and ϕ_{i} are Pfister forms of the same dimension, we have ϕ_{1} ≃_{k} ϕ ≃_{k} ϕ_{2}, which contradicts our hypothesis that Ram_{k}ϕ_{1} ≠ Ram_{k}ϕ_{2}.

6.9. Corollary If Ram_{k}ϕ_{1} and Ram_{k}ϕ_{2} are non-empty, disjoint sets and k_{0} is algebraically closed, then K/k_{0} is not rational.

Proof By 6.8, ϕ_{1} is an (even dimensional) unramified K-form, so 6.5 applies.

6.10. Proposition If k(\sqrt{f})/k_{0} is rational, then K(\sqrt{f})/k_{0} is rational.

Proof Since the k-form ϕ^{−} has a zero over k(\sqrt{f}), k(\sqrt{f})(ϕ^{−})/k(\sqrt{f}) is simple transcendental: k(\sqrt{f})(ϕ^{−}) = k(\sqrt{f})(t). But then

K(\sqrt{f}) = k(ϕ^{−})(\sqrt{f}) = k(\sqrt{f})(ϕ^{−}) = k(\sqrt{f}, t).

In particular, now we shall choose k = k_{0}(x, y) and f = xy. Then it follows from 6.10 that K(\sqrt{f})/k_{0} is rational, because k_{0}(x, y, \sqrt{f}) = k_{0}(x, \sqrt{f}). Thus, to obtain an example of a subrational (= unirational), non-rational 3-fold K/k_{0}, in view of 6.9 it only remains to check that there exist g_{1} and g_{2} in k_{0}[x, y] such that Ram_{k}ϕ_{1} and Ram_{k}ϕ_{2} are non-empty, disjoint sets.

The following choice of g_{1} and g_{2} will work. Pick 2 points on the x-axis and 2 on the y-axis in the (x, y)-plane, and let l_{1}, l_{2}, l_{3}, l_{4} be the 4 linear polynomials in k_{0}[x, y] whose equations l_{i} = 0 define the 4 lines different from the axes determined by these points. Let g_{1} = l_{1}l_{2}l_{3}l_{4}. Similarly, define g_{2} by choosing 4 additional points, two on each axis. Now an analysis of the centers of the DVRs of k/k_{0} on k_{0}[x, y] will show Ram_{k}ϕ_{1} and Ram_{k}ϕ_{2} are non-empty, disjoint sets.

For example, if

g_{1} = l_{1}l_{2}l_{3}l_{4} = (x + y − 1)(x + y − 2)(y + 2x − 2)(2y + x − 2),

then every element of Ram_{k}ϕ_{1} contains k_{0}[x, y] and is centered on k_{0}[x, y] at one of the 8 prime ideals

(l_{1}), (l_{2}), (l_{3}), (l_{4}), (x, y − 1), (x, y − 2), (y, x − 1), (y, x − 2).

Moreover, each of these prime ideals is the center of at least one element of Ram_{k}ϕ_{1}.

6.11. Remark The question of whether a subfield of a pure transcendental extension is itself pure transcendental has a long and interesting history. For k_{0} algebraically closed of characteristic 0 the answer is affirmative if n = 1 or 2, and the first counterexamples for n = 3 were discovered independently around 1972 by Artin - Mumford, Clemens - Griffiths, and Iskovskih - Manin. The Artin - Mumford example is the basis for the Colliot-Thélène - Ojanguren approach discussed here. See, for example, [9] for further references and remarks.

Finally, let us return to the example worked out in the introduction and see how it fits in to this quadratic form setting. Let k = k_{0}(x); suppose f = f_{1}f_{2}, with f, f_{1}, f_{2} ∈ k_{0}[x]; and consider the three linked Pfister k-forms

\{ \begin{matrix}ϕ & = & ⟨1, −f⟩, \\ ϕ_{1} & = & ⟨1, −f_{1}⟩, \\ ϕ_{2} & = & ⟨1, −f_{2}⟩. \end{matrix}

Further, let K = k(ϕ) := k_{0}(x)[Y]/(1 − fY^{2}) = k_{0}(x, \sqrt{f}).

Suppose we have chosen f_{1} and f_{2} such that Odd_{k}f_{1} and Odd_{k}f_{2} are disjoint nonempty sets. As noted in 6.3, Odd_{k}f_{i} = Ram_{k}ϕ_{i}, so then Ram_{k}ϕ_{1} and Ram_{k}ϕ_{2} are disjoint nonempty sets. Moreover, f_{1} = (f_{1}/\sqrt{f})^{2}f_{2} implies ϕ_{1} ≡_{K} ϕ_{2}, and therefore Ram_{K}ϕ_{1} = Ram_{K}ϕ_{2}. Since an element of Ram_{K}ϕ_{i} contracts to an element of Ram_{k}ϕ_{i}, we conclude Ram_{K}ϕ_{1} = Ram_{K}ϕ_{2} = ∅. But if K/k_{0} is rational, then Ram_{K}ϕ_{i} = ∅ iff \sqrt{f_{i}/c_{i}} ∈ K (for some c_{i} ∈ k_{0}) iff ϕ_{i} ≃_{K} ⟨1, −c_{i}⟩. Then, (in contrast to the 2-dimensional case, which requires that k_{0} be algebraically closed) a direct computation shows that the condition \sqrt{f_{i}/c_{i}} ∈ K contradicts the choice of the f_{i}.

Acknowledgement: I would like to thank W. Heinzer, J.W. Hoffman, A. Nobile, and the referee for helpful comments.

References

[1] H. Ahmad and J. Ohm. Funct fields of Pfister neighbors. J. of Algebra, 178:653–664, 1995.

[2] C. Chevalley. Algebraic functions of one variable. Amer. Math. Soc. 1951.

[3] J.-L. Colliot-Thélène and M. Ojanguren. Variétés unirationnelles non rationnelles: au-delà de l’exemple d’Artin et Mumford. Inventiones math., 97:141–158, 1989.

[4] R. Gilmer and J. Ohm. Integral domains with quotient overrings. Math. Annalen, 153:97–103, 1964.

[5] M. Knebusch. Generic splitting of quadratic forms I. Proc. London Math Soc., 33:65–93, 1976.

[6] T.Y. Lam. The algebraic theory of quadratic forms. Benjamin, 1980.

[7] S. Lang and J. Tate. On Chevelley’s proof of Lüroth’s theorem. Proc. Amer. Math. Soc., 3:621–624, 1952.

[8] J. Ohm. On ruled fields. Sém. Théor. Nombres Bordeaux (2), 1:27–49, 1989.

[9] J. Ohm. Function fields of conics, a theorem of Amitsur–MacRae, and a problem of Zariski. In C. Bajaj, editor, Algebraic Geometry and Its Applications, pages 333–363. Springer-Verlag, 1994.

[10] M. Ojanguren. The Witt group and the problem of Lüroth. Universitá di Pisa, 1988.

[11] P. Samuel. Algebraic theory of numbers. Hermann, 1970.

[12] W. Scharlau. Quadratic and Hermitian forms. Springer, 1985.