Fields Institute Communications Volume 00, 0000

A valuation-theoretic approach to non-rationality of the curves Y^{2} = f(X)

Jack Ohm

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

                                  jwho@earthlink.net

Abstract. An approach to non-rationality of the curves Y^{2} = f(X) using only elementary valuation-theoretic arguments is analyzed and compared with the classical approaches.

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)?

The question can be approached through the Riemann-Roch theorem or Hurwitz’s formula, but, for reasons of generalization, I want to decide it using only elementary valuation theory. In particular, I do not want to appeal to the notion of genus.

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)}.

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

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

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} 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.

The same argument proves:

1.1. Theorem Suppose k_{0} is an algebraically closed field of characteristic ≠ 2 and f ∈ k_{0}[x] is without repeated factors in k_{0}[x]. Then k_{0}(x, \sqrt{f})/k_{0} is not rational if (and only if) deg f ≥ 3.

This approach to the question of rationality is a primitive version of an idea of Colliot-Thélène and Ojanguren (cf. [4, Appendice] and, especially, [10]), who use the Witt group in place of etale cohomology to give (in my opinion) the first 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.)

We propose to make here a detailed analysis of the above argument, especially as it pertains to function fields of the form k_{0}(x, \sqrt{f})/k_{0}. It should be emphasized that it is the elementary valuation-theoretic methods rather than the results that are primarily of interest to us; sharper results can be obtained by the more sophisticated counting arguments of Riemann-Roch.

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 non-rationality 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 treats the classical approach to non-rationality via Hurwitz’s genus formula and §8 the classical approach via Riemann-Roch. Some preliminary results needed in §8, especially in characteristic 2, are developed in §7. The final section, §9, is devoted to some remarks on the possible degrees of a polynomial generator f for the function field k_{0}(x, \sqrt{f})/k_{0}.

1.2. Notation Fix a base field k_{0}, and let k be an overfield of k_{0} and x an element of k transcendental over k_{0}. We use V for both a valuation of 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 The non-rationality criterion

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). Also, 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. [5]), 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 non-rationality criterion.

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., [3, 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}. In particular, Theorem 1.1 follows from this remark and 4.6.

As for the separability assumption, the corollary holds without it (see §8), but I know of no way of treating the inseparable case without appealing to the notion of genus. 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 = k_{0}(x, \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 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.

5.2. 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. Now Theorem 4.3 becomes:

5.3. 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 view of 5.2 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}).

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 [6, 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 Ramification and the Hurwitz genus formula

Another approach to the non-rationality of the function field k_{0}(x, \sqrt{f})/k_{0}, is via Hurwitz’s genus formula, which requires a knowledge of the valuations of k_{0}(x)/k_{0} which ramify in K := k_{0}(x, \sqrt{f}).

Let ξ be a non-square in a field k and V be a valuation of k such that V (2) = 0 (i.e., such that the residue field of V has characteristic ≠ 2); let V′ be an extension of V to K := k(\sqrt{ξ}); and let ^{∗} denote image under the residue map of V′. Since [K : k] = 2, at most one of the following three possibilities occurs: either V splits in K (i.e., there exist two valuations of K which contract to V), the residue degree [K^{∗} : k^{∗}] is 2, or V ramifies in K (i.e., the index [G_{V′} : G_{V}] is 2).

6.1. Proposition Let ξ be a non-square in a field k and V be a valuation of k such that V (2) = 0. If V (ξ) is even, then either V splits in K := k(\sqrt{ξ}) or the residue degree of V is 2; and if V (ξ) is odd, then V ramifies in K.

Proof Let V′ be an extension of V to K = k(\sqrt{ξ}), and let y = \sqrt{ξ}, so that y^{2} = ξ.

Case (i): V (ξ) is odd. Since 2V′(y) = V (ξ) and V (ξ) is not even, V′(y) ∉ G_{V}. Therefore V ramifies in K.

Case(ii): V (ξ) = 0. Then V′(y) = 0 also. If y^{∗} ∉ k^{∗}, then [K^{∗} : k^{∗}] = 2. On the other hand, if y^{∗} ∈ k^{∗}, then there exists b ∈ k such that V′(y −b) > 0. But then V′(y + b) = 0, since V′((y − b) + (y + b)) = V′(2y) = 0; hence the automorphism of K/k taking y to −y takes V′ to a valuation ring different from V′, and V splits in K. Thus, V either splits in K or has residue degree 2 if V (ξ) = 0.

Case (iii): V (ξ) is even. Then there exists b ≠ 0 in k such that V (ξ/b^{2}) = 0. Since K = k(\sqrt{ξ}) = k(\sqrt{ξ/b^{2}}), by the previous case V either splits in K or has residue degree 2.

6.2. Corollary Let ξ be an element of k_{0}(x) and V be a set of valuations of k_{0}(x)/k_{0}, and assume char k_{0} ≠ 2. Then

Odd_{V}(ξ) = \{V ∈ V | V ramifies in k_{0}(x, \sqrt{ξ}).

6.3. Remarks i) Suppose f ∈ k_{0}[x] \ k_{0} is without repeated factors in k_{0}[x]. If deg f is odd, then V_{1/x} ramifies in K = k_{0}(x, \sqrt{f}), hence the extension of V_{1/x} to K has residue field k_{0}. In general, if f contains an irreducible factor f_{1}, then V_{f_{1}}(f) = 1 implies V_{f_{1}} ramifies in K, hence its extension to K has residue field k_{0}(α), where α is a root of f_{1}.

ii) The proof of 6.1 shows: If V (ξ) = 0, then V splits in K iff ξ^{∗} is a square in k^{∗}. Moreover, if V splits in K, then the two extensions of V are conjugates over k.

6.4. Theorem (Hurwitz genus formula) Suppose k/k_{0} is a 1-dim function field, K is a finite separable extension of k, and k_{0} is algebraically closed in K. Let γ (resp., Γ) denote the genus of k/k_{0} (resp., K/k_{0}). Then there exists a non-negative integer D such that

Γ − 1 = [K : k] · (γ − 1) + (1/2)D.  (11)

Moreover, if k_{0} is perfect and for every valuation V′ of K/k_{0}, char k_{0} does not divide [G_{V′} : G_{V′∩k}], then

D = ∑_{V′}([G_{V′} : G_{V′∩k}] − 1) · [K_{V′} : k_{0}],

where V′ runs over all valuations of K/k_{0}.

Proof See [12, p. 95], [3, pp. 106 and 69].

6.5. Corollary Let k_{0} be a perfect field of characteristic ≠ 2, let f ∈ k_{0}[x]\k_{0} be of degree n and without repeated factors in k_{0}[x], and let K = k_{0}(x, \sqrt{f}). Then

Γ(K/k_{0}) = \{ \begin{matrix}(n − 2)/2 if n is even, \\ (n − 1)/2 if n is odd. \end{matrix}

Proof We have [K : k] = 2 and k = k_{0}(x), hence the formula (11) becomes

Γ − 1 = 2 · (−1) + (1/2)D,

with

D = ∑_{V}[K_{V} : k_{0}],

where V runs over the valuations of k/k_{0} which ramify in K. In view of 6.2,

D = \{ \begin{matrix}deg f & if deg f is even, \\ deg f + 1 & if deg f is odd. \end{matrix}

Therefore

Γ = \{ \begin{matrix}(n − 2)/2 if n is even, \\ (n − 1)/2 if n is odd. \end{matrix}

6.6. Remark (on the V (2) = 0 hypothesis in 6.1) If V (2) > 0, then it can happen that V (ξ) = 0, yet V ramifies in k(\sqrt{ξ}). For example, let V_{0} = Z_{(2)}; let V be the extension of V_{0} to Q(x) defined by taking the value of a polynomial in x to be the minimum of the values of its coefficients; and let ξ = 2x + 1. If y = \sqrt{ξ} and V′ is an extension of V to K = Q(x, \sqrt{ξ}), then V′(y) = V′(y^{2}) = 0, and

V′((1 − y)(1 + y)) = V (1 − y^{2}) = V (2x) = 1,  (12)
V′((1 − y) + (1 + y)) = V (2) = 1.  (13)

If V does not ramify in K, then (12) implies, say, V′(1 − y) = 1 and V′(1 + y) = 0, which contradicts (13). Thus, V ramifies in k(y).

If V (2) > 0, it can also happen that V does not split in K, and the index and residue degree are both 1. An example of F.K. Schmidt runs as follows: Let k_{0} be a finite field of characteristic 2 and t be an indeterminate, and consider the power series field k_{0}((t)), endowed with the t-adic valuation. A cardinality argument shows that k_{0}((t)) contains an element u transcendental over k_{0}(t). Let k = k_{0}(t)(u^{2}) and K = k(u). Since we are working inside k_{0}((t)), k_{0} is the residue field and Z the value group of both k and K. Moreover, the valuation on k extends uniquely to K since every element of K has a square in k. (The henselian defect of this example is 2; see, e.g., [7] for some remarks and references on the different notions of defect.)

7 Minimal elements

In §8 we shall compute the genus of k_{0}(x, \sqrt{f})/k_{0}, for f an element of k_{0}[x] \ k_{0} without repeated factors in k_{0}[x]. In order to include the characteristic 2 case, we must first introduce the notion of minimal element

7.1. Definition An element f ∈ k_{0}[x] \ k_{0} will be called minimal in k_{0}[x] if for every nonzero f_{1} ∈ k_{0}[x] such that k_{0}(x, \sqrt{f}) = k_{0}(x, \sqrt{f_{1}}), deg f ≤ deg f_{1}.

If f ∈ k_{0}[x] \ k_{0} is minimal in k_{0}[x], then k_{0} is algebraically closed in k_{0}(x, \sqrt{f}); for otherwise there exists a ∈ k_{0} such that k_{0}(x, \sqrt{f}) = k_{0}(x, \sqrt{a}), a contradiction to the minimality of f. Moreover, a minimal f has no repeated factors in k_{0}[x], and by 2.6 the converse holds if char k_{0} ≠ 2:

If char k_{0} ≠ 2, then f is minimal in k_{0}[x] iff f has no repeated factors in k_{0}[x].

It is not always clear whether a given f ∈ k_{0}[x]\k_{0} is minimal or not when char k_{0} = 2. For example, we have seen in 5.4 that if the coefficients of f are squares in k_{0}, then k_{0}(x, \sqrt{f}) = k_{0}(x, \sqrt{x}) = k_{0}(\sqrt{x}); so such an f is minimal iff it has degree 1. For another example, f = 1+ax^{2} is not minimal because k_{0}(x, \sqrt{f}) = k_{0}(x, \sqrt{a}). The following lemma is useful in this respect.

7.2. Lemma Suppose char k_{0} = 2 and f ∈ k_{0}[x]. If deg f is odd and its derivative f′ is without repeated factors in k_{0}[x], then f is minimal.

Proof Suppose f_{1} ∈ k_{0}[x] is such that k_{0}(x, \sqrt{f_{1}}) = k_{0}(x, \sqrt{f}). Then there exist α, 0 ≠ β ∈ k_{0}(x) such that f_{1} = α^{2} + β^{2}f, hence f′_{1} = β^{2}f′. Since f is of odd degree, deg f′ = deg f − 1. Also, deg f′_{1} ≤ deg f_{1} − 1. Therefore it suffices to prove deg f′ ≤ deg f′_{1}. Since f′ is without repeated factors, this follows from unique factorization in k_{0}(x) and the equality f′_{1} = β^{2}f′.

7.3. Example ([6]) The polynomial f = x^{2n+1}+ax+b ∈ k_{0}[x], with \sqrt{a} ∉ k_{0}, is minimal (k_{0} of characteristic 2).

By 7.2, it suffices to show f′ = x^{2n} + a is without repeated factors. Suppose f′ = β^{2}g, β, g ∈ k_{0}[x]. Then (x^{n} + \sqrt{a})^{2} = β^{2}g implies g is a square in k_{0}(\sqrt{a})[x]. Therefore there exist γ, δ ∈ k_{0}[x] such that g = γ^{2} + δ^{2}a, and then x^{n} + \sqrt{a} = β(γ + δ\sqrt{a}). Since \sqrt{a} ∉ k_{0}(x), it follows that 1 = βδ; hence β ∈ k_{0}.

We shall need the following lemma in §8.

7.4. Lemma Let f ∈ k_{0}[x] \ k_{0} be minimal in k_{0}[x], let y = \sqrt{f}, and let \overline{k_{0}[x]} denote the integral closure of k_{0}[x] in K = k_{0}(x, y). Then \overline{k_{0}[x]} = k_{0}[x] + yk_{0}[x].

If char k_{0} ≠ 2, the lemma is a special case of the following:

7.5. Lemma Let D be a UFD with quotient field k, let f be a nonunit of D, let K = k(\sqrt{f}), and let \overline{D} denote the integral closure of D in K. If 2 is a unit of D and f is without repeated factors in D, then \overline{D} = D + D\sqrt{f}.

Proof The inclusion ⊇ is clear, so we must prove ⊆. Let ξ ∈ \overline{D}.

Case i): ξ^{2} ∈ k. Then ξ^{2} ∈ k ∩ \overline{D} = D. Since K = k + k\sqrt{f}, there exist a, b, 0 ≠ c ∈ D such that

ξ = (a/c) + (b/c)\sqrt{f}.  (14)

Moreover, if a = 0, we may choose b, c relatively prime in D. Then

c^{2}ξ^{2} = (a^{2} + b^{2}f) + 2ab\sqrt{f},  (15)

and since ξ^{2} ∈ k, this is equivalent to

\{ \begin{matrix}c^{2}ξ^{2} & = & a^{2} + b^{2}f \\ 0 & = & 2ab. \end{matrix}

Since 2 ≠ 0, either b = 0 or a = 0. If b = 0, then ξ ∈ k ∩ \overline{D} = D and we are done. If, on the other hand, a = 0, then c^{2}ξ^{2} = b^{2}f; and since b, c are relatively prime in D and f has no repeated factors in D, this implies c is a unit of D, and then ξ ∈ D\sqrt{f}.

Case ii): ξ^{2} ∉ k. Since ξ is integral over D and D is integrally closed in k, the monic irreducible polynomial over k for ξ has coefficients in D. Thus, there exist d, e ∈ D such that ξ^{2} + dξ + e = 0. Since 2 is a unit of D, we have (ξ + (d/2))^{2} + (e − (d^{2}/4)) = 0, and then by Case (i), ξ + (d/2) is either in D or in D\sqrt{f}.

It remains to prove 7.4 in the case that char k_{0} = 2.

Proof ([6]) We again have the equation (14). By the Euclidean algorithm in D, there exist η ∈ D + D\sqrt{f} and a_{1}, b_{1} ∈ D such that

ξ = η + (a_{1}/c) + (b_{1}/c)\sqrt{f}

and either a_{1}b_{1} = 0 or deg c > deg a_{1}, deg b_{1}. It will suffice to prove ξ_{1} = ξ − η is in D + D\sqrt{f}. Note that 2 = 0 implies ξ_{1}^{2} ∈ k ∩ \overline{D} = D; so if b_{1} = 0 or a_{1} = 0, we are done as in Case (i) above. Thus, it remains to observe that a_{1}b_{1} ≠ 0 leads to a contradiction. Since 2 = 0, cξ_{1} = a_{1} + b_{1}\sqrt{f} implies c^{2}ξ_{1}^{2} = a_{1}^{2} + b_{1}^{2}f, which implies k(ξ_{1}) = k(\sqrt{f}). Hence by the minimality of f, we have deg ξ_{1}^{2} ≥ deg f, and it follows that deg c^{2} ≤ deg a_{1}^{2}, deg b_{1}^{2}, a contradiction to our choice of a_{1}, b_{1}, c.

7.6. Remark Without the assumption that 2 is a unit of D, 7.5 is false, even with D = Z; cf. [11, p. 35, Theorem 1].

8 Non-rationality via Riemann-Roch

Let K be a finite algebraic extension of k_{0}(x) such that k_{0} is algebraically closed in K, and let V be the set of all valuations of K/k_{0}.

We regard the value group of an element V ∈ V as being a supgroup of the additive rationals in such a way that V (k_{0}(x)) is the group of integers. (In the case considered below, the index of a valuation from k_{0}(x) to K will be ≤ 2, hence our value groups will actually be subgroups of Z[1/2].)

A divisor A of K/k_{0} is a formal product

A = ∏_{V ∈V} V^{m(V)},

with m(V) ∈ Z, and at most finitely many m(V) nonzero. In other words, A is a map from V to Z which is nonzero on at most finitely many elements of V. We define the degree of A by

deg A := ∑_{V ∈V} m(V)[K_{V} : k_{0}],

and the k_{0}-vector space L(A) by

L(A) := {ξ ∈ K | for all V ∈ V, V (ξ) ≥ m(V)V (π_{V})},

where π_{V} denotes a generator of the maximal ideal of V. Moreover, for j ∈ Z, we define a new divisor A^{−j} := ∏_{V ∈V} V^{−jm(V)}. Thus,

L(A^{−j}) := {ξ ∈ K | for all V ∈ V, V (ξ) ≥ −jm(V)V (π_{V})}.

(Note: In the case below, where [K : k_{0}(x)] = 2, if V_{0} is a valuation of k_{0}(x)/k_{0} and V its extension to K, we have V (π_{V}) = 1/2 or 1, depending on whether V_{0} ramifies in K or not.)

8.1. Theorem (Riemann) For every divisor A of K/k_{0} the k_{0}-vector space L(A) is finite dimensional, and there exists a nonnegative integer Γ such that for every divisor A of deg > 0 and all sufficiently large j,

Γ := deg A^{j} − (dim L(A^{−j}) − 1).

The nonnegative integer Γ is called the genus of K/k_{0}. (See [3, p. 32, Corollary].)

We are now ready to prove the central theorem of §8. We remind the reader that if char k_{0} ≠ 2, then f ∈ k_{0}[x]\k_{0} is minimal in k_{0}[x] iff f has no repeated factors in k_{0}[x], and that, for all characteristics, f minimal implies k_{0} is algebraically closed in k_{0}(x, \sqrt{f}).

8.2. Theorem (genus theorem) Let f ∈ k_{0}[x] \ k_{0} be of degree n and minimal in k_{0}[x]. If K = k_{0}(x, y), with y = \sqrt{f}, then Γ(K/k_{0}) = (n − 1)/2 if n is odd and (n − 2)/2 if n is even.

Proof Let V be the set of all valuations of K/k_{0}. We write V = V_{x}∪V_{1/x}, where V_{x} (resp. V_{1/x}) denotes the set of elements of V containing k_{0}[x] (resp. k_{0}[1/x]). We consider first the case of characteristic 2, due to Lang-Tate [6].

Case i): n arbitrary and char k_{0} = 2. Since every element of K has a square in k_{0}(x), every valuation of k_{0}(x)/k_{0} extends uniquely to K. Let A be the divisor V_{∞}^{2}, where V_{∞} is the unique extension of V_{1/x} from k_{0}(x) to K. (We could also work with the divisor V_{∞}^{1}, but V_{∞}^{2} avoids some fractions.) Then deg A = 2 since [K_{V_{∞}} : k_{0}] = 1. It remains to compute dim A^{−j} for j large.

Since

\overline{k_{0}[x]} = \{ξ ∈ K | for all V ∈ V_{x}, V (ξ) ≥ 0\},

we have by 7.4

A^{−j} = {ξ ∈ k_{0}[x] + yk_{0}[x] | V_{∞}(ξ) ≥ −j}.  (16)

Now let ξ = a + by, a, b ∈ k_{0}[x]. Note that the minimality hypothesis on f implies that the leading term of f cannot be a square in k_{0}(x), hence ξ^{2} = a^{2} + b^{2}f implies deg ξ^{2} = max{deg a^{2}, deg fb^{2}}. Therefore,

V_{∞}(ξ) ≥ −j ⟺ V_{∞}(ξ^{2}) ≥ −2j
⟺ deg ξ^{2} ≤ 2j
⟺ deg a^{2} ≤ 2j and deg fb^{2} ≤ 2j
⟺ deg a ≤ j and deg b ≤ j − n/2.

Let s = n/2 if n is even and (n + 1)/2 if n is odd. Then

B := {1, x, …, x^{j}; y, yx, …, yx^{j−s}}  (17)

is a generating set for the k_{0}-space L(A^{−j}). Moreover, since 1, y are linearly independent over k_{0}(x) and the powers of x are linearly independent over k_{0}, B is a basis of L(A^{−j}). Thus, dim L(A^{−j}) = 1 + j + 1 + j − s = 2 + 2j − s, and Γ = 2j − (2 + 2j − s − 1) = s − 1, as stated in the theorem. This concludes Case (i).

The case of n odd is similar to that of Case (i) and will be taken up next. We retain the notation of Case (i).

Case ii): n odd and char k_{0} ≠ 2. We have V_{1/x}(f) = −n, so the extension V_{∞} of V_{1/x} to K has V_{∞}(y) = −n/2, which is ∉ Z because n is odd. Therefore V_{1/x} ramifies in K, hence V_{∞} is the unique extension of V_{1/x} to K and [K_{V_{∞}} : k_{0}] = 1.

As before, we take A to be the divisor V_{∞}^{2}, and we compute L(A^{−j}), which is again described by (16). If ξ = a + yb, with a, b ∈ k_{0}(x), then V_{∞}(a) ≠ V_{∞}(yb) since V_{∞}(a) ∈ Z and V_{∞}(yb) ∉ Z, and therefore V_{∞}(ξ) = min{V_{∞}(a), V_{∞}(yb)}. Thus,

V_{∞}(ξ) ≥ −j ⟺ V_{∞}(a) ≥ −j and V_{∞}(yb) ≥ −j
⟺ deg a ≤ j and deg b ≤ j − n/2.

The remainder of Case (ii) is the same as that of Case (i).

Case iii): n even and char k_{0} ≠ 2. We have f = a_{n}x^{n} + … + a_{0}, a_{i} ∈ k_{0} and a_{n} ≠ 0. If a_{0} = 0, i.e., if x is a factor of f, then f/x^{n} ∈ k_{0}[1/x] is of odd degree n − 1 in 1/x, and then we are in Case (ii); so we may assume x is not a factor of f in k_{0}[x]. Now factor f:

f = cp_{1} ⋯ p_{m}, c ∈ k_{0}, p_{i} monic irreducible in k_{0}[x],

and then p_{i} ≠ x. By 6.1, the V_{p_{i}} (i = 1, …, m) are the valuations of k_{0}(x)/k_{0} which ramify in K. Let V_{i} denote the unique extension of V_{p_{i}} to K, and let A be the divisor V_{1}^{2} ⋯ V_{m}^{2}. Note that V_{i}(f) = 1 and V_{i}(y) = 1/2. Also, deg A = 2n, since [K_{V_{i}} : k_{0}] = [k_{0}(x)_{V_{p_{i}}} : k_{0}] = deg p_{i}.

It remains to compute dim L(A^{−j}). Let W = {V_{1}, …, V_{m}}. Then

L(A^{−j}) = {ξ ∈ K | for all V ∈ W, V (ξ) ≥ −1 and for all V ∈ V \ W, V (ξ) ≥ 0. }

Let ξ ∈ L(A^{−j}). We have V_{i}(y^{2}) = 1 for V_{i} ∈ {V_{1}, …, V_{m}} and V (y^{2}) = 0 for V ∈ V_{x} \ \{V_{1}, …, V_{m}\}; so V (y^{2j}ξ) ≥ 0 for all V ∈ V_{x}, and therefore y^{2j}ξ ∈ \overline{k_{0}[x]} = k_{0}[x] + yk_{0}[x].

Similarly, let z = y/x^{n/2}. Then z^{2} = f/x^{n} ∈ k_{0}[1/x] and V_{i}(z^{2}) = 1 for V_{i} ∈ {V_{1}, …, V_{m}} and V (z^{2}) = 0 for V ∈ V_{1/x}\{V_{1}, …, V_{m}}. Therefore V (z^{2j}ξ) ≥ 0 for all V ∈ V_{1/x}; so z^{2j}ξ ∈ \overline{k_{0}[1/x]} = k_{0}[1/x] + zk_{0}[1/x], and then y^{2j}ξ ∈ x^{jn}k_{0}[1/x] + yx^{jn−n/2}k_{0}[1/x].

Then,

\begin{aligned}ξ ∈ L(A^{−j}) & ⟺ & y^{2j}ξ ∈ (k_{0}[x] + yk_{0}[x]) ∩ (x^{jn}k_{0}[1/x] + yx^{jn−n/2}k_{0}[1/x]) \\ & ⟺ & ξ ∈ (1/y^{2j}) < 1, x, …, x^{jn}; y, yx, …, yx^{jn−n/2} >, \end{aligned}

where < > denotes ‘k_{0}-space generated by’. Thus dim L(A^{−j}) = (1 + jn) + (1 + jn − n/2), so Γ = 2jn − (2 + 2jn − n/2 − 1) = (n − 2)/2, as required.

Incidentally, Case (ii) of the above proof is not really necessary, for we shall see in 9.2 below that for n odd, one can always find g ∈ k_{0}[1/x] such that deg g = n + 1 is even and k_{0}(x, \sqrt{f}) = k_{0}(1/x, \sqrt{g}).

8.3. Corollary Assume the hypothesis of 8.2 . If n ≥ 3, then K/k_{0} is not rational.

Proof If n ≥ 3, then by 8.2, Γ(K/k_{0}) > 0; so it remains to observe that K/k_{0} is rational implies Γ(K/k_{0}) = 0, which follows from 8.2 by writing k_{0}(t) in the form k_{0}(t^{2}, t).

8.4. Remark By 8.2, Γ(K/k_{0}) = 0 (with K as in 8.2) iff n = 1 or 2. If n = 1, then K/k_{0} is clearly rational; and if n = 2, it is easily seen (see, e.g., [9, p. 338, Theorem 1.4] or [1, p. 304, Theorem 7]) that K/k_{0} is rational iff the curve Y^{2} = f(X) has a k_{0}-point. It may also be seen from Riemann-Roch that every 1-dimensional function field of genus 0 and with exact constant field k_{0} is of the form k_{0}(x, \sqrt{f})/k_{0} (cf. [1, p. 304, Theorem 7]).

9 Defining pairs and degree

¿From the equalities n = 2Γ + 1 or 2Γ + 2 of Theorem 8.2, we obtain the following:

9.1. Corollary Suppose f_{i} ∈ k_{0}[x_{i}] \ k_{0} (i = 1,2) is of degree n_{i} and minimal in k_{0}[x_{i}]. If k_{0}(x_{1}, \sqrt{f_{1}}) = k_{0}(x_{2}, \sqrt{f_{2}}) and n_{1} is odd (resp., even), then n_{2} equals n_{1} or n_{1} + 1 (resp., n_{1} or n_{1} − 1).

The corollary brings up the question of just what values the degree n of f can have. To facilitate the discussion, call a pair (x, f), consisting of an element x transcendental over k_{0} and an element f ∈ k_{0}[x] \ k_{0}, a defining pair of degree n if f is without repeated factors in k_{0}[x] and deg f = n; in addition, if char k_{0} = 2, we require that f be minimal in k_{0}[x]. We shall also say that the defining pair (x, f) defines the function field k_{0}(x, \sqrt{f})/k_{0}. By 9.1, there are only two possibilities for the degree of a defining pair which defines a given function field K/k_{0}.

Let us call two defining pairs (x, f) and (x_{1}, f_{1}) equivalent if k_{0}(x) = k_{0}(x_{1}) and there exists an element β ∈ k_{0}(x) such that f_{1} = β^{2}f. Equivalent defining pairs clearly define the same function field. Moreover, by 2.6,

If char k_{0} ≠ 2, defining pairs (x, f) and (x_{1}, f_{1}) are equivalent iff k_{0}(x) = k_{0}(x_{1}) and k_{0}(x, \sqrt{f}) = k_{0}(x_{1}, \sqrt{f_{1}}).

Note that if x = x_{1}, then by the unique factorization in k_{0}(x), β is necessarily in k_{0}; and it follows from this that two equivalent defining pairs with the same first coordinate x have the same degree.

9.2. Proposition Every defining pair of odd degree n − 1 is equivalent to a defining pair of even degree n. If (x, f) is a defining pair of even degree n, then every defining pair equivalent to (x, f) is either of degree n or degree n − 1, and there exists a defining pair of degree n − 1 equivalent to (x, f) iff f has a linear factor in k_{0}[x].

Proof If (x_{1}, f_{1}) is a defining pair of odd degree n − 1, then (1/x_{1}, f_{1}/x_{1}^{n}) is a defining pair of even degree n equivalent to (x_{1}, f_{1}).

Now suppose (x, f) is a defining pair of even degree n and (x_{1}, f_{1}) is an equivalent defining pair. Let k = k_{0}(x) = k_{0}(x_{1}). Since f_{1} = βf^{2} for some β ∈ k, Odd_{V}(f_{1}) = Odd_{V}(f), where V is the set of all valuations of k/k_{0}; hence by equation (9) of §3, deg_{x_{1}} f_{1} = deg_{x} f (resp., = deg_{x} f − 1) if deg_{x_{1}} f_{1} is even (resp., odd).

If f has a linear factor x′ = ax+b in k_{0}[x], then (1/x′, f/(x′)^{n}) is an equivalent defining pair of degree n−1. Conversely, suppose there exists an equivalent defining pair (x_{1}, f_{1}) with deg_{x_{1}} f_{1} = n−1. Since n−1 is odd, Odd_{V}(f_{1}) = Odd_{V}(f) contains the valuation V_{1/x_{1}}, which has residue field k_{0}. Since deg_{x} f is even, there must then exist an irreducible factor p of f in k_{0}[x] of degree 1 such that V_{1/x_{1}} = V_{p}.

9.3. Remark Suppose (x, f) is a defining pair of degree n, and let K = k_{0}(x, \sqrt{f}). If n ≥ 5, (equivalently, if Γ(K/k_{0}) ≥ 2), then it is known (cf. [3, p. 74, Theorem 9], [1, p. 315]) by Riemann-Roch that K has a unique subfield k which is simple transcendental over k_{0} and such that [K : k] = 2. (Incidentally, this uniqueness implies K/k_{0} is not rational, for if there exists t ∈ K such that K = k_{0}(t), then k_{0}(t^{2}) and k_{0}(t + t^{2}) are two codeg 2 subfields of K.) Thus, if char k_{0} ≠ 2 and n ≥ 5, then a defining pair (x_{1}, f_{1}) of degree n_{1} is equivalent to (x, f) iff k_{0}(x, \sqrt{f}) = k_{0}(x_{1}, \sqrt{f_{1}}), i.e., then the condition k_{0}(x) = k_{0}(x_{1}) in (9) is automatic. If n = 4, the uniqueness of codeg 2 subfield fails, as the following example shows:

Let k_{0} = Q, f = x^{4} + 1, y = \sqrt{f}, and K = k_{0}(x, y). If x_{1} = y − x^{2} and y_{1} = 2x_{1}x, then y_{1}^{2} = −2x_{1}(x_{1}^{2} − 1) =: f_{1} and k_{0}(x, y) = k_{0}(x_{1}, y_{1}); but there is only one valuation which ramifies from k_{0}(x) to K and there are 4 which ramify from k_{0}(x_{1}) to K, so k_{0}(x) ≠ k_{0}(x_{1}).

If one modifies this example by taking f = −(x^{4} + 1), then K/k_{0} has no valuation with residue field k_{0}; and therefore K/k_{0} cannot have an equivalent defining pair (x_{1}, f_{1}) of degree n = 3, since whenever n is odd the extension of V_{1/x_{1}} to K ramifies, hence has residue field k_{0}.

If the requirement that k_{0}(x) = k_{0}(x_{1}) is dropped, then (in char ≠ 2) for every defining pair (x, y) of degree 4 such that the curve Y^{2} = f(X) has a k_{0}-point, there exists a defining pair (x_{1}, y_{1}) of degree 3 such that k_{0}(x, y) = k_{0}(x_{1}, y_{1}); cf. [2, p. 35].

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

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