数理解析研究所講究録
第 1794 巻 2012 年 70-72

70
Elementary equivalence of rational function fields
鹿児島国際大学国際文化学部 福崎 賢治 (Kenji Fukuzaki)
Faculty of Intercultural Studies,
The international University of Kagoshima

Abstract

Let $K|k$ be a function field over a field . We suppose that is an algebraically closed field or a finite extension of a prime field. We prove that$K\equiv k(x)$ implies $K\cong k(t)$ , whereis an indeterminate. For the case thatis algebraically closed, this was proven by Duret (1992), and for the case thatis a finite extension of a prime field, by Scanlon (2008). However we give asimple unified proof for both cases.
$k$

$k$

$k$

$t$

$k$

1Introduction

It appears to be an interesting question, whether for finitely generated fields, theelementary equivalence is the same as the isomorphism. In the beginning of $1980$ ’s
Sabbagh asked the following question: Let $K$ and be fUnction fields over with
td $(K|\mathbb{Q})=1$ and td $(L|\mathbb{Q})=2$ . Is it then possible that $K$ andare elementarilyequivalent?
The answer is no by the results of Pop (2002): The transcendence degree and transcendence bases of a function fields over number fields are definable in the languageof rings. It is already known in the positive characteristic case and in the geometriccase (that is, if the base field is algebraically closed). Furthernore, he showed thefollowings:$L$$\mathbb{Q}$$L$

Let $K$ and$L$

be function fields over prime fields with $K\equiv L$ . Then

1. there are embeddings $Karrow L$ and $Larrow K$ .
2. $Furthem\iota ore$ , if one of them is of general type then they are isomorphic.
is of general type if it is the function field of a projective smooth
We say thatvariety over of general type. It is known that smooth hypersurfaces of dimension
with degree $d>n+2$ are of general type. Roughly speaking, almost all varieties areof general type. We note that rational function fields and elliptic function fields are
not of general type.
$K|\kappa$$\kappa$$n$71
Non-general case remained an open question. However Scanlon (2008) proved
elementary equivalence implies isomorphism for all infinite function fields over prime
fields by using biinterpretability of such fields.
For geometric case, that is, for function fields overcosed fields, Dui et(1992) proved the folowing, using the facts on plane algebraic curves.
$c1^{\underline{|}gcb_{I}\cdot aically}$

Let $K$ and be function fields over an algebraically closed fielddence degree 1. Suppose $K\equiv L$ .
$L$$\kappa$with tmnscen-

1. If one of them has genus different from 1, then $K\cong L$ .

2. If

has characteristic and one of them is an elliptic function field with nocomplex multiplication, then $K\cong L$ .
$0$$\kappa$

Later Pop (2002) proved the followings:
Letfields$\kappa$andbe elementarily equivalent function fields over algebmically closed, respectively . Suppose $K\equiv L$ . Then
$K|\kappa$

1. td$L|\lambda$$\lambda$$(K|\kappa)$

2. Suppose
andequals td$K|\kappa$$(L|\lambda)$.

is of geneml type. Then there exist function subfields
such that $K=K_{0}\kappa$ and $L=L_{0}\lambda$ , and$L_{0}|\lambda_{0}\mapsto L|\lambda$

function fields.

In particular, if$K_{0}\mapsto K|\kappa_{0}$$K_{0}|\kappa_{0}\cong L_{0}|\lambda_{0}$$\kappa\cong\lambda$

are isomorphic, then

$K|\kappa\cong L|\lambda$

function fields.

are isomorphic as

We note that it remains open whether or not $K\equiv L$ implies that
It is unknown whether. (It is known that$\mathbb{Q}^{alg}(X)\equiv \mathbb{C}(X)$$\kappa\equiv\lambda.)$as$\kappa\cong\lambda$. (Note.

$\mathbb{Q}^{alg}[X]\not\equiv \mathbb{C}[X]$$)$2

Rational function fields in one indeterminate

We begin with the following simple lemma.
Lemma 1 Let $K|k$ be a function field $K=k(x, y)$ with $f(x, y)=0$ for an irreducible
polynomial $f(X, Y)$ over .
We let $\deg_{X}(f)=m$ and $\deg_{Y}(f)=n$ . $(\deg_{X}(f)$ and $\deg_{Y}(f)$ denote the degreeof $f(X, Y)$ with respect to $X$ and respectively. ) Let be an indeterminate. Then$K\cong k(t)$$s\in k[X]$ with $\deg(p),$ $\deg(q)\leq n$ andiff there exist polynomials$\deg(r),$ $\deg(s)\leq m$ such that $f(p/q, r/s)=0$ as polynomials.
$k$

$Y$

$t$

$p,$ $q,$ $r,$

72
Proof. The converse follows from L\"uroth theorem. We suppose that.
Let $xrightarrow\alpha=p(t)/q(t)$ and $yrightarrow\beta=r(t)/s(t)$ , where and (respectively and s)
are polynomials in $k[t]$ and have no common factors. We have $k(p/q, r/s)=k(t)$ . Of
course we have that $f(p/q, r/s)=0$ as polynomials.
We know that $q(X)-\alpha p(X)$ is an irreducible polynomial over, hence $[k(t)$ :$k( \alpha)]=\max(\deg(p), \deg(q))$ . Since $[K;k(x)]=n$ , we have $[k(t):k(\alpha)]=n$ , hence$\deg(p),$ $\deg(q)\leq n$ . Similarly we have $\deg(r),$ $\deg(s)\leq m$ .
$s$

$p$

$K\cong k(t)$
$r$

$q$

$k(\alpha)$$\square$Proposition 2 Let be an algebmically closed field or a finite extension of a pmme
field. Let be an indeterminate. Let $K$ be a function field over . Then $K\equiv k(t)$implies $K\cong k(x)$ .
$k$

$k$

$t$

Pmof. Suppose$K\equiv k(t)$. We easily have ch $(K)=$ ch $(k(t))$ . Since the transcendence

degree and transcendence bases of are definable, we have that td $(K|k)=1$ and theconstant field is definable.
Let $K=k(x, y)$ and $f(x, y)=0$ for some irreducible polynomial $f(X, Y)$ with$\deg_{X}(f)=m$ and $\deg_{Y}(f)=n$ . Suppose $K\not\cong k(x)$ . By the lemma, $f(X, Y)$ has
no rational parametrisations, $p(X)/q(X),$ $r(X)/s(X)$ with $\deg(p),$ $\deg(q)\leq n$ and$\deg(r),$ $\deg(s)\leq m$ .
If $K$ is defined over a finite extension of prime field, we have
$Yf(X, Y)=$0. In general it does not hold. So quantifying the coefficients, we have tliat in $k(t)$ ,there areand a polynomial $f’(X, Y)$ with $\deg_{X}(f’)=m$ and $\deg_{Y}(f’)=n$ such$f’(\alpha,\beta)=0$
that
holds and $f’(X, Y)$ has no rational parametrisations, $p(X)/q(X),$ $r(X)/s(X)$with $\deg(p),$ $\deg(q)\leq n$ and $\deg(r),$ $\deg(s)\leq m$ . By L\"uroth theorem, there exists$t’\in k(t)$ such that $k(t’)=k(\alpha, \beta)$ . If we write $\alpha=p(t’)/q(t’)$ and $\beta=r(t’)/s(t’)$ thenwe have $\deg(p),$ $\deg(q)>n$ or $\deg(r),$ $\deg(s)>m$ . From the proof of the lemma, wehave a contradiction.
$k$

$k$

$k(t)\models\exists X,$$\alpha,$$\beta$$s$$\square$

References[1] Jean-Louis Duret, “ Sur la th\’eorie \’el\’ementaire des corps des fonctions“, J. SymbolicLogic 51 (1988) 948-956.
[2] Jean-Louis Duret, “Equivalence \’el\’ementaire et isomorphisme des courbe sur uncorps alg\’ebriquement clos“, J. Symbolic Logic 57(1992) 808-823.
[3] David A. Pierce, “Function fields and elementary equivalence”, Bull. London
Math. Soc. 31 (1999) 431-440.

[4] Florian Pop, “Elementary equivalence versus isomorphism”, Invent. Math.
150(2002) 385-408.