Weak presentations of non-finitely generated fields

ANNALS OF PURE AND APPLIED LOGIC Annals of Pure and Applied

Logic 94 (1998) 223-252

Weak presentations of non-finitely generated fields’ Alexandra

Shlapentokh*

Department of Mathematics, East Carolina University. Greenville. NC 27858, USA

Abstract Let K be a countable field. Then a weak presentation of K is an isomorphism of K onto a field whose elements are natural numbers, such that all the field operations are extendible to total recursive functions. Given a pair of two non-finitely generated countable fields contained in some overfield, we investigate under what circumstances the overfield has a weak presentation under which the given fields have images of arbitrary Turing degrees or, in other words, we investigate Turing separability of various pairs of non-finitely generated fields. 0 1998 Elsevier Science B.V. All rights reserved. AMS

classijication:

Keywords:

03D20;

Presentations;

03D25;

03D35;

03C57

Turing Degrees; Recursive

1. Introduction introduced to formalize the notion of an algorithm over countable mathematical objects. The object under consideration was mapped into natural numbers and a function over the object was considered recursive if its translation over the presentation image was recursive. The original definition of presentations, as can be found in [2,6], required that all the operations of the object were translated by total recursive functions and required the image of the presentation to be recursive also. The objects that have such presentations are usually called recursive. On the other hand, there are naturally arising algebraic objects which do not have recursive presentations but are embedded in objects which are recursive. Therefore, the operations of these non-recursive objects can be represented by restrictions of total recursive functions. Existence of such objects can motivate one to define another class of presentations: weak presentations, which will not require the image of the presentation map to be recursive while requiring that all the operations associated with the object Presentations

* E-mail:

were originally

[email protected].

’ The research of this paper has been partially supported by NSA grant MDA904-96-1-0019. 016%0072198/%19.00 @ 1998 Elsevier Science B.V. All rights reserved PI1 SO 168-0072( 97)00074-2

224

A. Shlapentokhl Annals of’ Pure and Applied Logic 94 (1998) 223-2S2

are translated following

by the functions

definition

Definition

extendible

to total recursive

ones. Thus, we have the

which we will state for fields.

1.1. Let K be a countable

there exist total recursive

functions

field. Let j: K -+N be an injective

map such that

P+, P_, Px , Pi : N2 -+ N with the property that for all

~7.vcK> p+(j(x)&)) =j(x + YXP-(j(x)&)) =j(x - y)&(j(x),j(y)) =j(x x y), and if y #O, P/(j(x>,j(y))=j(x/~). Then j is called a tveak presentation of K as a field. Given

such a definition,

a recursive

one can consider

the following

object, what kind of weak presentations

ing a problem

of this type should illuminate

class of problems:

given

does such an object have? Solv-

the relationship

between

the algebraic

and “logic” structures of the object under consideration. The study of weak presentations of computable objects can also shed some light on the problems of Diophantine definability.

For more detailed discussion

In the preceding

work, Carl Jockusch

the weak presentations existence

of certain

following

algebraic

Definition

of various

of these issues see [7]. and the author of this paper have investigated

computable

weak presentations

fields (see [4,8-lo]).

of finitely

It turned out that

generated

fields depended

on the

contained

in some field F. Then

reducibility.

1.2. Let R,,Rz

be two integral

domains

we will say that RI is rationally separably less than R2 (RI drs R2) if both of the rings are finite or if there exist non-constant that for every XERI,

rational

for some 1
functions

Ht (T), . . . ,Hk( T) E F( T) such

H&x)ER~.

We would like to note here that one can show that this definition the field F containing p. 232). Furthermore, ing facts concerning

the integral

domains

from the definition rational

under consideration

is not dependent

on

(see [9], Lemma 2.2,

above one can also easily deduce the follow-

separability:

1. If RI CR2 then RI d,,Rz; 2. If RI Grs RZ and R2 is finite then RI is finite; 3. If RI is finite then RI 6,, R2; 4. The relationship “GTs” is transitive. The relationship “ < TS” is a generalization of well known algebraic notions of separable and inseparable polynomials and field extensions. (A discussion of these notions can be found in [5], pp. 176-182 and pp. 186-191.) Let F be a field and let P be a polynomial over F. Then P is called separable if all of its roots in the algebraic closure of F are distinct. Otherwise, the polynomial is called inseparable. If degree of P is greater than 1 and P has just one multiple root in the algebraic closure of F (in other words, in the algebraic closure of F, P = P(x) = (x - a)“), then P is called purely inseparable. If e/F2 is an algebraic field extension, then it is called separable if every element of Fl satisfies a separable irreducible polynomial over F2. Otherwise, the extension

225

A. Shlapentokhl Annals oj’ Pure and Applied Logic 94 (1998) 223-2.52

is called inseparable. polynomial

If every element

over F2 then the extension

useful facts concerning 1. If fi/fi

of F, satisfies a purely

the separable

is an inseparable

algebraic

such that Fz c G, the extension

is called purely

and inseparable

inseparable

inseparable.

irreducible

Here are some

extensions: then there exists a field G c FI

field extension,

G/F2 is separable

and the extension

Fl/G is purely

inseparable; 2. If the field extension fi/F~ is purely inseparable then fi,F? are fields of positive characteristic p and Ft\Fz consists of elements whose irreducible polynomials over F2 are of the form xJ” - a, where k is a natural number 3. If FL, F2 are fields of positive

characteristic

and a E F2;

p, fi /Fz is a separable

field extension,

and for some XEF,, xPEF~, Then XEF~; 4. If e/F2

is a separable

single element

and finite extension,

then it is simple (i.e. generated

of a larger field). This is not necessarily

true if the extension

by a is not

separable. If the field extension under consideration is not necessarily algebraic then the notion of separability is replaced by the notion of separable generation. That is, we say that F, is separably

generated

over Fz if FI contains

a transcendence

base C over F? (possibly

empty) such that the extension FI/F~(C) is algebraic and separable. These observations point to the following fact. If fi/Fl is a jinite purely inseparable extension (in this case Fl is finitely generated over F2), then F, < TSF2. In some sense this situation covers “most” of the occurrences of rational inseparability as described by the following

theorem.

Theorem 1.3. Let RI be an integral domain with a quotient field Fi and let F2 be another field such that both FI and Fz are contuined

in some field F. Then RI 6,, Fz

implies that either FI C F2 or F, is (I purely inseparable there exists

k EN with the property

the churacteristir

extension

of FI n F2 such that

that jar uny x E Fl, xp’ E F, n F2, where p > 0 is

of the Jiclds under consideration.

If RI and R2 share the same quotient

(See [93, Theorem

field determining

2.5, p. 234.)

under what circumstances

RI 6,, R2 is a more complicated matter outside the scope of this paper. (An interested reader can find a discussion of this situation in [lo].) In the case the quotient fields of rings under consideration are finitely generated, the described above is connected to weak presentations

algebraic separability relationship in the following manner.

Theorem 1.4. Let RI and R2 be two subrings of finitely respectively. l

l

Then the following

stutements

generated fields fi and F2,

are true:

If R2 & TsR, then jar any two r.e. degrees (I< b there exists a weak presentution of FlF2 such that the image of R2 is of degree b and the image of RI is of degree a; If R2 $,, RI and RI 6, R2 then for any two r.e. degrees a and b there exists a weak presentation of FlF2 such that the image of R2 is of degree b and the image of RI is of degree a;

A. Shlapenrokh I Annals oJ’ Pure und Applied Logic 94 (1998) 223152

226

If Rz
In general, we call algebraic weak presentation

objects contained

in a larger object which possesses

ing separable. Given the theorem above, the problem of classifying presentations

a

placing the two given objects into two different Turing degrees, Turall possible

weak

of finitely generated fields is reduced to the problem of determining

when

a subring of one finitely generated field is rationally separably less than a subring of another finitely generated field. On the other hand, the situation changes drastically when the fields under consideration

are not finitely

tions we will explore

of weak presentations

some aspects

generated.

In the following

of non-finitely

sec-

generated

fields.

2. The main constructions In this section distinguish finitely

we will discuss

two constructions

the case of finitely generated

generated.

(horizontal

and vertical)

which

fields from the case of fields which are non-

It is clear from the definition

of the weak presentations

that any

finitely generated object can have recursively enumerable weak presentations only. Considering fields which are not finitely generated opens a possibility for constructing weak presentations with images of arbitrary Turing degrees. In what follows we will present two constructions designed for this purpose. The horizontal construction will refer to the following

picture.

F

Fl

A F2

...

F,

...

\V K

Here we will assume that each F; has either an infinite transcendence or an infinite algebraic degree over K. The field F is assumed to contain each E and will satisfy certain conditions which are stated in Notations 2.1.

A. Shlapentokh I Annals of Pure and Applied Logic 94 (1998 I 223-252

The vertical construction

corresponds

Here we will assume that for iEN,

to the following

F;+, has an infinite

picture.

transcendence

algebraic degree over F;:. (We let K =Fo.) The precise assumptions will be described in Notations 2.13. We will start with the horizontal

227

or an infinite

for this construction

construction.

notations will be used in the construction below. F will extension of an infinite recursive field K. The fields fi, and

Notations 2.1. The following be a countable

recursive

Mi,t, where i, t E N, will have the following properties: 1. F; C F for all i EN. F is the smallest field containing 2. For each i E N, K = Mi.0 C A4i. 1 C 3. U~,M,,,

Ur,

fi;

C Mi,, . . . , where “C” means proper subset;

=F;.

Let xi,. . . ,xk E F. Then we will say that {Mi,,t,, . . . ,M,mt_,n,}is the minimal fields with respect to xi,. . . ,xk if the following 1.

XI,...,X~EMI,,I,

2.

If

XI ,...,

conditions

set of

are satisfied:

. ..Mi.,,t,,,; XkEhfj,J,

. ..M.J,,

then for each u = 1,. . . , m there exists

1 d u d r such

set of fields to be empty. In particular,

it is empty if

that i, = j, and t,
for the minimal

Remark 2.2. 1. It is clear from this definition that if x, y, z E F and z = field operation

op, then the minimal

x op _v for

some

set of fields with respect to z is a subset of the

union of the minimal set of fields with respect to x and _v. 2. It is also not hard to see that a minimal set of fields is defined with respect to any finite set of elements of F if and only if a minimal set of fields is defined with respect to any single element of F. Furthermore, given a finite collection of elements of F together with the minimal set of fields for each element of the collection, one can recursively construct a minimal set of fields for the collection as a whole. Indeed,

228

A. Shlapentokh I Annals of’ Pure and Applied Logic 94

(I 998) 223-252

is the minimal set of fields with respect suppose XI,. ..,xk EF, and M,,,,j,.,, . .vMi,,,,.j,,., to xl for some 1 d I d k. Let I be the finite set of natural numbers obtained from {it,1 ,. . .,irA,k} by removing all the repetitions from the set. For each iEZ, let j(i) be the largest natural number such that Mi,j(;) occurs as an element of the minimal set of fields with respect to some XI, I< 1 dk. x] ,..., xk is

Then the minimal

set of fields with respect to

iEl}.

{Mi.j(i),

Theorem 2.3 (The Horizontal Construction). Let F be as in Notations 2.13. Assume for any x E F the minimal subset of fields, as defined in Notations 2.1, exists. Furthermore, suppose that under some recursive presentation j of F there exists a recursive function ind: j(F) + {space of finite sequences of natural numbers}, such that for any xEF, ind( j(x)) = (n, il,. . . , in, tl ,...,tll), where {Mi,,,, , . . . ,Mi”,,,} is the minimal set of fields with respect to x. Then, for any sequence {Bi}iEN, of subsets of N, there exists a weak presentation J of F such that J(K) is recursive and for each i, Bi is Turing equivalent and enumeration reducible to J(fi). Proof. The set A! will be a set of pairs of the form (a, m) where a Ej(F) and m is an infinite dimensional

matrix whose entries will be in the set (0,1,2}. Only finitely many

entries of any matrix will be non-zero.

Below we will describe

how we assign values

to the matrix entries in a fashion that makes d a recursive set. We will also define functions 33+,K,YX’,,9’, : d2 -+ d which will be total recursive, and whose restrictions will eventually Let XEF,

represent

field operations

of F.

and let (nil ,..., in,tl ,..., tn) = ind( j(x)).

contain all the pairs (j(x), m) satisfying

the conditions

no pairs which do not satisfy the conditions 1. The matrix m contains

Then for each XEF, listed below, and d

d

will

will contain

listed below:

entries in the set (0,1,2} only;

2. If i@{il,..., i,,} then the ith row of m contains zeros only; 3. If i=ik for some k=l,...,n, then mir,u=O if and Only if U>tk. Let (a,m(a))

and (b,m(b))Ed.

tute a matching pair of elements

Then we will say that (a,m(a)) of d

if for all i,uEN

and (b,m(b))

consti-

such that m(a)i,u #m(b)i,,,

m(a)i,,m(b)i,, = 0. We will next define Y&, for opt {+, -, x,/}. Let (a,m(a))Ed and (b, m(b)) E XI be a matching pair. Then, assuming it is not the case that op = “/” and j-‘(b)=O, define $$((a,m(a))(b,m(b)))=(c,m(c)), where c= j(j-‘(a)opj-l(b)) and m(c) is constructed in the following fashion. Let ind(c) = (n, il,. . . , i,,, tl, . . . , t,). If if&, for any k= l,..., n or if for some k= I,..., n, i=ik, but u>tk, then let m(c)i,, = 0. Otherwise let m(c)i,* = max(m(a)i,U, m(b);,U). (We should note here, that by Remark 2.2, thus defined pair (c,m(c))~d.) If (am(a)) and (b,m(b)) do not constitute a matching pair or if op = “/” and j-‘(b) = 0 then define $$,((a, m(a))(b, m(b))) = (j(O), O), where 0 is the zero matrix.

229

A. Shlapentokhi Annals of‘ Pure and Applied Logic 94 (1998) 223-252

Given the definitions 0pE {+, -, x,/},

above, it is not hard to see that .d is recursive,

and for each

9& is total recursive.

Let {Bi} be a sequence

of subsets of natural numbers

as described

above. We will

now define J : F -+ SI’. Let x E F be given. Then define J(n) = (j(x), m), where for all i, u E N, either mi,U = 0, or mi,, = 1 and u E B,, or mi.U = 2 and u $Bi. Again, it is not difficult to see that for every x E F, J(x) that if (a,m(a)),(b,m(b))EJ(F), pair, and for any opt{+, J-‘(c) =J-‘(a)opJ-‘(b). j_‘(b) = 0.) Furthermore, J(F;).

Indeed,

is defined and it is defined uniquely. then (a,m(a))

It is also clear

constitute

a matching

-, x,/}, .~,~((a,m(a)),(b,m(b))) = (c,m(c))EJ(F), (Here, as before, we exclude the case of up=“/”

we claim that Bj is Turing suppose the characteristic

of all, we can compute

and (b,m(b))

equivalent

reducible

of Bi is given. Let (a,m)~d.

function

ind(a) to determine

and enumeration

where while

whether

a =j(x)

to First

for some XEF;.

(It

is clear that a~j(F;) if and only if ind(a) = (1, i,t) for some 1.) Secondly, if for j=l , . . . , t, m;,j is equal to 1 if j~Bi and to 2 otherwise, (a,m)~J(fi). Otherwise, (a,m)@J(F;). Conversely,

suppose the characteristic

function

for J(F;:) is given. Let HEN. First of

all, find a pair (a,rn)~J(F;) such that ind(a) =( l,i,n). If this entry is 1 then n EBi, otherwise n @B;. Finally,

suppose we have a recursive

this listing will contain function

signifies

Next check the m entry mi.,.

listing of J(F;).

For every natural

number

it,

a pair (a, m), where ind(a) = (1, i, n). (This value of the index

that a =j(x)

for some x EA4i.n\Mi,n_l.

complement is not empty for any natural numbers listed.) If mi,n = 1 then list n.

By assumption

on F, this

n and i, and thus such an a will be

The above construction was carried out for fields but an analogous construction will, of course, apply to any non-finitely generated object satisfying the appropriate recursiveness its infinitely

conditions. generated

In particular, recursive

Next we will prove technical sentations

2.3 applies to rings, such as Q, and

results which we will use later to describe

of a class of non-finitely

Theorem 2.4. Let F,c,Mi,,,ind,

Theorem

subrings. generated

weak pre-

fields.

j be us in Notations 2.1 and Theorem 2.3. Assume

all the fields are of characteristic p>O. Furthermore, the form

assume that any extension of

is separably generated. Let G be a completely inseparable extension of F such that there exists u recursive extension jo of j with the property that jG(F) is recursive in jc(G). Let Li = {XE G 13 k EN, xp’ EE}. Then, for any countable collection {Bi)iENy of subsets of N there exist a presentation JG of G such that JG(K) is recursive and for each i, Bi is Turing equivalent and enumeration reducible to JG(Li) ET JG(F; ).

A. Shlapenrokhl Annals of Pure und Applied Logic 94 (1998) 223-252

230

Proof. First of all, we will show the following. Indeed,

let x be an arbitrary

certainly

contain x4 Thus if

ind(j(x))=(k,ii

element

If x EF, then ind(j(x))

= ind(j(xP)).

of F. Note that any field containing

x would

,..., &,ti ,..., tk)

then

ind(j(xP))=(m,Z

zm,r ,,..., r,),

I,...,

where mdk, {II,..., Z,}C{il,..., ik}, and whenever lj=i,, rj
of fields of length greater than 1 such that

To=M,,r, . . . MI”,,,!CTI C

C

..’

Tu=M,,t, ...MA,,

and for all v = 1,. . . , u, T,+l = TzMj,,,,. By Assumption 2.1, every extension T,+l/T= is separably generated. On the other hand, for some z, T, contains xp but not x, while Tz+l contains is algebraic x@T,(C,).

both. Let Cz be the transcendence and separable.

Since

base of T,,l/T,

the extension

Thus, the fact that XE T,+, contradicts

separable. We can now expand

the definition

ment of the theorem. The mnction

Tz(C,)/Tz

such that T,+,/T,(C,)

is purely

the assumption

of index to the field G described

will be defined as follows. For any XEF, ind&c(x))= XE G, ind&G(x)) = ind(j(xp’ )), where x P’ E F. By the argument index function indG is well defined and is recursive on jo(G). xcG,

set of fields with respect to x to t,,). Note that this definition = (n, ii,. ..,i,,tl,..,,

2.2 applies to minimal

2.3. Furthermore,

such that xp’, yp‘ E F. (Existence

given

sets of fields computed

let x, y,z E G be such that z =x op y for some field operation number

of natural

ind(j(x)). For any above, the extended

we can now define the minimal

where ind(j(x)) be M;I,l,,...,n/li,,f, is consistent with the one used in Theorem part I of Remark

is

in the state-

indc :jo(G) + {space of finite sequences

numbers},

Given

transcendental,

that Tz+l/Tz(Cz)

of s follows

this definition,

using indc. Indeed,

op. Let s be a natural

from the assumption

that the

G/F is purely inseparable.) Then z P’ E F and the minimal set of fields with respect to zp ’ is contained in the union of the minimal set of fields with respect to xp‘ and yp‘. However, the minimal set of fields with respect to z and zp’ is the same. The analogous statement applies to x and xp’ and y and yp’. Thus, the minimal set extension

of fields with respect to z is contained respect to x and y. We can now use indc to define do,

in the union of the minimal a recursive

extension

of d.

sets of fields with The first element of

the pair in && is going to be an element of jo(G) and the corresponding matrices Will be constructed using indG as in Theorem 2.3. Using indc, one can extend P+, Z, gx, 4 to do so that cdo and the extended functions are recursive. Thus, we can define

A. Shlapentokhi Annals of Pure and Applied Logic 94 (1998)

Jo : G 4 &o using Theorem

a procedure

2.3. Finally,

is an extension

231

223-252

to the one used to define J: F -+ & in

analogous

it is clear that Jo is an isomorphism

of J, Jo(LI) S_TJo(E)=J(F,),

from G onto Jo(G) which

and Jo(Li) + Jo(E) = J(F;).

Next we prove a theorem which is a different version of Theorem

2.4.

Theorem 2.5. Let K,F,F;,Mi.,, ind, j he us in Notutions 2.1 und Theorem 2.3. Furthermore, assume that any extension of the form

(2.2)

Ml.11...~~,I~M~,+,,I,,,/M~,.,, . ..M.,,t,

is purely transcendental. Let H be a recursive algebraic extension of F such that there exists a recursive function deg : H -N with [F(x): F]=deg(x). Let H, be the algebraic closure of F; in H. Then, for any countable collection {Bi}lcN, of subsets of N there exist a presentation Jn of H such that Jn(K) is recursive and for each i, Bi is Turing equivalent and enumeration reducible to Jn(Hi) E_TJn(F;). Proof. As in the proof of the Theorem

2.4, we will start with extending

the index

function to H. Let x E H and let As + . . . + A,,_ 1T”-’ + T” be the manic irreducible polynomial of x over F. Then define the minimal set of fields with respect to x to be the minimal in Theorem

sets of fields with respect to Ao, . . . , A,_ 1, and define the index function as 2.3. (Note that x is algebraic over the smallest field containing the fields

in its minimal

set.) We need to show the following:

1. The minima1 set of fields is well defined for all the elements 2. The extended

index function

can be computed

of H;

recursively;

3. If z =x op y, where x, y, z E H and op is a field operation, then the minimal set of fields with respect to z is contained in the minimal set of fields with respect to x and _v. To show that the minimal

set of fields is well defined with respect to all elements

H we need to show the following:

of

Let M;,.,,,. . .,Mi,.(, be the minimal

set of fields

with respect to x as defined above. Suppose Mj,,u,, . . ,MjV,, is another

set of fields

such that x is algebraic

over Mj,. ,,, . . . Mj, ,U,. Thenfors=l,...,k,thereexists

such that i, = jm and &
11, p. 280, and Assumption

l
. ..Ml..,,Mj,,,, . . . Mj,., and note that by [l],

(2.2), x has the same manic irreducible

polyno-

mial over all the three fields involved. Thus, Ao, . . . ,‘4,_ l EM,, ,u, . Mj,,, and the first assertion follows from the properties of the minimal set of fields as defined over F. The fact that the index function can be computed recursively follows from our assumption that we can recursively determine the degree of every element of H over F. Finally, the third assertion follows from the fact that if x and y are algebraic over fields Nt,l&, respectively. Then z is algebraic over NIN2. Now the rest of the proof can proceed in the same fashion as in Theorem

2.4.

We will now apply Theorems 2.3, 2.4, and 2.5 to describe possible weak presentations of an arbitrary “sufficiently” recursive non-finitely generated field.

232

A. Shlapentokh

I Annals of Pure and Applied Logic 94 (1998) 223-252

Corollary 2.6. Let F be a recursive field, let K be a recursive subjield of F such that the following conditions are satisfied l There exists a recursive set of elements (z,,.. .,.zk,.. .) algebraically independent over K such that F is algebraic over T = K(zl,. . .,zk,. .). This set can be empty or finite; l There exists a recursive set {Ml,. . , c(k, . .} such that F = K(zl, . . . ,a,, . . .). The set can be finite or empty; {@I>. ’ .I l There exists a recursive function deg from N into N such that deg(k + 1) = [T(ai,. . . , &+I ) : T(crl,. . . , ak)], where (al,. . . , ak,. . .) is a fixed recursive listing of the set {Q}; l Either the transcendence degree of F over K is injinite or the algebraic degree of F over T is injnite. Then for any set B c N, there exists a weak presentation J of F such that J(K) is recursive and B is Turing equivalent and enumeration reducible to J(F). Proof. First assume the algebraic that the conditions the above

of Theorem

mentioned

theorem.

degree of F over T is infinite. 2.3 can be satisfied

We have to show

with F = F, in the notations

of

Let Mi, =M, = T(al,. . . , ccl). We have to show that

under a given recursive presentation of F, the index function described in the proof of Theorem 2.3 will be recursive. (It is clear that it is well defined, assuming, of course, that all the values be the recursive

of the degree function

presentation

are greater than 1.) Let j : F --f N

of F under which the sets and functions

statement of the corollary are recursive.

discussed

in the

Suppose n EJ’(F) is given. We want to compute

ind(n). First of all, we note the following. Since K is recursive and the set {zt,. . .} is recursive we can generate an effective listing of T by listing systematically all the rational functions an effective

listing

in zt , . . . over K. Given a listing of T, we can locate ~(1 and generate of linear

combinations

of 1,. . . , a?”

generate an effective listing of MI. Furthermore, at the coefficients

of the linear combination

)-I

over T, that is we can

given an element in that list by looking

we will be able to tell whether the element

is in T. It is clear by induction that we can effectively generate listings for every Mi. Since n will have to appear in one of the lists, its index can be computed. Thus, we can apply Theorem 2.3 to reach the desired conclusion. Suppose now F is of finite degree over T. We will first construct the desired weak presentation of T and then extend it to F. In this case we will let Mi, = MI = K(zl, . . . , z,) and F, = T. Let j be as above. Then given nEj(M) we need a recursive procedure to determine whether n Ej(T), and if so compute its index, i.e. the smallest t such that nEA4,. First of all, by generating the lists described above we can determine effectively whether n EJ’(T). If n cj(T), that means we have a representation of j-‘(n) as a rational function over K in zi , . . . ,z,. The only remaining question is whether we can eliminate any of the Zi’s from this representation. This can be accomplished using the method described in [8], Theorem 2.2, part 3, p. 739. Thus we can again apply Theorem 2.3 to obtain the desired presentation J of T. Next we note, that by [7],

A. Shlapentokh I Annals sf Pure and Applied Logic 94 (1998I 223-252 Lemma 2.12, p. 1075, .I can be extended to F so that J(T)

233

is Turing and enumeration

equivalent

to J(F).

Corollary

2.7. Let M be a recursive field. Let K be a recursive subfield of M such

that K has a splitting algorithm. Further, let {tl,

. , tk,. . .} be a recursive set of ele-

ments of M algebraically independent over K such that M is algebraic and separable over K(tl, . .). Assume furthermore that K( tl, . . .) is u recursive field, Then assuming either that {tl, .} is infinite or that A4 is of injinite degree over K(tt,. . .), for any set B c N, there exists a weak presentation J of M such that J(K)

is recursive and

B is Turing equivalent and enumeration reducible to J(M). Proof. We will first consider the case when M is of infinite degree over K(tl, . .). By Corollary 2.6, it is enough to produce a set {at,. .} generating M over K(t,, .) so that the degree function defined in Corollary 2.6 is recursive. The construction of such a set will proceed

inductively.

Assume we have an effective listing of M and assume

we have already constructed CI, , . . . , tq as well as the irreducible polynomials of tlk over K(t,, . . . , sll,. . . , &-I) for k = 1,. . . , i. We will then consider the first, so far unprocessed /?EM.

element

Via a systematic

search we can construct

a manic

polynomial

P(T)

satisfied by p over K(tl,. . . , &,,a~, ,ai). From [2], Chapter 17, one can deduce that we can construct effectively a splitting algorithm for K( tl,. , t,, ~(1,.. . ,ai). Thus, we can factor P(T)

over this field. If P(T)

has any linear factors and j3 is one of the

roots, a new p is selected. Otherwise, we set ai+\ = p and record the degree of the irreducible factor of P(T) over K(tl,. . . ,I,,c(I,_ . . , cti) satisfied by p as the value of deg(i + 1). Suppose now M is of finite degree over K(tt,. .). In this case the set {tt,. . .} is infinite and we can proceed exactly in the same fashion as in the analogous case of Corollary

2.6.

Corollary

2.8. Let M be a field of characteristic 0 such that M is non-jinitely gen-

erated over Q and the largest purely transcendental extension of Q contained in M is recursive and contains a recursive transcendence base over Q. Then for any set of natural numbers B there exists a weak presentation J of M such that B is Turing equivalent and enumeration reducible to J(M). Proof.

The proof of this corollary follows from the fact that Q has a splitting algorithm

and Corollary

2.7.

Corollary 2.9. Let G be any recursive field of positive characteristic p ,iqith a nonfinitely generated recursive subfield M such that G/M is purely inseparable and M is separably generated (with a recursive transcendence base C) over a finite field K. Assume also that K(C) is recursive. Then for any set B c N, there exists a weak presentation J of G such that B is Turing equivalent and enumeration reducible to J(G).

234

A. Shlapentokhl

Proof. By assumption, and separable

Annals of Pure and Applied Logic 94 (1998) 223-252

G contains

a recursive

and either A4 is of infinite

set C such that M/K(C)

degree

assume that M is of infinite degree over K(C).

over K(C)

is algebraic

or C is infinite.

Since K has a splitting

algorithm,

First and

K(C) is recursive, using the same method as in Corollary 2.7 we can construct a chain A40= K(C) C MI C . . . of subfields of M such that M = Ui Mi, and the index function satisfying

the requirement

of Theorem 2.3. We can then use Theorem 2.4 to reach the

desired result. Suppose now that M is of finite degree over K(C). and K(C) are recursive, Corollary

2.3. Furthermore,

index function

as described

using a method similar to the one used in the proof of

2.7 we can determine

can apply Theorem

Since, C

using the same method as in the second part of the proof of

2.6, one can ascertain that K(C) has a recursive

in Theorem Corollary

Then C is infinite.

the degree of any element

of G over K(C). Thus, we

2.5 to reach the desired conclusion.

The results described above did not use the full strength of the construction described in Theorem 2.3. Below we will describe an example of a field over which we can utilize

the construction

by all the roots of unity. well-known

facts

completely.

Before

concerning

This

we can proceed,

cyclotomic

field will be the field generated however,

extensions

and

we need to state some algebraic

extensions

in

general.

Lemma 2.10. Let M/L be a simple algebraic field extension. Let M = L(a) and let X” +A n_ IF-1 + .. . -I- A0 be the manic irreducible polynomial of CI over L. Let O#x= Crzi aiui, y= CylJ biUi. Then YOPX= Ci&,t ciGli, where OP is a jield OPb,_ 1) is a jixed rational function eration and ci=Ri(AO,...,An-,,ao,...,a,-l,bl,..., over Q or a jinite field (depending on the characteristic of L) in the listed arguments. Proof. The cases of addition

and subtraction

are clear. Next we note the following.

For k> 1, n-1

&k=

c

&(Ao, . . . ,An_, )a’,

i=O

where & is a polynomial over Q or a finite field in the listed variables. This can be seen by induction. The claim is certainly true for k
k>n ak = _

A

n

_,&’

_ . . . _ Aoak-“,

and our claim is true. Furthermore, since xy = c:co2 xk+_,, ukbsam, the lemma is clearly true for multiplication. To show that the lemma holds for division, it is enough to show that x-’ can be written in the desired form. We need to solve the

I Annals

A. Shlapentokh

following

of Pure and Applied

Logic 94

( 1998) 223-252

235

equation:

Zn-2

akd&’ = 1, cc m=O !++s=m a-2

n-l

c in=0

c

akh

k+s=m

n-l

2n-2

1=0

m=O

c c c

c i=O

1,

Pm.j(Ao,...,A"-,)&=

akd,P,,,(Ao,...,A,-I)ff’=

1.

k+s=m

a-2

cc

m=O

akd&,.j(Ao,

. . , A,-I I= &,

k+s=m

where bo= 1, and fii=O

- 1

for i= I....,n

am-Xm,i(Ao....,An-

1)

Thus, we have a linear system in do,. . . ,d,_ I whose coefficients are polynomials in ao, . . , a,,_ I, Ao, . . . , A,_, . Since we know that this system must have a unique solution, we know its determinant

is non-zero

listed above. Thus, we can conclude Lemma 2.11. Let n,mEN,

The following Iet t,,<,

and must again be a polynomial

statements

he primitive

in the variables

that the lemma is true. are true: nth and mth roots of unity, respectively.

Then

Q(i:,)nQ(~m>=Q(4(m.n,), while

where lcm(m,n) Let

denotes

p he a rationul

he natural

[Qtt,d

numbers.

the least common

prime

relatively

multiple

prime

of m and n.

to a natural

number

n. Let k> 1

Then [Q( {r~,,) : Q( ~5,) = p”-’ ( p - 1 )]. On the other hand,

: Q(&,d = pk-‘I.

(See [3], 1.9, pp. 39-45. ) Proposition 2.12. Let F he the infinite extension of Q generated by all the roots of unity. Let E c F be the field generated over Q by the pkth primitive roots of unity, where pi is the ith rational prime under the standard enumeration of primes

A.

236

Shlapentokh I Annals of’ Pure and Applied Logic 94 (1998) 223-252

and k rums through all the natural numbers. Then for any countable collection of of subsets of N there exists a weak presentation J of F such that B, is B1,B21... Turing equivalent and enumeration reducible to fi.. Proof. Let Mi,, = Q(CP;), where show that F has a recursive Theorem

2.3 is well defined

&; is a pith

presentation

and recursive.

defined for this field is a consequence

primitive

root of unity.

under which the index function

We need to described

The fact that the index function

in

is well

of the Lemma 2. Il.

Indeed, let XEF. Let m be the smallest natural number such that x E Q(g,). (Such a field exists by Lemma 2.11. Here we will let l=rl.) If m#l then m= π. We claim that either m = 1 and ind(x) = (0) or id(x) = (r, i,, . . . , ir, tl,. . . , t,.). Indeed, suppose not. Then for some n, x E Q(&), while some prime in factorization of m does not occur in the factorization

of n or the exponent

corresponding

to some prime

occurring in both factorizations is lower in the factorization for n. In this case, m does not divide n and I= gcd(m,n) CM. This leads to a contradiction of our assumption on m, since as has been mentioned above, x~Q(tl). Thus the index function is well defined for F. We next turn our attention to constructing of a recursive presentation of F under which the index function Let [ik denote a pfth primitive {il,kl,.

..,ir,kY,ir+19k,+l}

will be recursive.

root of unity. Then, by Lemma 2.11, for any

CN

we can determine

using the following

rule: - 1). In this case the manic irreducible ik} then d = pi,+, rA+‘-‘(P~L+, 1+1-I - 1). polynomial of &l+,.m+, over K([j,,l-,,. . .,[jL.r,) is (Pi::: - l)/(TFJ~+l 2. Next without loss of generality, assume that ik+l = il. If ~1 >~k+l then d = 1. Otherwise, d = p’l+l -II, and the manic irreducible polynomial of [jl+,,rA+, over K(
ii,,?,) is Tp”+‘-” - [i,,r, =O. Thus, if {(&,r,)}, s= l,... is an effective listing of pairs of natural numbers, using Lemma 2.10, we can construct F = U,“=,T,, where T, = T,_I(&,,), and any a~ T, c K is presented as a linear combination of powers of ci,,,, over T,_ 1. Furthermore, given any element CIE K, presented as described above, we can determine effectively the minimal polynomial of Mover Q and a finite set {(it, r-1), . . . , (i,, I-,)}, where for k # I, ik # iI As we have shown above, if (s,il,rl , . . . , is, r,) is not such that c1E Q( ii, +s,, . . . , &,). the index of g, then CI is contained in a proper subfield of Q([i,,r,, . . . , [i,,r, ) generated by roots of unity. There are finitely many such subfields and their generators can be determined effectively. By Lemma 2.11, each subfield of Q(ii,,r,, . . . , (i,,r, ) is generated by the dth primitive root of unity td, where d is a divisor of p:; . . . p;;. By factoring the minimal polynomial of CI over Q over those fields we can determine if c1 belongs to any of them.

237

A. Shlapentokhi Annals of Pure and Applied Logic 94 (1998) 223-252

We will conclude

this section with the description

main difference between of the index function, Notations

the horizontal

of the vertical construction.

and vertical constructions

as will be explained

below.

2.13. Below we will use the following

will be collections

of recursive

countable

K,F, {E}~cN,

notations.

fields satisfying

the following

1. F= ur,F,. 2. For all natural numbers i, F; c fi+,, and this inclusion is strict. 3. For all natural numbers i, Uz., M,., = 4. For all natural numbers sion Mi,,+i/M,,,

The

lies in the definition

is not trivial and is generated

by a single element

{M;.I}r.r~~

conditions.

i and t, the exten~i.~+l.

4. Ma.0 = K, Ml.0 = F;_ 1. The field K is infinite. Definition 2.14. Let F, fi;;:, M,, be as described

in Notations

let Gen(x) (the generator set of x) be a subset of elements inductively.

2.13. Let x ~fi\l’$l.

Then

of F; which we will define

1. If x E K then Gen(x) = 0. 2. If x~Mi,i\Mi,j_l,

j # 0 and x = ~~=, a,&J, where 1,. . . , XL, are linearly independent over Mi,,j-t), at,. . .a, EMi3(j_l,, and Mi+j is algebraic over Mi,(j_i I, then Gen(x) = lJi=, Gen(a,) U Gen(ai,,), where Gen( CC~,J)is the union of the generator sets of the coefficients

of the manic

irreducible

polynomial

of 2i.i over M,,,;_I and

{%il.

3. If X E Mi,j\Mi,i_ 1, j # 0, and x = relatively prime polynomials in

P(ai,j)/Q(tL,,j), where P( ai.j), Q(cx~,, ) are manic

Clr,j

Ml,(j_l), then Gen(x) is the union polynomials P and Q and {qi}.

over M;,(j_i,

and M,.J is transcendental

of the generator

sets of the coefficients

Let ind(x) = (n, ii, tl,. . . , i,, t,,), where x E F;,,\F;,,_ 1, for all j = 1,. . . , n, (i
over of the and

Lemma 2.15. The following statements are true. 1. Let x, y E F and let op be any field operation such that z =x op y is dejned. Then

Gen(z) 2 Gen(x) U Gen(y). 2. For any XEF, Gen(x) is finite. 3. Function ind is well dejined for all x. 4. In the notations above, [f ij,tj occur in ind(z) then this pair occurs in ind(y ) or ind(x). Proof. It is enough to show that the first statement

of the proposition

holds. The other

statements will follow from the first one by induction and definitions. First of all we would like to note the following. Suppose the first assertion is true for some subfield E of F such that E contains K. Then it is easy to see that the following is true. Let z = R(xl ,...,x,) for some z,xi,. . .,x, E E, where R(xl, . . ,x,) is a rational function in listed variables over K. Then Gen(z) 2 U:=, Gen(x;).

238

A. Shlapentokh I Annals of Pure and Applied Logic 94 (1998)

We will show that the first statement It is clear that the statement

is true by the double

223-252

is true for {x, y} c K. Suppose now that the statement

true for {x, y} c fi:- I, and show that the statement is true for {x, y} c Mi,,_l,

{x, y} EMi,l. Without

10s~ of generality

We have to consider

t>

Assume

fkther

0, and show that the statement

we can now assume that

two cases: Mi,t/Ml,,_I

is

is true for {x, y} c E;;. If the state-

ment is true for {x, y} C fi- I, it is true for {x, y} C Ml,0 = 4-1. the statement

on i and t.

induction

algebraic

{x,y}

that

is true for

EMi,l\Mi,t-1.

and transcendental.

We will

start with the algebraic case. Since XEM~.(, x = C,‘__, u,$~, where 1,. . .,all are linearly independent over Mi,(,_i), al,. . . , a, EMi.(t-1). Similarly, y = ~~==, b&“, where boy..., b, EMi.(l-I 1. Let Ao, . . . ,A, be the coefficients of the manic irreducible polynomial of qr over Mi.t_1. Then, by Lemma 2.10, z = c:=, R#,, where R, is a rational in a~, . . . , a,, bo, . . . , b,, A 0,. . . ,A,.. Then Gen(z) 2 Gen(Ro) U Gen(RI ) U . . . U Gen(R,) U Gen(c+) C Gen(ao) U.. . U Gen(a,) U Gen(bo) U .. . U Gen(b,.) U Gen(cci,,)& Gen(x) U Gen(y) by the observation at the beginning of this proof, induction hypothesis

function

and the assumption We next consider

that {x, y} c M~,,\kfi,~-l. the case of Mi,t/Mi.,-I

being a transcendental

extension.

In this

case,

where PI and Q,, and 9 and Q2 are manic relatively

prime pairs of polynomials

in tqf

over Mi%,_1. Then z = P(c+)/Q(E~,~) where P, Q are manic relatively prime polynomials in cl;.! over Mi,,_, and all the coefficients are rational functions over Q or a finite field (depending

on the characteristic

of K) in coefficients

of PI, Ql, 4, Q2. Thus, by obser-

vation at the beginning of the proof and the induction hypothesis, Gen(P(ai,t)/Q(ai,t)) will be a subset of the union of {Cli,l} and the generator sets of all the coefficients of

PI, Ql, 9, Q2. By assumption on i and t, Cli,l is an element of the generator set of x or y. Thus, again we obtain Gen(z) c Gen(x) U Gen(y). Theorem 2.16 (The Vertical Construction). Let K, F,fi,Mil be as in Notations 2.13 and let ind be as in Definition 2.14. Assume there exists a recursive presentation j of F under which all the fields listed above are recursive and ind is a recursive function. Let B1
2.3 using the index function

from Definition 2.14. Under our assumptions on the index function and j, d is again a recursive set, while 4, K, gX, 9, are total recursive functions. Suppose now a characteristic function for Bi is given. Let (j(x), m)~ &‘. We want to determine whether this pair is an element of J(fi). First of all, we compute the index of j(x) to determine whether j(x)~j(F;:). If it is so, then

ind(j(x)) = (n, il, tl , . . . , in, t,),

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A. Shlapentokh I Annals of’ Pure and Applied Loyic 94 (1998) 223-252

where i,
functions

for BI, . . . , B,_ 1.

suppose the characteristic

function

(j(x),m)~J(fi)

of all, we locate an element pair (i,n).

if and only if for s = 1,. , n, Y = 1,. . . , ts, mi,,, = 1 Note that for all j = 1,. . , i, Bj
Index of any element

for J(F;)

is given. Let n EN.

such that ind( j(x))

of j(Mi,,\Mi.,_i)

has occurrence

will have an occurrence

First of a

of such

a pair. Once such a pair has been located, check the entry m;.n. If it is 1, n~Bi, and n $!B; otherwise. To show enumeration reducibility, consider the following. For any natural number n, any enumeration of F, will contain a pair (j(x), m), where ind( j(x)) has an occurrence of the pair (i,n). Using an argument able to decide if n belongs

similar to the one above, at that point we will be

in the listing of Bi.

3. Turing separability of non-finitely generated fields We shall next address the issue of obtaining results in the spirit of Theorem As above, we start with some technical preliminaries.

1.4.

Lemma 3.1. Let G/F be an arbitrary field extension. Let h(z) be a rational function over G. Assume for infinitely many aEF, h(a)E F. Then h(z)EF(z). (See [9], Lemma 2.3, p. 233.)

Lemma 3.2. Let No c NI ‘. . c Nk c ‘. . be a chain of field extensions, where each extension is generated by a single element. Assume N = IJF,, Ni is recursive and we have the following recursive functions: l For each x EN, ind(x) = t such that x E N,\N,_ 1. l

For each natural number &g(t) is the generator of Nt over N,_I, d(t) is either 0, if NtIN,_, is transcendental, or it is equal to [N1 : Nr_ I].

Let R(zl,. . . ,z,.) be a rational function over N presented as a ratio of two polynomials over N. Then there is a recursive procedure to compute the smallest t such that R is a rational function over Nt.

Proof. Let R(zl , . . . ,zr ) be presented as described in the statement of the lemma. Then using the index function we can determine an upper bound for a desired value of t. Thus the problem is reduced to the following question. Suppose R is a function over N1. Can we determine (in a recursive manner) whether or not R is a function over N,_i? If d(t)>0 then we can use [8], Theorem 2.2, p. 738, part 2, to answer the question. If d(t) = 0, then using g(t) we can rewrite all the coefficients of R as rational functions in g(t) over NIP 1 and then consider R as a function in variables g(t ),zI, . . ,z,. Thus, the question we will need to answer is whether or not R is a function of g(t) over Nt_i(zl,.. part 3.

..zr).

To answer this question

we can use [8], Theorem

2.2, p. 738,

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A. Shlapentokhl Annals of‘ Pure and Applied Logic 94 (1998) 223-2S2

Corollary 3.3. Let N be as above. Assume further that G is an extension of N generated by a single transcendental element T of G or by a single element a of G which is algebraic and separable over N. In case G is algebraic over N, let to be a natural number such that [N(a): N] = [N,,(a): Nl,]. Let R(zl,. . .,z,.) over G be a rational function over G. Then the following statements are true. 1. There is a recursive procedure to determine whether R is a rational function over N. 2. If R is not a rational function over N and G/N is transcendental, there exists a recursive procedure to determine the smallest natural number t such that R is a rational function over N1(T). 3. If R is not a rational function over N and G/N is algebraic, there exists a recursive procedure to determine the smallest natural number t > to such that R is a rational function over N,(a). Proof. The first statement

of the corollary

follows

directly

from [8], Theorem

p. 738, parts 2,3. To see that the second and third parts of the corollary that, assuming t > to in the algebraic case, the extensions will be transcendental if N,/N,_r was a transcendental Nt( T)/N,_i (T), N,(a)/N,_I (a) will be algebraic algebraic. Furthermore, we can use a method

N,(T)/N,_,(T), extension.

of the original

2.2,

are true, note

N,(a)/N,_I (a)

On the other hand,

degree if N,/N,_ 1 was

the generators will remain the same in either case. Thus, similar to the one used in Lemma 3.2 to reach the desired

conclusion.

Lemma 3.4. Let R1 be an injinite integral domain with the quotient field fi. Let F2 be another field of the same characteristic. Furthermore, assume that it is not the case that RI G~~Fz, and let {Hl(zt ,..., z,.),..., H&z, ,..., z,.)} be a finite set of rational functions over F,Fz such that for each i = 1,. . . ,m, Ht(z1,. . . ,z,) is not a rational function in ~2,. . . ,z,. Then there exists infinitely many elements b E RI \F2 such that for each i, Hi(b,. . . , zr) will be well defined and will not be a rational function over F2. (See [9], Lemma 2.4, p. 234.) Corollary 3.5. Let M/N be a field extension, where N is an infinite jield. Let y @M be such that it is either algebraic and separable or transcendental over M. Let z,.)} be a finite set of rational functions over M(y) such {HI (ZI,...,zr),...,Hm(z~,..., that each function in the set is not a rational function in ~2,. . . ,z,. Then there exist infinitely many a E N(y)\M such that for all i = 1,. . . , r, Ht(a,zl, . . . ,z,.) is a well-defined rational function over M(y) but not over M. Proof. We would like to apply Lemma 3.4 to prove this corollary. In our case RI = Fj = N( y), F2 = M, and Fi F2 = M( y). To see that this identification is valid, it is enough to show that N(y) is not rationally separably less than M. By Theorem 1.3, N(y) drsM would imply N(y) is a purely inseparable extension of N(y) nM CM. However, since y

241

A. ShlapentokhI Annals of Pure and Applied Logic 94 (1998 i 223-252

is either separable or transcendental

over h4, it is not purely inseparable

Thus, N(y) is not a purely inseparable

extension

over N(y )rlM.

of N(J~) 0 M.

Theorem 3.6, Let K c E c F be recursive fields. Then the following statements are true: 1. Assume E is not finitely generated over K, F is not finitely generated over E, and extensions K - E - F satisfy the conditions of Theorem 2.16 with E playing the role of F, and F = F2. Then for any two subsets A T A there exists a weak presentation J of F such that J(F) ZT B and J(E) q A. 3. Assume there exist infinitely many fields MI c Ml . . . such that K = MOc MI and E = Ur, Mi and the chain of fields Mi satisjies the assumptions of Lemma 3.2. Assume F is finitely generated over E, and it is not the case that F G,, E. Then for any B c N and for any r.e. A c N, there exists a weak presentation J of F such that J(E) q B and J(F) z-T join(A, B). 4. If E and F are both jinitely generated over K, and F is not rationally separably less than E then for any two r.e. sets A
required here is a combination

and the construction used in [4]. We will proceed in the following which the index function

manner.

and the procedure

of the construction

in Theorem

Let j be a weak presentation for determining

2.3

of F under

the smallest constant

field

for rational functions are recursive. We will construct a recursive set D of the following pairs. The first element of the pair will be a natural number and the second element of the pair will be a finite sequence Let L(n) be a recursive will describe

consisting

for even numbers

and “2” ‘s.

A. In the course of the construction we function f from j(F) into natural will also define a function inv from natural numbers into the space of in countably many variables over j(F). Initially inv will be defined only. At the “end” of the construction, inv will be defined for all We will initially set inv(2n) =xn. For those natural numbers whose function

below, we will construct

numbers and we rational functions

of “1”‘s

listing

a recursive

natural numbers. inv has been defined, we will also define finitely many finite sequences

so that the pairs

consisting of the above mentioned natural number and one the sequences will become elements of a. Given a natural number m whose inv has been formed already, the corresponding set of sequences will be defined in the following fashion. First of all, by Lemma 3.2 we can compute the smallest t such that inv( m) is a rational function (or an element of) over M,. (At the same time we will make sure that the inv under

242

A. Shlapentokhl Annals qf’ Pure and Applied Logic 94 (1998) 223-252

consideration

does not contain

Theorem

any extraneous

variables.

2.2, p. 738, part 3. The same procedure an inu has become an element

after a substitution corresponding

to inu(m)

This can be done using [8],

can be used below to determine of j(F).)

will consist of all the possible

if

Then the set of sequences sequences

of length t whose

entries are “1”‘s and “2” ‘s. For each n, we will initially

add a pair (2n,c) to 9?‘, where s will denote the sequence

of length 0. We will also define &‘s

for op E {+, -, x,/}

functions on a, whose restrictions will be translations of the construction will involve the following steps: Step 6k + 1: Choose an element Pick the smallest far. Set f(j(b)) possible

odd number SI ,

The first part

b E F such that f (j( b)) has not been defined yet.

2r + 1 which has not been used in the construction

= 2r + 1, inu(2r + 1) = j(b).

sequences

which will be total recursive

of field operations.

t = ind(j(b)),

Assuming

. . . , s21 of length t consisting

of “1”‘s

construct

so

all the

and “2” ‘s. Add all pairs

(2r+ l,SZ!) to a. (2r+ l,Sl),..., Steps 6k + 2,6k f 3,6k + 4,6k + 5: Given a pair (nl,st),(n2,s2)

E 99, we will call

this a matching pair under the following circumstances. Let 1 be the minimum of the sequences’ lengths. Then for all i = 1,. . . ,I, the ith entries of the sequences match. Let be such that for some op, $,P((nlr~l),(n2,~2))

(nt,sr),(n2,~2)~~

has not been defined

yet while inu(n,),inu(nz) have been defined already. If (nt,st),(n2,s2) E ?J do not constitute a matching pair or if op = “/“, while inu(n2) =j(O), then set 90J(nt,si ),(Q,s~)) = (0, E). Otherwise, compute U = inu(nl )op ino( Let t be the smallest possible natural number such that U E MI. Then let s be the sequence of length t whose ith entry is equal to the ith entry of sI or ~2. (At least one of the sequences has the length

greater

or equal to t.) If 9

then set ~~~((nl,sl),(n2,s2))=(n,s). not been used in the construction (n, w), where w is a sequence Set 90J(nl,si),(n2,~2))

= (n,s),

contains

a pair (n,s),

= U,

Otherwise, let n be an odd number which has so far. Set inu(n) = U. Add all pairs of the form

of length t whose entries where s was described

are “1”‘s above.

J’(U)=n. Step 6k + 6: Let n = L(k). Then we need to find an element the construction (a) Substitution

SI or 92

with h(n)

and “2”‘s

If U Ed

to g.

then set

a E j(E), not used in

so far satisfying the following requirements: of a for ~2~ would not make any two previously

unequal

rational

functions

in the range of inu equal. not make any rational function over (b) Substitution of a for XX,, would j(M,)\j(M,_r ) in the range of ino, a rational function over j(A4,_1). (c) Substitution of a for ~2~ would not result in a zero in the denominator of any rational function in the range of inu. (d) After the substitution no variable different from ~2~ will disappear from any rational function in the range of inu. By Lemma 3.1, we know that only finitely many a gj(E) will fail to satisfy the second requirement. An argument similar to the one found in the proof of [4], Lemma 2.2, p. 202, will demonstrate that only finitely many a Ed can fail to satisfy the other three requirements. Furthermore, Lemma 3.2 assures us that we can determine

A. Shlapentokh I Annals oj’ Pure and Applied Logic 94 (1998

when Requirement into all the rational after substitution,

243

i 223-252

(b) is satisfied. Once the desired a has been located, functions

in the range of inv which have an occurrence

for some m, inu(m)=j(b)

Ed,

It is not difficult to see that S is a recursive tions on 2. Furthermore,

substitute

using arguments

then define f(j(b))=

it

of x2,,. If m.

set, and Y(,,,‘s are total recursive

func-

similar to the ones used to prove [4], Theo-

rem 2.1, p. 201, one can show that j’ o j is a weak presentation

f(E)=T A. (foj(b),s) E 8,

of F and

Furthermore, due to Requirement (b) in the steps 6k+6, for all b E F, implies the length of s is equal to the index of b in F.

The second part of the construction is very similar to the constructions in Theorems 2.3 and 2.16. That is, for b E F we define J(b)=( f oj(b),s), where for i less or equal to the length of s, the ith entry of the sequence and it is 2 otherwise. theorems

Arguments

will show that J(F)

similar

is equal to 1 if i E B

to the one used in the above mentioned

=_TB.

3. The construction required here is similar to the one described above, the difference being that values of F\E will be assigned to the variables whose indices were enumerated

in some effective listing of A.

First of all, without loss of generality

we can assume that F is generated

over E by

a single element which either transcendental or separable over F. Indeed, since it is not the case that F d TsE, by Theorem 1.3 the extension F/E is not purely inseparable. Thus, we can assume that over E, F is generated

by {XI,. . ,x,, x,/31,. . , /3,.}, where

Xl,. .,x,, m 20, are algebraically independent over E, ct is separable and of degree greater or equal to 0 over E(xi, . . . ,x,), and ,91,. . . , fir with r > 0 are purely inseparable over E(xl , . . . ,x,, r). Furthermore, the rational separability condition implies that either m > 0 or r is not of degree 0 over E. Thus, we can construct for E(xl ) or E(cc) (if m = 0) and then extend the one described

in [7], Lemma

of F = E(T).

or F = E(cY) where x is separable the index function

Let z = P( r)/Q( T), where P, Q are manic

over E. Then define i&(z)

similar

2.12, p. 1075. From now on we will assume

F = E( T), where T is transcendental, Our first task will be to extend

the required presentation

it to F using a procedure

to be the maximum

over E.

to F. First consider relatively

of the indices

to that

the case

prime polynomials

of the coefficients

of

P and Q. In the case, F = E(a), first let to = &d(a), where M1,, is the field containing all the coefficients of the manic irreducible polynomial of CIover E. If z = a0 + at c( + . . + a,_] CC-~‘, where n = [F : E], then the index of 2 will be the maximum indices

of a~, . . , a,_~, sl. Using an argument

similar to the one used in the proof of

Lemma 2.15, one can show that for any field operation ind(z) d max( i&(x), ind( y)). We will now proceed in the following

of the

manner.

op and x, y,z E F,z =x op y +

Let j be a presentation

of F under

which the index function and the procedure for determining the smallest constant field for rational functions, as described in Lemma 3.2 and Corollary 3.3, are recursive. Further, let %‘,L(n), f,inv, &,, play the same roles they played in the proof of the preceding part with some modifications described below. Again, we will start with setting inv(2n)=x,. Given a natural number m whose inv has been formed already, the corresponding set of sequences will be defined in the following fashion. First of

A. ShlapentokhlAnnals

244

of Pure and Applied Logic 94 (1998) 223-252

all, by Lemma 3.2 and Corollary is a rational

function

3.3, we can compute the smallest

over Mg or M,(T)

the smallest t > to, where to = ind(a)),

or M,(a) depending

(in the last case we are looking for on whether inu is a rational

over F\E

or over E and the nature of the generator

or if F/E

is a transcendental

inu(m) will consist

extension,

of all the possible

t such that inu(m)

then the set of sequences sequences

function

of F over E. If inu(m) E j(F) corresponding

to

of length t whose entries are “1”‘s

and “2” ‘s. If F = E(a) and inu(m) @j(F)

then the set of sequences

will consist of all the possible

of length mux(to, t) whose entries are “1”‘s

sequences

and “2”‘s. We will call the length of the sequences of inu(m). If F/E is transcendental,

corresponding

for each n, we will initially

corresponding to h(m)

to m

the index

add a pair (2n,&) to J%?‘, where

E will denote the sequence of length 0. If F/E is algebraic, for each n, we will initially add pairs (2n, si)i = I,_,,,zooto 8, where $1,. . , s210are all the possible sequences of length to consisting

of “1”‘s

and “2” ‘s. The first part of the construction

following steps. Step 6k + 1: Choose an element

b E F such that f( j(b))

will involve

the

has not been defined yet.

Pick the smallest odd number 2r + 1 which has not been used in the construction so far. Set f( j(b)) = 2r + 1, inu(2r + 1) = j( b). Compute t = ind(j(b)). Next construct all the possible

sequences

~1,. . . ,sp of length t consisting

“1” ‘s and “2” ‘s. Add all

pairs (2r+ 1,st) ,..., (2r+ l,s2!) to B. Steps 6k + 2,6k + 3,6k + 4,6k + 5: Given a pair (nr,sr),(n2,~2)~B inv(nl),inu(nz) have been defined already assume that for some op,

has not been

defined

yet. If (nt,sr ), (Q,s~) E 9J do not constitute

such that

a matching

pair

or if op = “/“, while inv(n2) = j(O), then set YO,,((nl ,sr ), (122,s~)) = (0, E). Otherwise, compute U = inu(nl)opinu(n2). Let t be the index of U. Then let s be the sequence of length t whose ith entry is equal to the ith entry of sI or ~2. (As in the preceding construction, (n,s),

with h(n)

odd number U ej(F)

either s1 or s2 is of length greater or equal to t.) If .B contains = U, then set 9$,((nl,sl),(n2,s2))

= (n,s).

which has not been used in the construction

then set f(U)

let n be an

so far. Set inu(n) = U. If

=n. Add all pairs of the form (n,w),

of length t whose entries are “1”‘s

Otherwise,

a pair

where w is a sequence

and “2”‘s to %?. Set 9$,((n~,s~),(n~,sg,))=(n,s),

where s was described above. Step 6k+6: Let n =L(k). Then we need to find an element a Ej(K( T))\ j(K)

in the

case F/E is transcendental or element a E j(M,,(a))\ j(M,,) in case F/E is algebraic, not used in the construction so far satisfying the following requirements. (a) Substitution of a for ~2~ would not make any two previously unequal rational functions in the range of inv equal. (b) Substitution of a for xzn will make every rational function in the range of inu where substitution took place a rational function over j(F)\j(E). (c) Substitution of a for ~2~ will not decrease the index of any rational function in the range of inv.

A. Shlapentokhl Annals of Pure and Applied Logic 94 (1998) 223-252

of a for xzn would

(d) Substitution rational (e)

not result in a zero in the denominator

of any

in the range of inv.

function

After the substitution

no variable different from ~2,~will disappear from any rational

in the range of inv.

function

By Corollary

3.5, we know that there are infinitely

which will satisfy the second requirement. many a’s in j(K(T))\j(K)

finitely

245

requirement.

As before,

many a’s in the specified

On the other hand, by Lemma

or j(MlO(a))\j(Mt,,)

an argument

similar

set

3.1, only

will fail to satisfy the third

to the one found in the proof of [4],

Lemma 2.2, p. 202, will demonstrate that only finitely many a’s in j(F) can fail to satisfy the remaining three requirements. Furthermore, Lemma 3.2 and Corollary 3.3 assure that we have a recursive

procedure

to determine

satisfied. Once the desired a has been located, substitute in the range of inv which have an occurrence k(m)

=j(b)

when all the requirements

are

it into all the rational functions for some m,

of x2,,. If after substitution,

cj(F),

then define f(j(b)) =m. As in the previous proof, it is not hard to see that 98 is a recursive

are total recursive

functions

on ~8. Also as above, using arguments

set, and POP’s

similar to the ones

used to prove [4], Theorem 2.1, p. 201, one can show that f oj is a weak presentation of F, f( j(F))

=TA,

and f( j(E))

for some b E F, (f o j(b),s)

is recursive,

is a part of Theorem

The next

of results

will

K <,, M, under the assumption

1.4.

consider

weak

generated Auf(M)

of a pair of fields

is drastically

over a finitely

gen-

different from the one we

when both fields were finitely generated.

Notations 3.7. For the remainder algorithm,

and possessing

presentations

that both of them are algebraic

erated field. Here we will see that the situation

a splitting

(c), if

is identical to the second part of the construction

4. This statement

have observed

due to Requirement

E A?, then the length of s is equal to the index of b in F.

The second part of the construction above.

group

Furthermore,

algebraic

a recursive

of the section, A4 will denote a recursive

and normal

set of generators

over a finitely

the automotphisms

field with

recursive

field T

over T. Note that T as a field finitely

over Q or finite field will have a splitting will denote

generated

algorithm

also.

group of M. K will denote

a recursive

subfield of M with a splitting algorithm, containing T. AM(K) will denote the subgroup of Aut(M) containing the automorphisms of M restricting to the automorphisms of K. j will denote a recursive presentation of M under which the splitting algorithms and fields mentioned notations between of M.

above are recursive.

for the elements

For the sake of convenience

of At(M)

these two sets is clear.) Finally,

We will start with a brief discussion fields into their algebraic

closures.

we will use the same

and Aut( j(M )). (The natural .I will denote an arbitrary

of well-known

correspondence

weak presentation

facts concerning

embeddings

of

246

A. ShlapentokhlAnnals of Pure and Applied Logic 94 (1998) 223-252

Lemma 3.8. The following statements are true. 1. Let k be a jield. Let k be its algebraic closure. Let u E k and let f(X) manic irreducible polynomial of CLover k. Let o: k-k

be the

be an embedding. Let

p E k be a root of a( f ). Then o can be extended to an isomorphism from k(a) to o(k)(P)

sending CIto /?. (See [5], pp. 170-171.)

2. Let k be a field, E an algebraic extension of k, and o : k + L an embedding of k into an algebraically closed field L. Then there exists an extension of o to an embedding of E into L. (See [5], Theorem 2, p. 171.) 3. Let T be a jield, let k be an algebraic extension of T, let M be an algebraic extension of k, normal over T. Then M is normal over k. Further, let B be an embedding of k into its algebraic closure keeping T ftxed. Then o(k) c M and a can be extended to an automorphism of M. (See [5], Theorems 4,5, pp. 175-176).

Lemma 3.9. There is an efSective procedure to determine, given ~(1,.. , , c(kEM, and B ,, . . . , /Z$EM, whether there exists a E Am(M)

Proof. First of all, assuming the characteristic a a as described

such that a(a;) = /?; for i = 1,. . . , k.

of M is p > 0, we note that there exists

in the statement

of the lemma if and only if for any ~1,. . . , rk E N, there exists a z E Aut(M) such that $a:” ) = pi”‘. Indeed, suppose r(~li) = pi. Then, of course, r( CL:”) = /?p” . Conversely,

suppose t( c$ ’ ) = /?:‘I and r( ~1,)= I+. Then yp” = @’

or (yJ/$)J”’ = 1. In characteristic

p >O this equation

Yi = Pi. Thus without

we can assume

T, replacing can determine elements

loss of generality

them if necessary the necessary

has only one solution:

that al,. . . , j3k are separable

by their p’th powers with a sufficiently

r by examining

the irreducible

over T, its manic

above. If clp’ is separable

1. So

polynomials

irreducible

over

large r. (We over T of the

polynomial

over T

will not have any multiple roots.) Furthermore, al can be mapped to /?I over T if and only if these elements are conjugate over T, i.e. have the same manic irreducible over T. This can be determined by examining their irreducible over T. Assuming we can map ~1,. . . , cti to /?I,. . . , pi, we can extend

polynomial

phism of T(al,...,

ai) onto T@,..

.,/Ii) to an isomorphism

polynomials any isomor-

of T(cc~,.. .,c~i,tli+l) onto

T(P 1,. . . ,/?i,/&+l) if and only if the irreducible polynomial of tli+i over T(al,. . .,ai) is mapped to the irreducible polynomial of pi+, over T(p,, . . . , pi). Since all the extensions are separable, we can effectively

factor polynomials

over T( ~1,. . . , ai+1 ) and over

T(fi1,. ,/$+I), and thus determine if this requirement is fulfilled (For a discussion of splitting algorithms under separable extensions reader to [2], Chapter

for cl;+1 and pi+,. we again refer the

17.)

Before we proceed with the next theorem we would like to discuss some aspects of normal extensions and normal closures. (For a detailed discussion of normal extensions see [5], Chapter VII, Section 3.)

A. ShlapentokhlAnnals

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247

223-252

Definition 3.10. Let T c F c K be fields such that K is algebraic over T. Then the normal closure G of F in K over T is the intersection of K and the normal closure of F over T. Lemma 3.11. Let T, F, K, G be as above. Then the jollowing statements are true. 1. IfF is of .finite degree over T then G is also of ,finite degree over T. 2. Lf o : K -+ K is an automorphism of K leaving T $x-ed. Then (r restricted to G is also an automorphism. 3. !f z is an embedding of K into its algebraic closure leaving T jixed and z restricted to G is not an automorphism of G, then T is not an automorphism of K. Proof. 1. This follows from the fact that if F/T is finite, then the normal closure of F over T is also of finite degree over T. 2. Let F be the normal closure of F over T. Next note the following o(G) c P and o(G)ca(K)=K. Thus, o(G)cpnK=G. Similarly, a-‘(G)cG, so that Gso(G). Therefore,

G = CJ(G).

3. This statement

follows from the preceding

one.

Theorem 3.12. Suppose Aut(K)c Aut(M) is of ,$nite index in Am(M). Then for any weak presentation J of M, J(K) 6-r J(M). Conversely, zf Aut(K) c Aut(M) is of injnite index in Aut(M) then there exist inhnitely many weak presentations J of M such that J(K) is not Turing reducible to J(M). Proof. First assume Aut(K) of representatives

is of finite index in Aut(M). Let {al,. , .,cJ~} be a set of the left cosets of Aut(K) in Aut(M), excluding Am(K) it-

self. We claim there exists i c K such that K/T is a finite extension, a~Aut(M)

and for any

such that a(K)#K,a(K)#K.

cr,( xi) $ K. Let I’? be the normal

For each i=l,...,k, let U,EK be such that closure of T(a1 , . . , c(k) in K. Next let r E Aut(M)\

Am(K). Then, for some i, T = cr;of, where f E Aut(K). Then r(E) = oi(f (K)) = a;(K) @K. Here we have used the fact that f(K) =k, since E is normally closed in K. Since I? is finitely generated, it possesses a splitting algorithm, and furthermore h4 is normal over Z?. Next note the following. Let czE K. Let 0 EM be a conjugate of a over I?. Then fl E K. Indeed, suppose not. Then consider an automorphism (T of M over I? which sends a to /?. (Such an automorphism exists because M is normal over I?.) Then a(i) =x, but a(K) # K. This is a contradiction of the argument above, and thus the statement is true. So suppose J is a weak presentation a polynomial

satisfied

by n over J(K).

of M and n E J(M)

is given.

This can be done effectively

First we find because

I? is

finitely generated. We can then determine what irreducible polynomial P is satisfied by J-‘(n) over E? and find effectively all the roots of P in K. If there are none, 17$2J(K). Otherwise, n E J(K), by the argument above. Suppose now Aut(K) is of infinite index in Aut(A4). Then for any subextension F of K which is of finite degree over T, there exists rs E Au?(M) and fi E K such

248

A. Shlapentokh I Annals of Pure and Applied Logic 94 (1998) 223-252

F is the identity. Indeed, since F is of finite degree over T,F has only finitely many embeddings into its algebraic closure which leave T fixed. Since AZ&(K)has infinitely many left cosets, there exist ol, c2 E Aut(M)\Aut(K) that a(/?) #K

and cr restricted

such that they represent embedding

to

two different

of F into M keeping

left cosets of At(K)

and restrict to the same

T fixed. Since, ~1 and CQ do not belong to the same of K, but the restriction

left coset, cr = a;’ o ~1 is not an automorphism

of cr to F is

identity. Furthermore,

we note the following.

over T and normally automorphism

Let F be a subfield

closed in K. Then any automorphism

of F. Thus, any two embeddings

is not an automorphism

of K of finite degree of K will restrict to an

61, c-12of F into A4 such that cry I o 01

of F will extend to embeddings

of M which are in different

left cosets of A&(K) in Aut(M) only. Let t be any embedding of K into M. Let rF be the restriction and cr be as above. Finally,

let G be the normal

closure of F(P)

of r to F. Let /I over T in K. Then

r and r o CJ are extensions of rF sending G to different images. Indeed, since 0 does not move F, TO o is an extension of r,V. Furthermore, consider r-’ o (r o a) restricted to G. This composition of G, because

is equal to cr. But r~ restricted

0 sends at least one element

zoo(G). In view of the above discussion

consider

to G is not an automorphism

of G, namely

/?, out of K. Thus r(G) #

the following

full binary

tree of embed-

dings. Let (j(F~),zo,~), where Fo is a normally closed in K finite extension of T, and r. is an embedding of j(F0) into j(M), be the root of this tree. Let the ith level consist of pairs ( j(F;: ), Zij ), j = 0,...,2’ - 1, where F; is a finite non-trivial normally closed in K. each

extension

ri,2, and ri,zj+i be two extensions of ri- r,j, such that rTij+, 0 ri92j = oi_i, is the identity on j(fi_i ), and ci-1 (j(fi)) @j(K). Further, let lJ fi = K. Given the tree above, we can construct sequences

Indeed, let {si} be a sequence satisfying

a one-to-one

of O’s and l’s and a set of distinct the following

l

r restricted

l

Assume

to j(F0)

r restricted

of 6-i

j(M).

correspondence

left cosets of Aut(j(K))

Let

where oi_r

between

all the

in Aut(j(M)).

of O’s and 1‘s. Then let z be an automorphism

of j(M)

conditions. is equal to ~0. to j(F;-i)

is ri_i,j.

If Si = 0 then r restricted

Otherwise, r restricted to j(F;:) is ri,zj+i. First of all, since U I;;: = K and for all i, j, Zi,zj,Zi.2j+l are extensions

to j(F;)

is ri,2j+

of ti,j, there exists

an embedding of K into M satisfying the requirements described above. Further, since M is normal over T, this embedding can be extended to an element r E Aut(j(M)). IS in the same left coset of Aut(j(K)) as r. Then Secondly, suppose + E Aut( j(M)) II/ = r o f, where f induces automorphism on K, and therefore on fi for every i. Thus, for every i, $(E) = z(E). Furthermore, suppose $ belong to a different coset of Aut(j(K)). Then for some i,T-‘$ is not an automorphism of fi. Hence, for this i, $(F;)#z(F;:). Finally, let {Ui} b e a sequence of O’s and l’s different from {si}. Let 4 be an element of Aut(j(M)) corresponding to this sequence. Let k be the smallest index such that Sk # Uk. Then, by construction of the tree, restrictions of t and 4 to

A. ShlapentokhIAnnals

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above, z and 4 must

to different left cosets of Aut(j(K)).

Consider

now a recursive

are recursive

presentation

fields with a splitting

j : M -tN

algorithm.

Lemma 3.9, for every i, we can effectively phism tree described 1. Let PO Ed

above following

under which j(T),j(K)

Given

construct

the first i-levels

Given

an element

and

of the automor-

the steps below.

such that /& has a conjugate

over j(T)

which

know such a @s exists and it can be located by a systematic of j(K).

and j(M) on K,M,

our assumption

of j(K)

determine

its manic

is not in K. We search of elements

irreducible

polynomial

over j(r). (This can be done since T is recursive and has a splitting algorithm.) Since M is normal over T all the roots of this polynomial must be in j(M). Find all the roots and determine

if any lie outside

of j(K).

Assuming

we found the

desired PO, let yet = JO,. . . , yor be all the j( T)-conjugates

of ,/I0 in j(K).

j(Fo)= j(T)(yol.. ..,yo,.) and note that j(Fo) is normally

closed in j(K).

be a conjugate

of fro not in j(K).

Using

EM such that a map sending aIn,..., CXO~ automorphism of j(M). 2. Assume

we have constructed

Then let Let ~(01

the procedure from Lemma 3.9, find yoi ----)clg;, i = I,. . .,Y will extend to an

Fo,. . . , F;_ I together with the appropriate

embeddings y/i,. . . , y/,$ of F,

~0,. . . , zi_-l,zJ. Assume further that we have a set of generators over F,_1 for all 1 = 1, . . . , i - 1 and the embeddings are presented by their actions on these generators. Our next task is to find pi E j(K) such that there exists a ci E Aut(j(M))

sending

PI outside j(K)

and keeping j(&_ 1) fixed. Since we know

such a fli E j(K) exists, using Lemma 3.9 again we can locate the desired element by a systematic search. Once bi is found, we locate its K-conjugates over j(T) to be included

in j(4).

Finally,

to make sure that Uz,

next element b in some recursive conjugates

listing of j(K)

over T in K to the set of generators

j = 0,2’ - I, we let zi,zj+ I to be any extension

F;: = K, we will also locate the

and add this element and all of its for j(F;)

over j(fi_ I ). Further, for

of zl.j which is a restriction

element of Aut( j(M)). This extension can be constructed using Lemma 3.9. Finally, let ri,zi = ri%zj+t o Gi.

of some

via a systematic

search

This tree, of course, is not unique. Its’ construction depends on the choices of extensions of ti-1.i to F; for each i and j. The choice of F; and consequently of 0i-t is also not unique.

On the other hand, given an effective listing of j(M)

choices completely

deterministic

by requiring

we can make these

that at every step the smallest

suitable

codes are utilized. Let B be any set of natural numbers, and let z be the automorphism of j(M) corresponding to the graph of the characteristic function of B. For any x E M, define J(x) = T o j(x). Note that J(M) = j(M) is still a recursive set. We claim that J(K) =_TB. First suppose we have the characteristic

function

of B.

Let m EJ(M). We can identify all the K-conjugates of j-‘(m) over T. If there are no such conjugates then m $!J(K). Assume the opposite. We can then identify an i such that all the j(K)-conjugates of m belong to j(E). (We have a recursive set of generators for all j(fi)‘s. Thus, an appropriate i can be found by a dove-tailing search of all j(l;;.)‘s). Finally, using the characteristic function of B, we can construct the

A. Shlapentokh I Annals of Pure and Applied Logic 94 (1998) 223-252

250

restriction

of T to j(F;)

and determine

whether

of m are

any of the j(K)-conjugates

mapped to m. Conversely,

suppose we have the characteristic

i E B. Indeed,

assume

mined whether

function ) in J(K),

of j(fi ) over j(&t

ing the images of generators

that using the characteristic

function

or not 1,. . . , i - 1 E B and thus determined

of J(K).

Then by examin-

we can determine of J(K),

whether

we have deter-

that r restricted

to fi-i

is

equal to ri-i,j for some j E N. Let yil,. . . , yir be the generators of fi over 6-i. If for at least one s, either ri,zj+i( j(yis)) $ZJ(K) or ri,u( j(y,)) +!J(K), we can determine Z'S restriction

to 6 and whether or not i E B. So suppose that this is not the case, and for

,...,r,Pso=zl.?j+i(j(~is)),Psi =~i,2j(j(~i~))~J(K)=5oj(K). On the otherhand, J-‘(/&t ) should be a conjugate of yis over 7’ for all s = 1,. . . , Y and all 1 = 0,l. Thus, J-‘(fist) is a K-element with a conjugate in fi. But fi is normally closed in K, and s=l

thus J-‘(&)EF, contains

and r-‘(/&)~

both ri,Zj+i( j(E))

ri.Ij+i( j(E))

C: ri,zj( j(E))

j(4).

Therefore,

and ri,zj( j(E)). or ri,zj(j(F;.))

the image of r restricted

This would

2 ri92J-i( j(E)).

however

imply

to j(fi)

that either

We know by construction

that

we cannot have t;,2j+i (j(6)) = t,,lj( j(e)). Thus, either ri,zj+i (j(E)) C ri,zj( j(e)) or ri,zj( j(E)) c zi,2,+l (j(fi)),where inclusions are strict. Neither strict inclusion is possible, however,

since both images must be of the same finite degree over j( T).

Next we have the following

lemma and proposition.

Lemma 3.13. Let K be a jield of finite transcendence degree over a finite field of characteristic p>O.

Then K is finitely generated over K*.

Proof. Let F be a finite field. Let E = F(tl , . . . , tk) be a purely transcendental extension of F such that K is algebraic and separable over E. (Such tl , . . , tk can always be found. See the proof of [2], Lemma 17.7, p. 234.) Then K = K*(tl, . . , tk) = K*E. Indeed, note and separable over E *. Let UEK*. Let Ao+AtT+...+T’ be

that K* is algebraic the manic extension

irreducible E/E*

of CI over E*. Since u is separable

polynomial

is purely inseparable

and finite, by [5], Theorem

same degree over E as over E*. By assumption * . In E, each Ai =B,?,

Ai are rational

over E* and the

6, p. 177, c( has the functions

over F in

for some B; E E. Thus, Bl + Bpcl + . . . + cc’= 0. Let

t;,....tr /I E K be such that fi* = X. Then p is of degree at most I- over E, but E(p) contains c(. Since [E(E) : E] = Y, E(a) = E(B). H ence, E(a) contains pth root of CI and thus EK* contains

pth roots of all the elements

of K*. Hence, EK* contains

K.

Proposition 3.14. Let K be algebraic over a finitely generated field T. Let A4 be a purely inseparable finitely generated extension of K. Assume M and K are recursive fields with splitting algorithms. Then for any weak presentation J of M, J(M) -_TJ(K). function of J(K) is given. Let m E N be also compute the code of the pkth power of J-‘(m),

Proof. Suppose that the characteristic given.

Assuming

J-‘(m)

exists,

251

‘4. Shlapentokh I Annals of‘ Pure and Applied Loyic 94 ( 1998) X3-252

where Mph c K. Let r be this code. If r is not in J(K) Otherwise,

find a polynomial

this polynomial generated.)

satisfied by r over J(T)

under J. (Both tasks can be performed

Assume

since T is finitely

is of the form Aa + A 1T + .

+ T”. Then factor

+ T”J” over M. If the polynomial

does not have any

linear factors then m is not in J(M ). Otherwise Find the codes of all the solutions characteristic M is finitely Conversely.

above, m E J(M) suppose

for some k EN,

by generating

otherwise

the characteristic

a listing of J(M)

in M.

using the

it is not in J(M ).

function

of J(M)

is given.

Next note that

KP’ c MP’ c K. Furthermore,

Lemma 3.13, K is finitely If m $2J(M)

polynomial

in J(M)

let k be the number of solutions

function of J(K). This can be done effectively because by Lemma 3.13, generated over Mp” and consequently over K. If m matches one of the

codes mentioned

given.

image of

effectively

this polynomial

A0 + A1 TP’ + .

the polynomial

then m is not in f (M ).

and find the inverse

since transcendence degree is finite, by over KP' , and therefore over MP' . Let m E N be

generated

then obviously

m @J(K ). So assume m EJ(M).

P satisfied by m over J(T).

Then compute

a

This, as before, can be done effectively because

T is finitely generated. Since T is finitely generated we can compute J-‘(P) and using the splitting algorithm over K find all the roots of J -l(P) in K. Next we can use the fact that K is finitely generated over M ph to produce an effective listing of J(K) using the characteristic function of M. Thus, we can compute J-images P in J(K). If none of these images is equal to rn then m@J(K). otherwise. We summarize

the last two propositions

in the following

of all the roots of and it is in J(K)

theorem.

Theorem 3.15. Let F, M, F r?M he recursive jields with splitting ulgorithms algehruic over a finitely generated field T and such that F < TLM. Assume M is as described in Notations

3.7. Then for ever?, weak presentation

of MF, J(F)


lf und onI)>

(f the subgroup Aut(F n M) is of finite index in Aut(M ). Proof. The proof of the theorem note the following. 1.3 and Lemma

can be easily derived

from the results above if we

Since F and M are of finite transcendence 3.13, F 6,, M implies

F is finitely

generated

degree, by Theorem over F f? M.

Next

we can use Theorem 3.12 to get the desired results for the pair of fields M and F n M. Finally, we apply Proposition 3.14 to connect weak presentations of F n M and F.

References [I] [2] [3] [41

E. Artin. Algebraic Numbers and Algebraic Functions, Gordon and Breach, New York. 1967. M. Fried, M. Jarden, Field Arithmetic, Springer, New York, 1986. G.J. Janusz, Algebraic Number Fields. Academic Press, New York, 1973. C. Jockusch, A. Shlapentokh, Weak presentations of computable fields. J. Symbolic Logic 60 (1995) 199-208. [5] S. Lang, Algebra, Addison-Wesley, Reading, MA, 1971.

252

A. Shlapentokhl

Annals of Pure and Applied Logic 94 (1998) 223-252

[6] M. Rabin, Computable Algebra, Trans. Amer. Math. Sot. 95 (1960) 341-360. [7] A. Shlapentokh, Diophantine equivalence and countable rings, J. Symbolic Logic 59 (1994) 1068-1095. [8] A. Shlapentokh, Non-standard extensions of weak presentations, J. Algebra 176 (1995) 735-749. [9] A. Shlapentokh, Algebraic and turing separability of rings, J. Algebra 185 (1996) 229-257. [IO] A. Shlapentokh, Rational separability over global fields, Ann. Pure Appl. Logic 79 (1996) 93-108.