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Recursive properties of euclidean domains
Annals of Pure and Applied North-Holland
RECURSIVE
Logic 29 (1985) 59-77
PROPERTIES
Leonard
SCHRIEBER
Department
of Mathematics,
OF
EUCLIDEAN
University, Ithaca, NY 14853,
Cornell
Communicated by A. Nerode Received 1 January 1983; revised
59
1 October
DOMAINS
USA
1983
1. Introduction A recursively presented ring is a ring which is a recursively enumerable set of integers, and which has recursive operations (+, -, *). The notion of r.p. rings was introduced by Frohlich and Shepherdson [3] (where they are called ‘explicit’ rings) and by Rabin [8] (‘computable’ rings). Since then the subject has been pursued by many authors, including Seidenberg [l], Tennenbaum [6], and Metakides and Nerode [7]. In this paper we consider two properties of recursively presented Euclidean domains: E:
The domain has a recursive, finitely valued Euclidean function.
U: The set of units in the domain is recursive. The main theorem is 1.1. Theorem. Properties E and U of recursively presented Euclidean
domains are
consistent with and independent of each other. That is, there are Euclidean satisfying E&U, E & -U, -E &U, -E & -U.
domains
2. Notation We denote the class of ordinals by ‘on’. A hierarchy is a function {R},,,, on the ordinals with the property that p c (Y3 R, c &. We write R,, for UDcu R, The rank function 4 : R, + on corres%, and R,, for R,\R<,. R, for U-, ponding to the hierarchy is defined by 4(x) = (~.a (x E k). For elements a, b of any domain R we write a - b to mean that a and b are associate. We write a 1b to mean that a divides b. We denote R \{O} by R”. For any set S c R we denote by (S) the closure of S under multiplication. When I is an ideal in a ring R and D a multiplicatively closed subset of R, we 0168-0072/85/$3.30
0
1985, Elsevier
Science
Publishers
B.V.
(North-Holland)
60
L. Schrieber
denote by Q(R, D, I) the ring whose elements are equivalence classes [a/b], with a E R and b E 0, where [a/b] = [c/d] iff ad - bc E I. When I = {0}, we denote this ring as Q(R, D) and its elements as a/b; it is the same as the fraction ring with denominators from D. If R is recursively presented, D is r.e. and I is recursive, then Q(R, D, 1) is r.p. We denote by k an arbitrary field of characteristic zero. In Section 10 and later k is assumed to be recursively presented. In an expression such as k[V] or k(V), we usually mean V to be a set of variables. A monomial in the polynomial domain k[V] is an element of the form X1 * . . X,, where X,, . . . . , X, are variables in V. We denote the degree of an element p of k[V] as deg(p). The e-th partial recursive function is denoted 4e. The set K is the diagonal set, {e 1&(e) halts}. The phrase ‘not recursive, but recursively enumerable’ is abbreviated as ‘n.r.e.‘.
3. Euclidean domains For the reader’s convenience we review some of the basic facts concerning Euclidean domains. Samuel [9] presents an account of these and other interesting results about Euclidean rings. A survey of the field, including many construction principles, can be found in Lenstra [4]. A domain R is said .to be Euclidean on R” satisfying, for all d, a in R”: if d 1 a, We then call the remainder
if it possesses an ordinal valued function
then +(a + dq) < 4(d)
4
for some q E R.
4 a Euclidean function for R. Informally we may regard a + dq as left by the division of a by d.
We say that 4 is a normal Euclidean function if in addition d ( a 3 4(d) s 4(a). The function 4 is not uniquely determined by the Euclidean domain R. For example, in the ring 2 of integers, the process of long division shows that the absolute value function is Euclidean for 2. Another Euclidean function is obtained by defining 4(d) to be the number of digits in the base 2 representation of d. Still other pathological examples can be found (see [9]). It is well known that every Euclidean domain is a principal ideal domain, hence it is a unique factorization domain, hence it possesses a greatest common divisor function. What is not so well known is that Euclidean domains can be characterized without reference to any Euclidean function given in advance. Suppose R is an arbitrary domain; we attempt to build a ranking scheme on R based upon a principle of inexactness of division. At rank zero we place the set R” of units, since division by a unit is always exact. Next, at rank 1 we place all nonzero elements d which are not units and which have the property that division by d always is either exact or leaves a unit as remainder. In general for each ordinal (Y we place at rank CYall nonzero elements d which are not at any rank less than OL
Recursive properties of Euclidean
domains
61
and which have the property that division by d always is either exact or leaves a remainder of rank less than (Y. Eventually this ranking process must stop, since the ordinals form a proper class. When it does, either every nonzero element of R has become ranked or else some elements are left over. It can be shown that in the former case R is Euclidean and the rank function 4 determined by the ranking scheme is the least possible Euclidean not Euclidean.
function for R ; and in the latter case R is
Thus we have:
3.1. The Transfinite inductively by:
Construction
R,={dER’j(V
a E R’)(d
[ 10,9].
Given
a domain R, define {R,},,,,
1 a + (3q E R)(a +dq E R,,))!
Then: is a hierarchy. We denote by & : R, ---, on the corresponding rank (1) R],,,, function and call it the Euclidean rank function on R. (2) Each R, is closed under multiplication by units. Hence if a E R, and x E R”, then &(a~) = &(a). (3) R is Euclidean iff R,= R”. (4) If R is Euclidean, then & is the smallest Euclidean function for it. This means that if 4 is any Euclidean function for R, then #+.(d)<+(d), all d E R”. 3.2. m&&ion.
The least (Y for which R, = R,,
is called the Euclidean
rank of
R. In particular, R is said to be of rank o or finitely ranked if R, = R,,. AU Euclidean domains constructed in this paper are finite&reed and all Euclidean functions are finitely valued.
4.Thefustcase:E&U This case is immediate. The most familiar example is 2. A recursive Euclidean function for it is the absolute value function, and the set of units (-1, +l} is recursive, since it is finite. Another example is given by a polynomial domain k[X] in one variable X (assuming k is recursively presented). Here a Euclidean function is the degree function; the set of units is the field k, which is a recursive subset of k[Xj. An even simpler example is k itself; the constant zero function is Euclidean for it.
5. Prime in Euclidean
domains
A Euclidean domain is a unique factorization domain, and consequently the primes play a fundamental role. In this section we consider some properties of the primes in a Euclidean domain and reduce the Transfinite Construction to a hierarchy on the set of products of primes.
62
5.1.
L. Schrieber
Definition.
Let R be a UFD and P c R a set of primes. We call P a set of
distinct primes if for all p and q in P, p - q + p = q. We call P a canonical set of primes if P is a choice set for R’/--. This says that P is a set of distinct primes and for every prime q E R there is p E P with q - p. 5.2. Definition.
Let R be a UFD and P c R a set of distinct primes. We denote:
F=F(P)={p,*.*p,\s~w G=G(P)={~ER’I~~
andp,,...,p,EP) b for all PEP}.
Given I E R”, we can write r = ab for some uniquely determined a E F and by f(r)=a and b E G. We define onto maps f = fp: R” +F and g=g,:R’+G g(r) = b. We call f the P-content function. As a multiplicative monoid R” is isomorphic to the direct product F X G. The functions f and g are the projection maps onto F and G and hence they satisfy the following simple properties: (1) r = f(r)g(r), for all r E R”. (2) f and g preserve multiplication. (3) ff=f, gg=g and fg=gf=l. (4) f is the identity on F(P) and f = 1 on G(P). (5) g is the identity on G(P) and g = 1 on F(P). In view of (4), we may characterize the elements of G by saying that their P-content is 1. We can also characterize G in the following way. Extend P to a canonical set Q of primes. Then G(P) = (R” U (Q\P>). Thus G is the set of finite products of units and primes not ‘in’ P. In particular, if P is a canonical set of primes, then G(p) = R”. 5.3.
Rema&.
essentially
When
P is a canonical
set of primes,
then the function
fp is
the same thing as a theory of divisors (see, e.g., [2]).
5.4. Lemma. Let R and S be unique factorization domains and let P c R, Q c S be sets of distinct primes. Let y : R + S be a map which takes F(P) into F(Q) and G(P) into G(Q). Then far = -yfp and gay = ygp. In short, if the diagram below exists at all, then it is commutative.
F(P) -L
T
fP
F(Q)
T
fo
R&S
I G(P)
L
go
G(Q)
Recursive propertiesof Euclidean domains
Proof.
To show for = rfp we calculate, far(a)
63
assuming a E R”:
= f&fF44g,(u)) = (forfP(u))(faYg~(u)) =
fc?TfPb)
because
y&(a)
=
YfPb)
because
-y&(u) E F(Q).
The calculation
to show go-y = -ygp is similar.
E G(Q)
q
a ER’ we can write f(a)= p1 - * - pS for some uniquely determined Pl, . . .? ps E P, We call the number s the P-degree of a and denote it as deg,(u). Obviously deg, =degJ. In case P is a canonical set of primes, then s is determined independently of the particular choice of P; we call it the R-degree of a and denote it as deg,(a). It is clear that If
deg,(ab)
= degp(a) +degp(b).
Note that in a polynomial domain R = k[V] the ordinary polynomial degree of an element is, in general, larger than the R-degree. 5.5. De6nition. Let R be a UFD and P c R a set of distinct primes. We denote the domain Q(R, G(P)) of fractions as R(P). For A c R we denote by A/l the image {u/l ( a E A} of A in R(P). Elementary considerations show that R(P)* = {u/b 1a, b E G(P)}, that P/l is a canonical set of primes for R(P) and that deg,(,,(u/b) = deg,(a). If D c P, then D/l is a set of distinct primes in R(P), and deg,,,(u/b) = deg,(a). 5.6.
Example. Let R = k[V],
where V is a set of variables.
The V-content
f(p)
of a polynomial p E k [ V]” is the largest monomial which can be factored out of p ; hence we call f the monomial content function. Elements of G(V) are polynomials divisible by no individual variable. mial in k[V]
Observe that the V-degree
of a mono-
is the same as its polynomial degree.
We denote k[ V]( V) as k(V). Each element of k(V)’ can be written as mplq, where m is a uniquely determined monomial and p and q are polynomials of monomial content 1. Thus every element of k(V) may be regarded as a monomial, up to multiplication by units. Hence we call k(V) a monomial domain. In general, if R is any UFD and P c R a set of distinct primes, we can define a homomorphism of multiplicative monoids from R(P)’ to R” by u/b +-f(u). This map is onto and its kernel is R(P)*. It follows that u/b 1c/d (in R(P)) iff f(u) 1f(b) (in R). 5.7.
De&nition.
Let R be a UFD
and PC R a set of distinct primes.
Define
64
L. Schrieber
{F I P
alEon
inductively by: F,={dEFI(VaEF) (d x a 3 (3q E R)(sq,
We call P a Euclidean
E G)(f(aq,
+ dq) E F,,))).
set of primes for R if F, = F.
Plainly F,, = F fl R” = (1). A simple induction shows that {F,} is a hierarchy. We denote by & : F, + on the corresponding
rank function.
Observe that if P is a canonical set of primes for R, then G = R* so the definition of the F-hierarchy simplifies to: F, = {d E F 1(va E FM
1 a =+ (34 E R)(f(a
+ dd E F<,))).
5.8. Example. In Z, the positive primes form a canonical Euclidean set of primes. In this case f is the absolute value function, and g is the ‘sign’ function. The set F is a free commutative monoid on the set P of generators, and as such it is a considerably simpler object than the UFD R in which it is contained. The usefulness of the F-hierarchy lies in the way it mirrors the hierarchy on R given by the Transfinite Construction. 5.9. Theorem. If P is a canonical a. %. = f-l[FJ
and
set of primes in a UFD R, then for all ordinals
f[RJ = F,.
Proof.
Induct on a. To show fel[Fa]= R,, suppose d E fpl[Fa]. Given a E R”, assume d/a. Then f(d)/f( a > since P is canonical. Since f(d) E F,, there is q E R E such that f (f(a) + f (d)q) E F,,. By the induction hypothesis then f(a)+f(d)q R
= a and f(d) = dg(d)-’
we obtain
To show R, c fpl[Fa], suppose d E R,; we must show assume f(d) 1 a. Then d 1 a, so there is q E R such that a + induction hypothesis and noting that d = f(d)g(d) we have Thus f(d) E F,. That f[K] c F, follows trivially. To see that F, c f[R,], the identity. Hence F, = f-‘[F,] = k, so F, c f[&]. q 5.10. &f.
COPOUT.
If P is a canonical
a + dg(d)-‘qg(a)
E R,,.
f(d) E F,. Given a E F, dq E R,,. Applying the f(a + f(d)g(d)q) E F,,. observe that f on F, is
set of primes in a UFD
R, then & = &f
=
If P is also Euclidean, then R is a Euclidean domain.
Proof. For a E R” note that a = f(a)g(a)
and g(a)
is a unit. Hence,
applying
Recursive
Theorem
5.9, we have
Euclidean,
5.11.
then
Theorem. F(P/l),
&(a)
F = F,,
SO
propertiesof Euclidean
= &(f(a)g(a))
R” = f-‘[F]
If P is a set of
65
domains
=&(f(a))=
= fel[Fm] = R,.
&(f(a)).
If P is also
0
distinct primes in a UFD R, then for all ordinals
CY:
= F(P)&
Proof. It is obvious that the canonical embedding of R in R(P) given by x + x/l takes F(P) into F(P/l). It takes G(P) into G(P/l) because if a E G(P) and p/l ) a/l for some p E P, then u/l = (p/l)(q/r), for some q/r E R(P), whence ar = pq, so p divides ar which is impossible because both a and r are in G(P). Applying Lemma 5.4, we then have a commutative diagram:
F(P)
il
F(P/l)
T
f, T
fn 1
R
7
R(P)
To establish the stated equality we induct on a. To show F(P/l), c F(P),/l, suppose d/l EF(P/~),; we must show d E F(P),. Given a EF(P), assume d I( a. Then d/l 1 a/l, so there is q/r E R(P) such that f,l,(a/l + dqlr) E F(P/l),,. Noting that fpll(l/r) = 1, and applying the diagram above and the induction hypothesis we obtain fP(ar + dq)/l E F(P),,/l, i.e., fP(ar + dq) E F(P),,; and r E G(P). Thus d E F(P),. To show F(P),/1 cF(P/l),, suppose d EF(P),; we must show d/l EF(P/~),. Given a/l E F(P/l), assume d/l] a/l. Then d 1 a, so there are q E R and q1 E G(P) such that f,(aq, + dq) E F(P),,. The induction hypothesis together with the diagram then shows that fP,l(uqI/l + dq/l) E F(P/l),,; and q,/l E G(P/l). Thus d/l E F(P/l),. 5.12.
Cl Corollary.
If P is a set of distinct primes
R(P): ~R~P~bIb) = 4F(pjfp(ahIf domain.
Proof.
Applying &&a/b)
Corollary
5.10 and Theorem
= &(PlljfPll(a/b)
If P is also Euclidean,
P
in a UFD
R, then for all a/b in
is also Euclidean, then R(P)
= &o&(a)ll)
is a Euclidean
5.11, we have = d+(pjfp(a).
then
F(P/l), = F(P)dl
= F(P)/1 = F(P/l),
hence P/l is Euclidean for R(P). Since P/l is also canonical Euclidean b:l Corollary 5.10. q
for R(P), then R(P)
is
66
L. Schrieber
6. Discrete
sets of primes
In an arbitrary Euclidean
domain the rank of an element is determined
by its
relation to every other element of the ring. It is this global nature of the rank which makes the construction of Euclidean domains such an arduous task. In this section we introduce a property which localizes the determination of rank, and thereby allows for the construction of a class of Euclidean domains of a particularly simple structure. 6.1. De6nition. Let R be a UFD and P c R a set of distinct primes. We say that R is discrete with respect to I?, or that P is a discrete set of primes for R, if: and
a, b EF(P)
In case P is a canonical,
gcd(a, b)discrete
a + b - gcd(a, b) for a, b E F(P).
1 +
a+ b E G(P).
set of primes for R, it is easy to see that
If P is discrete
for R, then P/l is discrete for
R(P). 6.2.
Prop&&n.
R is a Euclidean
If R is a UFD and R has a canonical,
discrete set of primes, then
domain and & = deg,.
Proof. Let P be a canonical, discrete corresponding F-hierarchy satisfies F,, = {d E F(P) 1deg,(d)
G n},
set of primes for R. We show that the for all n E o.
For n = 0 it is trivial, since F. = (1). Assume n >O. Suppose d E F,,. If d = 1, then deg,(d)
< n. If df 1, then p 1d,
for some PEP. Write d = d,p. Then d,EF and d EF,, so f(d,+dq)EF,,, some q E R, since d 1 dl. Applying the induction hypothesis and replacing d,p, we have n > degR(f(4
+ 4pq))
= degR(4f(l
for d by
+ pq)) 2 deg,(dJ.
Hence deg,(d)
= deg,(d,)+
1 G n.
Conversely suppose d E F and deg,(d)G n. Suppose a E F and d ,/ a. Then degR(f(a + d)) = deg,(gcd(a, d)) < II. By the induction hypothesis, then f(a + d) E F,,. Thus d E F,. Since deg,(d) is always finite, it follows that F, = F. Hence R is a Euclidean q domain by Corollary 5.10, and & = &$ = deg,f = deg,. If P is a discrete set of primes for a UFD R, then R(P) is Euclidean P/l is canonical and discrete for R(P). We then have &&a/b)
= deg,,,,(a/b)
= deg&).
because
67
Recursive properties of Euclidean domains
This says that the rank of u/b in R(P) is the same as the P-degree the rank of a/b is completely restricted
determined
by the characteristic
of a in R. Thus function
to the set of primes which occur in a. We call this property
of P
‘prime-
independence’. 6.3.
Example.
A monomial
domain k(V)
is Euclidean,
because
V is a discrete
set of primes in k[V]. The Euclidean rank of an element p/q in k(V)’ is the same as the polynomial degree of the monomial content of p. In particular, if m is a monomial in k[V]‘,
then 4kcv,(m/l> = deg(m).
7. The second case: E & -U We begin with an r.p. Euclidean
domain which possesses a recursive,
normal
Euclidean function. We choose an n.r.e. set of distinct primes and declare them to be units. The resulting fraction ring has an n.r.e. set of units and, as noted by Motzkin [lo], is Euclidean. However, our construction of a Euclidean function for the ring must differ from his in order that it be recursive. 7.1. De6nition. Let R be an r.p. Euclidean domain and 4 a Euclidean function for it. We call q a partial recursive division algorithm for 4 if q is a partial recursive function defined on a subset of R” x R”, and for all d, a in R”: (1) If d 1 a, then q(d, a) halts. (2) If q(d, a) halts, then ~$(a + dq(d, u))<+(d). 7.2. Lemma. Let R be an r.p. Euclidean domain. Each function C$for R has a partial recursive division algorithm. Proof.
Let C$ be a recursive
Euclidean
function
recursive Euclidean
for R. We can then define a
partial recursive division algorithm q for C$ as follows: Given d and a in R”, search for x in R such that 4(a + dx) <4(d). If such an x is found, then set q(d, a) = x. If no such x is found, then q(d, a) remains undefined. It is clear that q satisfies the requisite properties. 0 7.3. Proposition. Let R be an r.p. Euclidean domain and 4 a Euclidean function for it. If 4 has a partial recursive division algorithm q, then R has a recursive greatest-common-divisor function. Proof. We compute a gcd of a pair do, dI of elements of R” by passing through a sequence of stages s (s = 0, 1,2,. . .). At stage s we simultaneously perform operations A, and J3, : A, : B,:
Search for x E R” such that d, = dS+Ix. Compute
q(d,+,, d,>.
68
L. Schrieber
By definition of q, one or the other must halt. If B, halts first, then set d st2 = 4 + 4+1d4+1, d,) and go to stage s + 1. If A, halts first, then the procedure ends. At each stage s we have #r(d,+,)<&(d,+,) and so the procedure must eventually halt by finding an x such that d, = d,,, x. The usual argument then shows that
d,+l is a gcd of d,, and d,.
0
7.4. CoroIIary. If an r.p. Euclidean then it has a recursive gcd function. 7.5.
timma.
domain has a recursive Euclidean
Let R be an r.p. Euclidean
function,
domain, let Cc R be an r.e. set of distinct
primes, let P be a canonical set of primes extending C, and set S = R(P\ C). Then: (1) S is recursively presented. (2) Every normal Euclidean function b, for R induces a Euclidean for S. (3) If do is recursive, then & is recursive.
function &
S is recursively presented because S = Q(R, (R* U C)), and (R” U C) is r.e. Define & on So by +,(a/b) = &(a ,), where a = a, * gcd(a, b). It is a simple Proof.
matter
using the normality
of 4
to show that &
is well-defined.
Note that
+&lb) s+(a). It is clear from Corollary 7.4 that & is recursive if 4 is recursive. To see that & is Euclidean for S, suppose p/pl, d/d, E So and d/d, 1 p/pl. We may presume gcd(d, d,)- 1. Write 1$(pfdq)<4(d), some q E R. Then &(P/P~ + (dld,)(d,qlpJ)
= &((P + d&p,) s4(p+dq)
q
7.6. Theorem. There is an r.p. domain which has a recursive, finitely valued Euclidean function and a nonrecursive set of units. Proof.
Let
p :o + 2
be a one-to-one
enumeration
of the set P of positive
primes. Let Cc o be any n.r.e. set and set C = p[C]. Then S = Z(P\ C) is an r.p. domain which by Lemma 7.5 has a recursive Euclidean function. The recursive 0 function from w to S given by e + pJ1 shows that C s1 S”. Hence S” is n.r.e.
8. Monomial
maps
The construction of a Euclidean domain in which property E fails requires the ability to defeat each recursive function. The possibility of doing so is suggested by the existence of the minimum Euclidean function. What is needed is a method by which the Euclidean ranks of certain elements can be increased to arbitrarily high values. Monomial maps provide such a means. A monomial map is a map on
Recursive
properties of Euclidean
domains
69
a polynomial domain k[V] which takes each variable X to a positive power Y’ of another variable. Upon passing to the corresponding
monomial domain, the rank
of X/l changes, in effect, from 1 to t. A power map is a monomial map which maps distinct variables to powers of distinct variables. Under suitable conditions a monomial map can be approximated by power maps. Power maps preserve divisibility of monomials
and they
commute with the monomial content function. These properties will be useful in showing that the kernel of the monomial map to which the power maps belong is recursive. In this section we establish the basic properties of monomial and power maps. 8.1. De&&ion. Let P and V be sets of variables. We call a map y : k[P] -+ k[ V] a monomial map if y fixes k and there are functions h : P -+ V and t : P + 2’ for which yX = (AX)“, all X E P. We call A and t the defining functions of y. If A is one-to-one, then we call y a power map. It is clear that any monomial map y : k[P] * 8.2. Proposition. Let y : k[P] + k[V] X and t. Then (1) deg(ya) “deg(a), for a E F(P). (2) deg&r(p))
~deg(f,(p)),
for P E
k[ V] takes F(P) into F(V).
be a monomial map with defining functions
Wl\ker(y).
Proof. (1) is clear: since the values of the defining function t are assumed to be positive integers. (2) We calculate:
dedfdp)) = &d.fv~fp(p))+~4fmp(~)) 2 dd_f~rf&))
= deg(yfAp))
3 deg(f,(p)).
q
For a positive integer t denote by k’ the set {a* ) a E k}. 8.3.
Lemma.
Let P be a set of variables, t : P+
Z’
a function and p E k[P]O.
There is a valuation (a,, . . . , a,) of the variables (Yl, . . . , Y,) occurring in p for which p#O and aiEk’9, i=l,.. .,s.
Proof. Induct on the number s of variables occurring in p. If s = 0, then pf 0 is constant. For s+l, write Y for YS+l and write
p=peY”+.*.+po,
e>O,
p,#O,
PO, * * * 3 pe E W\WH. Applying the induction hypothesis, assign a value 4 E kiYc to each variable Yi occurring in pe so that pe # 0; assign zero to the remaining variables, except Y.
70
L. Schrieber
Then p becomes a nonzero polynomial in k[Y]. but kty is infinite. 0 8.4. Proposition. Let y : k[P] ---, k[V] and t. ‘Then: (1) y is one-to-one, (2) U]yp+XIp,
forXEPand
It has only finitely many roots,
be a power map with defining functions
A
pEkD?l,
(3) ya 1rb + a 1b, for a, b E P(P), (4) Y KW)I c GWI,
fvr = rfp and gvy = m.
(5) Proof.
(1) Given p E k[P]‘, to show -yp# 0 write p = p(X,, . . . ,X,). Then yp = PO.x,F, - . . , (AX,)“~). By Lemma 8.3 we may choose a,, . . . , a, E k for which p(aixl,. . . , a”) # 0. Assigning & to AXi, then yp# 0. (2) Assume AX divides yp. Write p = qX+ r, for some q E k[P] and rE
k[P\{X)]. Then yp = (yq)(U?Jti+ yr. Since AX divides -yp then AX divides yr. But MC does not occur in yr, because X does not occur in r and A is one-to-one. Hence yr = 0, so r = 0. Then X divides p. (3) Assume ya divides yb. Write a = X;ll. and nj may be zero). Then ya = (AXJmltX, . . . (xX,)“sas
and
**Q
and b = X;l-
yb = (~l)“~‘%
- *m
(some mi
. . . (~s)“sas.
Since A is one-to-one we may conclude that mitXi c n,tX, for each i = 1, . . . , s; hence mi s Q since tXi a 1. Consequently a divides b. (4) If p E G(P), then p is not divisible by any variable in P. To show that ?/pis not divisible by any variable in V, note that the only variables which could possibly divide yp are those variables which occur in yp. Every such variable is AX for some X E P. But X does not divide p, so AX does not divide yp by (2). (5) By Lemma 5.4. Cl
9. Renaming systems Throughout the remainder content function. 9.1.
Definition.
of this paper, by f we always mean the monomial
Let A c P c V be sets of variables,
Let A : A + V and t : A +
Z+. The map on k[P] determined by A and t is the map y : k[P] --j k[V] which fixes k and P\A and maps XE A to (AX)%. It is the same as the monomial map with extensions A and t of h and t as defining functions, where A is the identity and t equals 1 on P\ A. Note that if A is one-to-one and h[A] is disjoint from P, then A is one-to-one and so y is a power map.
71
Recursive properties of Euclidean domains
9.2. Definition. Let V be a set of variables.
We call (A, t, A) a renaming system
in Vif
and t:A+Z+.
For
h:A+V\A
AcV,
isone-to-one,
by yA, the map on k[Vl
any AOc A we denote
determined
by the
restrictions of X and t to AO. We denote yA simply as y. For A,,c A and PC V the restriction of yPnAo to k[P] is the map on k[P] determined
by the restrictions
of A and t to P rl AO. It is clear that yPnA, agrees
with yA, on k[P]. Hence if h[P n A,] is disjoint from P, then the restriction of y,+ to k[P] is a power map. 9.3.
Lemma. Let (A, t, A) be a renaming system in V. For A0 c A, c A: (1) ?*,?/a,= 3/A,- In particular, y?/a, = y.
(2) deg(fy,,p)==degvy&p),
for P E
Wl\k4yAI).
Proof. (1) We show yA,yAoX = yA,X for each variable X E TJ. If XEA~, then AXE V\A, so y,,hX=hX. Then YA,x4$=
If X$AO,
~Alwa*)=@X)*=YA,X.
then -y&X=X,
so ya,ybX=
(2) degCfyAlp) = dcgUyAlyhp)
yA,X.
sdcg(fy,p)
by Lemma 8.2.
9.4. Definition. Let (h, t, A) be a renaming {X E P n A 1hX E P} the internal set of P.
0
system in V. Given
P c V, call
It is plain that the internal set of P increases as P increases. If A,, is the internal set of P, then yA, maps k[P] into k[P\A,], and the internal set of P\A, is empty. If the internal set of P is empty, then the restriction of y to k[P] is a power map. 9.5.
Remark. If A, is the internal set of P, then combining Proposition
Lemma 9.3 we have the following commutative
8.4 with
diagram:
WI
%I \ W’\AoI be-to-one) A f
I
F(P\Ao) y,
kW\Al
I
f
WV-U
A similar diagram holds with g in place of f. In particular, for p E k[P] we have the useful formulas yfy&p = fyp and ygy&p = gyp.
72
L. Schrieber
9.6.Proposition. Let (A, t, A) be a renaming system in V and let A,
be the
internal set of P c V. Then for p E k[P]: (1) P E MyI (2)
P E
$7 Y&P = 0
y-l[G(V\A)I
ifi ~Y&P = 1.
Proof. Applying the diagram above, we compute: (1) poker iff ?/p=O, iff iff
YY~P=&
y&p=O.
(2) P E Tl[G(V\A)I
iff iff
YP E G(V\A), f’yp=l,
ifi iff
yf~&p = 1, fy&p=l.
Cl
10. Recursive renamingsystems Throughout presented.
the remainder
of this paper
we assume
k to be recursively
10.1.Definition. Denote by W the countable set {X,},,, of variables. We call (A, t, A) a recursive renaming system if it is a renaming system in W, h[A] is a recursive set, and A and t are partial recursive functions. It is clear that if A, c A is recursive, then rA, : k[ W] + k[ W \ A,] is a recursive function, determined uniformly from A,. In particular this is true if A0 is finite.
10.2.Lemma. Suppose A, B c W and A : A * B is a bijective partial recursive function. Assume B is recursive. Then given any finite set PC W, the internal set of P can be computed uniformly from P. Proof. Let {Ai}iEo be a recursive enumeration of all the finite subsets of A. We can then find i for which A[A,] = B n P, since A is onto. Since A is also one-to-one, it follows that Ai = A-‘[B n P]. The internal set of P is then P n Ai.
q
10.3.Lemma. If (A, t, A)
is a recursive ypl[ G( W \ A)] are recursive sets.
Proof.
renaming
system,
then
ker(r)
and
Given p E k[ W] we can compute the internal set A,, of the set of variables
occurring in p. Then, by Proposition 9.6, p E ker(y) iff r&p = 0, and p E q r-l[G( W\ A)] iff fy&p = 1, and both of these conditions are recursive.
Recursive
properiiesof Euclidean
73
domains
10.4. Definition. Let (A, t, A) be a renaming system R(h, t, A) the domain Q(k[W], y-l[G(W\A)], ker(y)).
in W. We denote
by
We see that y induces an isomorphism R(X, r, A) = k( W \A) given by [u/b] + -&yb, hence R(X, t, A) is a Euclidean domain. The isomorphism, which we also denote by y, shows that {[X,/l] ( e E W\ A} is a canonical, discrete set of primes for R(h, t, A). We summarize the results of this section. 10.5.Theorem. Let
(h, t, A)
be
a recursive
R(h, t, A). Then: (1) R is a recursively presented Euclidean
renaming
system
and
let R =
domain.
(2) & (Mql) = deg(fyp), for all [dsl E R”. (3) In particular, if XE W then: (a)
& ([x/11) = tX
if X E A,
(b)
&([X/ll)
if X# A.
= 1
(4) The map from W\A[A]
to R given by X-,[X/l]
is one-to-one.
proof. (1) follows from Lemma 10.3. (2) is by applying the isomorphism y between R and k(W\A), and Example 6.3. (3) follows directly from (2). For (4), suppose X and Y are in W\ A[A] and [X/l] = [Y/l]. Then yX = +yY. But y restricted to k[ W\ h[A]] is a power map, hence one-to-one, because the internal set of W\I\[A] is empty. So X= Y. Cl
11. The third case: -E&U 11.1.Lemma. Let r be a recursively presented Euclidean domain. Suppose w : w -+ R” is a recursive function
and &w(e)
recursive, finitely valued Euclidean
> @e(e) for all e EK.
Then
R has no
function.
Proof. Define a recursive function h : w + w to satisfy 4,,(e) = c#+w,all e in w. Suppose +,e : R” + o is any recursive function. Then &h(e) is total, so h(e) E K. men &w(We)) >h,dh(e)) = &NdeN. Thus it is not the case that 4e ~4~ Consequently 4= cannot be a Euclidean function for R.
0
11.2.Theorem. There is a recursively presented Euclidean domain R which has no recursive, finitely valued Euclidean
function.
Proof. Let B c o be any infinite, co-infinite recursive set. Set B = {X, 1b E B}. Let i: o + o \B be any one-to-one recursive function. Set A = i[Kj. Then A is r.e. and infinite. Set A = {X, ] a E A}.
L.
14
Schrieber
Let h : A + B be any partial recursive bijection. Define t : A ---, Z+ by tX& = &(e) + 1. Then (h, t, A) is a recursive renaming system, Euclidean domain. Define w : w + R” by w(e) = [X&/l]. &w(e)= BY Lemma
11.3.
tX;ce,>4e(e)
so R = R(h, t, A)
Then
for all e EK.
11.1 it follows that R has no recursive Euclidean
m-rem.
is an r.p.
function.
[7
Let R be the domain of the previous theorem. We then have:
(1) R” and Ral are recursive. (2) R,, is recursively isomorphic to K, for n 22. (3) 7’he set of primes in R is recursively isomorphic to w \K. Proof. Enumerate the set A as {Ye}_,, and set A, = {Yi 1i
if degCfy4,-,vc,,(p))
divergent,
2 n, for some e,
otherwise.
The function
4 is partial recursive, since the polynomial degree function is recursive and ya. nv(P) is a recursive function determined uniformly from e and p. If degCfyp)an, then 4(p) converges because ~~-,~(~)p = yp as soon as % n V(P) = A n V(P). If degCfyp)
diverges because then deg(fy%,,(,,p)
by Lemma 9.3. Thus R,, = {[p/q] 1p E domain(+)},
because
C n for all e,
&([p/q]) 2 n iff deg(fyp) 2 n iff
4(p) converges. Hence R,, is r.e. In particular, Ral is r.e. Since R* is always r.e. and R”= R” 6 Ral, we conclude that R* and Ral are both recursive. If n 32, then R,, $ K since R,, is r.e. Conversely to show that K s1 R,,, define a one-to-one
4dx)
=
recursive function v : w + o by &.(e)+n-1,
if eeK,
divergent,
otherwise
for all XEO. Consider the recursive function from o to R given by e + [XiuceJl]* Since i and v are one-to-one and i maps into w\B, Theorem 10.5 shows that the function is one-to-one. If e EK, then v(e) E K SO Xiuclv(e)E A. Then &([Xi”(,)/lI) = %,ce)= &,,(v(e)) + 15 n. Hence If e$K, then v(e)$K R-1.
[Xiu,e,/lI ER,,. SO
X,,,,$A.
Then 4,([Xiu,,/l])=
1. Hence
[Xiu(eJ1IE
15
Recursive properties of Euclidean domains
The function e + [Xioce,/l] thus shows that K s1 R,,. This same function
also shows that w\K
12. Further properties 12.1.
corollary.
of the domain R
The predicate &(x> < +R(y) is not decidable.
Proof. Suppose it is. Choose any variable X$ A. Then r([Xll]) = X/l, so the rank of [X/l]’ is exactly t. Given any [plp,]~ R”, decide successively &([p/pJ)< &([X/l]t) for t = 1,2,3, . . . . The first t for which it is true tells us that the rank of [p/p,] is t- 1.
Cl
12.2. Definition. A Euclidean function for a recursively presented domain is said to have a recursive division algorithm if it has a partial recursive division algorithm (Definition 7.1) and if the divisibility relation in the domain is decidable. We show that 12.3.
Lemma.
4R has a recursive division algorithm.
Let R be the domain of Theorem 11.2. For [d/d,], [p/pJ~ R”:
(1) Cd/d111[P/PJ $7 frd 1fry. (2) If P = W, A0 is the internal set of P and p, d E k[P], fyhd
then [d/d,] ) [p/p11 ijj
1fyhp.
Proof. (1) [d/&l 1[PIPJ
iff
r&r&
1YP~YPI,
iff frd 1fry. (2) By Remark frd
I frp
9.5 and Proposition iff
rfy&
iff fs$i
8.4:
I yfyhpy fybp.
17
12.4. Theorem. The following operations are effective, for [d/d,], [plp,]~ R”: (1) To decide whether [d/d,] ( [p/p,]. (2) If [d/&l
Proof.
1 [P/PJ,
to find
[s/sJE
R
for which
(1) Compute the internal set A0 of the set of variables occurring in p or in
76
L. Schrieber
d. It is then simply a question of deciding whether fy&d 1fy&p, which can be done by explicitly computing them and comparing exponents. (2) Compute the internal set A, of the set of variables occurring in p or in d. Set q = dlgy4p 9.5:
and q1 = pIgybd.
Note that frd 1 fyp. Then,
applying
Remark
4~ ([P/PII+ [d4l[qld)
= &~~~\/~YP/YPI)+ (rdrdd(r4 = d&(~p
* mphp,
- gd)
- gyd + w’ * gyp))
= hd..fCfyp + fr4) < dedfd
= 4+([d/dJ.
0
13. The fourth case:-E & -U To construct a Euclidean domain in which both of the conditions E and U fail, we apply the technique of Theorem 7.6 to the domain of Theorem 11.2. The property of prime-independence possessed by discrete domains allows the set of units to be introduced without collapsing the ranks of elements elevated by the renaming system. 13.1. Theorem. ‘There is a recursively presented Euclidean domain which has no recursive, finitely valued Euclidean function, and in which the set of units is not recursive. Proof. Let R be the proof of that Let Cc w be D = W\ (A U C).
the Euclidean domain of theorem. For any set E any n.r.e. set disjoint This gives a partition of
Theorem 11.2. We use the notation of c W, denote [E/l] = {[X/l] 1XE E}. from AUB. Set C = {X, 1c E C} and W into A, C and D; and B c D. Since
D is disjoint from A, [D/l] is a set of distinct primes in R and y maps [D/l] oneto-one onto D/l. We have seen that [C/l] U [D/l] is canonical and discrete for R ; hence the smaller set [D/l] is also discrete for R. Set S = R([D/l]). Writing S = Q(R, (R*U[C/l])) makes it clear that S is recursively presented, because C is r.e. and so (R* U [C/l]) is also r.e. Proposition 6.2 shows that S is Euclidean. Denote the element [X,/l]/[l/l] of S as x,. Then, applying the isomorphism y between R and k(W\A), we see that &(x,)
= deg,(yX,)
for all e E w.
We show C s,,, S* via the recursive function from o to S given by e + x,. If e E C, then yX, = X,, so deg,(yX,) = 0, since X, 6 D. If e$ C, then yX, is either a positive power of hX, E B c D (if X, E A) or X, itself (if X, ED). In either case deg,(yX,)>O. Thus e EC iff &(x,) = 0 iff x, E S*. Hence S” is not recursive.
Recursiue properties
To see that S has no recursive 11.2. Define
of Euclidean domains
Euclidean
function,
77
we proceed
as in Theorem
w : w + So by w(e) = x+_). If e EK, then:
&(w(e))
= degD(+G(,,) = deg, ((XXi(ej)*q(=l)
(since Xicej E A)
= txiCe)
(since
AXice, E B C 0)
=+e(e>+l. By Lemma
14. An open Church
11.1, S has no recursive
Euclidean
0
function.
question
and
Kleene
showed
how
infinite ordinals as values (see [5]). ordinal-valued recursive functions?
to define
recursive
Do
Theorems
Nerode
for
functions
11.2
and
which
take
13.1
hold
for
problem
and
for
Acknowledgements The
author
is indebted
to Anil
posing
the
supplying assistance and encouragement throughout the writing of this paper. He also thanks Jeff Remmel for drawing his attention to the fourth case of the main theorem and for suggesting a method of proof.
References [ll A. Seidenberg, Constructions in algebra, Trans. Amer. Math. Sot. 197 (1974) 273-313. [2] Z.I. Borevich and I.R. Shafarevich, Number Theory (Academic Press, New York, 1966). [3] A. Frohlich and J.C. Shepherdson, Effective procedures in field theory, Phil. Trans. Royal Sot. London (ser. A) 284 (1955) 407-432. [4] H.W. Lenstra, Jr., Lectures on Euclidean Rings (Bielefeld, 1974). [Sl H. Rogers, Theory of Recursive Functions and Effective Computability (McGraw-Hill, New York, 1967). [6] J.B. Tennenbaum, A Constructive Version of Hilbert’s Basis Theorem, Ph.D. Dissertation (1973), Univ. of Calif., San Diego. [7] G. Metakides and A. Nerode, Effective content of field theory, Ann. Math. Logic 17 (1979) 289-320. [81 M.O. Rabin, Computable algebra, general theory and theory of computable fields, Trans. Amer. Math. Sot. 95 (1960) 341-360. [9] P. Samuel, About Euclidean rings, J. Algebra 19 (1971) 282-301. Cl01 Th. Motzkin, On the Euclidean algorithm, Bull. Amer. Math. Sot. 55 (1949) 1142-1146.
RECURSIVE
Logic 29 (1985) 59-77
PROPERTIES
Leonard
SCHRIEBER
Department
of Mathematics,
OF
EUCLIDEAN
University, Ithaca, NY 14853,
Cornell
Communicated by A. Nerode Received 1 January 1983; revised
59
1 October
DOMAINS
USA
1983
1. Introduction A recursively presented ring is a ring which is a recursively enumerable set of integers, and which has recursive operations (+, -, *). The notion of r.p. rings was introduced by Frohlich and Shepherdson [3] (where they are called ‘explicit’ rings) and by Rabin [8] (‘computable’ rings). Since then the subject has been pursued by many authors, including Seidenberg [l], Tennenbaum [6], and Metakides and Nerode [7]. In this paper we consider two properties of recursively presented Euclidean domains: E:
The domain has a recursive, finitely valued Euclidean function.
U: The set of units in the domain is recursive. The main theorem is 1.1. Theorem. Properties E and U of recursively presented Euclidean
domains are
consistent with and independent of each other. That is, there are Euclidean satisfying E&U, E & -U, -E &U, -E & -U.
domains
2. Notation We denote the class of ordinals by ‘on’. A hierarchy is a function {R},,,, on the ordinals with the property that p c (Y3 R, c &. We write R,, for UDcu R, The rank function 4 : R, + on corres%, and R,, for R,\R<,. R, for U-, ponding to the hierarchy is defined by 4(x) = (~.a (x E k). For elements a, b of any domain R we write a - b to mean that a and b are associate. We write a 1b to mean that a divides b. We denote R \{O} by R”. For any set S c R we denote by (S) the closure of S under multiplication. When I is an ideal in a ring R and D a multiplicatively closed subset of R, we 0168-0072/85/$3.30
0
1985, Elsevier
Science
Publishers
B.V.
(North-Holland)
60
L. Schrieber
denote by Q(R, D, I) the ring whose elements are equivalence classes [a/b], with a E R and b E 0, where [a/b] = [c/d] iff ad - bc E I. When I = {0}, we denote this ring as Q(R, D) and its elements as a/b; it is the same as the fraction ring with denominators from D. If R is recursively presented, D is r.e. and I is recursive, then Q(R, D, 1) is r.p. We denote by k an arbitrary field of characteristic zero. In Section 10 and later k is assumed to be recursively presented. In an expression such as k[V] or k(V), we usually mean V to be a set of variables. A monomial in the polynomial domain k[V] is an element of the form X1 * . . X,, where X,, . . . . , X, are variables in V. We denote the degree of an element p of k[V] as deg(p). The e-th partial recursive function is denoted 4e. The set K is the diagonal set, {e 1&(e) halts}. The phrase ‘not recursive, but recursively enumerable’ is abbreviated as ‘n.r.e.‘.
3. Euclidean domains For the reader’s convenience we review some of the basic facts concerning Euclidean domains. Samuel [9] presents an account of these and other interesting results about Euclidean rings. A survey of the field, including many construction principles, can be found in Lenstra [4]. A domain R is said .to be Euclidean on R” satisfying, for all d, a in R”: if d 1 a, We then call the remainder
if it possesses an ordinal valued function
then +(a + dq) < 4(d)
4
for some q E R.
4 a Euclidean function for R. Informally we may regard a + dq as left by the division of a by d.
We say that 4 is a normal Euclidean function if in addition d ( a 3 4(d) s 4(a). The function 4 is not uniquely determined by the Euclidean domain R. For example, in the ring 2 of integers, the process of long division shows that the absolute value function is Euclidean for 2. Another Euclidean function is obtained by defining 4(d) to be the number of digits in the base 2 representation of d. Still other pathological examples can be found (see [9]). It is well known that every Euclidean domain is a principal ideal domain, hence it is a unique factorization domain, hence it possesses a greatest common divisor function. What is not so well known is that Euclidean domains can be characterized without reference to any Euclidean function given in advance. Suppose R is an arbitrary domain; we attempt to build a ranking scheme on R based upon a principle of inexactness of division. At rank zero we place the set R” of units, since division by a unit is always exact. Next, at rank 1 we place all nonzero elements d which are not units and which have the property that division by d always is either exact or leaves a unit as remainder. In general for each ordinal (Y we place at rank CYall nonzero elements d which are not at any rank less than OL
Recursive properties of Euclidean
domains
61
and which have the property that division by d always is either exact or leaves a remainder of rank less than (Y. Eventually this ranking process must stop, since the ordinals form a proper class. When it does, either every nonzero element of R has become ranked or else some elements are left over. It can be shown that in the former case R is Euclidean and the rank function 4 determined by the ranking scheme is the least possible Euclidean not Euclidean.
function for R ; and in the latter case R is
Thus we have:
3.1. The Transfinite inductively by:
Construction
R,={dER’j(V
a E R’)(d
[ 10,9].
Given
a domain R, define {R,},,,,
1 a + (3q E R)(a +dq E R,,))!
Then: is a hierarchy. We denote by & : R, ---, on the corresponding rank (1) R],,,, function and call it the Euclidean rank function on R. (2) Each R, is closed under multiplication by units. Hence if a E R, and x E R”, then &(a~) = &(a). (3) R is Euclidean iff R,= R”. (4) If R is Euclidean, then & is the smallest Euclidean function for it. This means that if 4 is any Euclidean function for R, then #+.(d)<+(d), all d E R”. 3.2. m&&ion.
The least (Y for which R, = R,,
is called the Euclidean
rank of
R. In particular, R is said to be of rank o or finitely ranked if R, = R,,. AU Euclidean domains constructed in this paper are finite&reed and all Euclidean functions are finitely valued.
4.Thefustcase:E&U This case is immediate. The most familiar example is 2. A recursive Euclidean function for it is the absolute value function, and the set of units (-1, +l} is recursive, since it is finite. Another example is given by a polynomial domain k[X] in one variable X (assuming k is recursively presented). Here a Euclidean function is the degree function; the set of units is the field k, which is a recursive subset of k[Xj. An even simpler example is k itself; the constant zero function is Euclidean for it.
5. Prime in Euclidean
domains
A Euclidean domain is a unique factorization domain, and consequently the primes play a fundamental role. In this section we consider some properties of the primes in a Euclidean domain and reduce the Transfinite Construction to a hierarchy on the set of products of primes.
62
5.1.
L. Schrieber
Definition.
Let R be a UFD and P c R a set of primes. We call P a set of
distinct primes if for all p and q in P, p - q + p = q. We call P a canonical set of primes if P is a choice set for R’/--. This says that P is a set of distinct primes and for every prime q E R there is p E P with q - p. 5.2. Definition.
Let R be a UFD and P c R a set of distinct primes. We denote:
F=F(P)={p,*.*p,\s~w G=G(P)={~ER’I~~
andp,,...,p,EP) b for all PEP}.
Given I E R”, we can write r = ab for some uniquely determined a E F and by f(r)=a and b E G. We define onto maps f = fp: R” +F and g=g,:R’+G g(r) = b. We call f the P-content function. As a multiplicative monoid R” is isomorphic to the direct product F X G. The functions f and g are the projection maps onto F and G and hence they satisfy the following simple properties: (1) r = f(r)g(r), for all r E R”. (2) f and g preserve multiplication. (3) ff=f, gg=g and fg=gf=l. (4) f is the identity on F(P) and f = 1 on G(P). (5) g is the identity on G(P) and g = 1 on F(P). In view of (4), we may characterize the elements of G by saying that their P-content is 1. We can also characterize G in the following way. Extend P to a canonical set Q of primes. Then G(P) = (R” U (Q\P>). Thus G is the set of finite products of units and primes not ‘in’ P. In particular, if P is a canonical set of primes, then G(p) = R”. 5.3.
Rema&.
essentially
When
P is a canonical
set of primes,
then the function
fp is
the same thing as a theory of divisors (see, e.g., [2]).
5.4. Lemma. Let R and S be unique factorization domains and let P c R, Q c S be sets of distinct primes. Let y : R + S be a map which takes F(P) into F(Q) and G(P) into G(Q). Then far = -yfp and gay = ygp. In short, if the diagram below exists at all, then it is commutative.
F(P) -L
T
fP
F(Q)
T
fo
R&S
I G(P)
L
go
G(Q)
Recursive propertiesof Euclidean domains
Proof.
To show for = rfp we calculate, far(a)
63
assuming a E R”:
= f&fF44g,(u)) = (forfP(u))(faYg~(u)) =
fc?TfPb)
because
y&(a)
=
YfPb)
because
-y&(u) E F(Q).
The calculation
to show go-y = -ygp is similar.
E G(Q)
q
a ER’ we can write f(a)= p1 - * - pS for some uniquely determined Pl, . . .? ps E P, We call the number s the P-degree of a and denote it as deg,(u). Obviously deg, =degJ. In case P is a canonical set of primes, then s is determined independently of the particular choice of P; we call it the R-degree of a and denote it as deg,(a). It is clear that If
deg,(ab)
= degp(a) +degp(b).
Note that in a polynomial domain R = k[V] the ordinary polynomial degree of an element is, in general, larger than the R-degree. 5.5. De6nition. Let R be a UFD and P c R a set of distinct primes. We denote the domain Q(R, G(P)) of fractions as R(P). For A c R we denote by A/l the image {u/l ( a E A} of A in R(P). Elementary considerations show that R(P)* = {u/b 1a, b E G(P)}, that P/l is a canonical set of primes for R(P) and that deg,(,,(u/b) = deg,(a). If D c P, then D/l is a set of distinct primes in R(P), and deg,,,(u/b) = deg,(a). 5.6.
Example. Let R = k[V],
where V is a set of variables.
The V-content
f(p)
of a polynomial p E k [ V]” is the largest monomial which can be factored out of p ; hence we call f the monomial content function. Elements of G(V) are polynomials divisible by no individual variable. mial in k[V]
Observe that the V-degree
of a mono-
is the same as its polynomial degree.
We denote k[ V]( V) as k(V). Each element of k(V)’ can be written as mplq, where m is a uniquely determined monomial and p and q are polynomials of monomial content 1. Thus every element of k(V) may be regarded as a monomial, up to multiplication by units. Hence we call k(V) a monomial domain. In general, if R is any UFD and P c R a set of distinct primes, we can define a homomorphism of multiplicative monoids from R(P)’ to R” by u/b +-f(u). This map is onto and its kernel is R(P)*. It follows that u/b 1c/d (in R(P)) iff f(u) 1f(b) (in R). 5.7.
De&nition.
Let R be a UFD
and PC R a set of distinct primes.
Define
64
L. Schrieber
{F I P
alEon
inductively by: F,={dEFI(VaEF) (d x a 3 (3q E R)(sq,
We call P a Euclidean
E G)(f(aq,
+ dq) E F,,))).
set of primes for R if F, = F.
Plainly F,, = F fl R” = (1). A simple induction shows that {F,} is a hierarchy. We denote by & : F, + on the corresponding
rank function.
Observe that if P is a canonical set of primes for R, then G = R* so the definition of the F-hierarchy simplifies to: F, = {d E F 1(va E FM
1 a =+ (34 E R)(f(a
+ dd E F<,))).
5.8. Example. In Z, the positive primes form a canonical Euclidean set of primes. In this case f is the absolute value function, and g is the ‘sign’ function. The set F is a free commutative monoid on the set P of generators, and as such it is a considerably simpler object than the UFD R in which it is contained. The usefulness of the F-hierarchy lies in the way it mirrors the hierarchy on R given by the Transfinite Construction. 5.9. Theorem. If P is a canonical a. %. = f-l[FJ
and
set of primes in a UFD R, then for all ordinals
f[RJ = F,.
Proof.
Induct on a. To show fel[Fa]= R,, suppose d E fpl[Fa]. Given a E R”, assume d/a. Then f(d)/f( a > since P is canonical. Since f(d) E F,, there is q E R E such that f (f(a) + f (d)q) E F,,. By the induction hypothesis then f(a)+f(d)q R
= a and f(d) = dg(d)-’
we obtain
To show R, c fpl[Fa], suppose d E R,; we must show assume f(d) 1 a. Then d 1 a, so there is q E R such that a + induction hypothesis and noting that d = f(d)g(d) we have Thus f(d) E F,. That f[K] c F, follows trivially. To see that F, c f[R,], the identity. Hence F, = f-‘[F,] = k, so F, c f[&]. q 5.10. &f.
COPOUT.
If P is a canonical
a + dg(d)-‘qg(a)
E R,,.
f(d) E F,. Given a E F, dq E R,,. Applying the f(a + f(d)g(d)q) E F,,. observe that f on F, is
set of primes in a UFD
R, then & = &f
=
If P is also Euclidean, then R is a Euclidean domain.
Proof. For a E R” note that a = f(a)g(a)
and g(a)
is a unit. Hence,
applying
Recursive
Theorem
5.9, we have
Euclidean,
5.11.
then
Theorem. F(P/l),
&(a)
F = F,,
SO
propertiesof Euclidean
= &(f(a)g(a))
R” = f-‘[F]
If P is a set of
65
domains
=&(f(a))=
= fel[Fm] = R,.
&(f(a)).
If P is also
0
distinct primes in a UFD R, then for all ordinals
CY:
= F(P)&
Proof. It is obvious that the canonical embedding of R in R(P) given by x + x/l takes F(P) into F(P/l). It takes G(P) into G(P/l) because if a E G(P) and p/l ) a/l for some p E P, then u/l = (p/l)(q/r), for some q/r E R(P), whence ar = pq, so p divides ar which is impossible because both a and r are in G(P). Applying Lemma 5.4, we then have a commutative diagram:
F(P)
il
F(P/l)
T
f, T
fn 1
R
7
R(P)
To establish the stated equality we induct on a. To show F(P/l), c F(P),/l, suppose d/l EF(P/~),; we must show d E F(P),. Given a EF(P), assume d I( a. Then d/l 1 a/l, so there is q/r E R(P) such that f,l,(a/l + dqlr) E F(P/l),,. Noting that fpll(l/r) = 1, and applying the diagram above and the induction hypothesis we obtain fP(ar + dq)/l E F(P),,/l, i.e., fP(ar + dq) E F(P),,; and r E G(P). Thus d E F(P),. To show F(P),/1 cF(P/l),, suppose d EF(P),; we must show d/l EF(P/~),. Given a/l E F(P/l), assume d/l] a/l. Then d 1 a, so there are q E R and q1 E G(P) such that f,(aq, + dq) E F(P),,. The induction hypothesis together with the diagram then shows that fP,l(uqI/l + dq/l) E F(P/l),,; and q,/l E G(P/l). Thus d/l E F(P/l),. 5.12.
Cl Corollary.
If P is a set of distinct primes
R(P): ~R~P~bIb) = 4F(pjfp(ahIf domain.
Proof.
Applying &&a/b)
Corollary
5.10 and Theorem
= &(PlljfPll(a/b)
If P is also Euclidean,
P
in a UFD
R, then for all a/b in
is also Euclidean, then R(P)
= &o&(a)ll)
is a Euclidean
5.11, we have = d+(pjfp(a).
then
F(P/l), = F(P)dl
= F(P)/1 = F(P/l),
hence P/l is Euclidean for R(P). Since P/l is also canonical Euclidean b:l Corollary 5.10. q
for R(P), then R(P)
is
66
L. Schrieber
6. Discrete
sets of primes
In an arbitrary Euclidean
domain the rank of an element is determined
by its
relation to every other element of the ring. It is this global nature of the rank which makes the construction of Euclidean domains such an arduous task. In this section we introduce a property which localizes the determination of rank, and thereby allows for the construction of a class of Euclidean domains of a particularly simple structure. 6.1. De6nition. Let R be a UFD and P c R a set of distinct primes. We say that R is discrete with respect to I?, or that P is a discrete set of primes for R, if: and
a, b EF(P)
In case P is a canonical,
gcd(a, b)discrete
a + b - gcd(a, b) for a, b E F(P).
1 +
a+ b E G(P).
set of primes for R, it is easy to see that
If P is discrete
for R, then P/l is discrete for
R(P). 6.2.
Prop&&n.
R is a Euclidean
If R is a UFD and R has a canonical,
discrete set of primes, then
domain and & = deg,.
Proof. Let P be a canonical, discrete corresponding F-hierarchy satisfies F,, = {d E F(P) 1deg,(d)
G n},
set of primes for R. We show that the for all n E o.
For n = 0 it is trivial, since F. = (1). Assume n >O. Suppose d E F,,. If d = 1, then deg,(d)
< n. If df 1, then p 1d,
for some PEP. Write d = d,p. Then d,EF and d EF,, so f(d,+dq)EF,,, some q E R, since d 1 dl. Applying the induction hypothesis and replacing d,p, we have n > degR(f(4
+ 4pq))
= degR(4f(l
for d by
+ pq)) 2 deg,(dJ.
Hence deg,(d)
= deg,(d,)+
1 G n.
Conversely suppose d E F and deg,(d)G n. Suppose a E F and d ,/ a. Then degR(f(a + d)) = deg,(gcd(a, d)) < II. By the induction hypothesis, then f(a + d) E F,,. Thus d E F,. Since deg,(d) is always finite, it follows that F, = F. Hence R is a Euclidean q domain by Corollary 5.10, and & = &$ = deg,f = deg,. If P is a discrete set of primes for a UFD R, then R(P) is Euclidean P/l is canonical and discrete for R(P). We then have &&a/b)
= deg,,,,(a/b)
= deg&).
because
67
Recursive properties of Euclidean domains
This says that the rank of u/b in R(P) is the same as the P-degree the rank of a/b is completely restricted
determined
by the characteristic
of a in R. Thus function
to the set of primes which occur in a. We call this property
of P
‘prime-
independence’. 6.3.
Example.
A monomial
domain k(V)
is Euclidean,
because
V is a discrete
set of primes in k[V]. The Euclidean rank of an element p/q in k(V)’ is the same as the polynomial degree of the monomial content of p. In particular, if m is a monomial in k[V]‘,
then 4kcv,(m/l> = deg(m).
7. The second case: E & -U We begin with an r.p. Euclidean
domain which possesses a recursive,
normal
Euclidean function. We choose an n.r.e. set of distinct primes and declare them to be units. The resulting fraction ring has an n.r.e. set of units and, as noted by Motzkin [lo], is Euclidean. However, our construction of a Euclidean function for the ring must differ from his in order that it be recursive. 7.1. De6nition. Let R be an r.p. Euclidean domain and 4 a Euclidean function for it. We call q a partial recursive division algorithm for 4 if q is a partial recursive function defined on a subset of R” x R”, and for all d, a in R”: (1) If d 1 a, then q(d, a) halts. (2) If q(d, a) halts, then ~$(a + dq(d, u))<+(d). 7.2. Lemma. Let R be an r.p. Euclidean domain. Each function C$for R has a partial recursive division algorithm. Proof.
Let C$ be a recursive
Euclidean
function
recursive Euclidean
for R. We can then define a
partial recursive division algorithm q for C$ as follows: Given d and a in R”, search for x in R such that 4(a + dx) <4(d). If such an x is found, then set q(d, a) = x. If no such x is found, then q(d, a) remains undefined. It is clear that q satisfies the requisite properties. 0 7.3. Proposition. Let R be an r.p. Euclidean domain and 4 a Euclidean function for it. If 4 has a partial recursive division algorithm q, then R has a recursive greatest-common-divisor function. Proof. We compute a gcd of a pair do, dI of elements of R” by passing through a sequence of stages s (s = 0, 1,2,. . .). At stage s we simultaneously perform operations A, and J3, : A, : B,:
Search for x E R” such that d, = dS+Ix. Compute
q(d,+,, d,>.
68
L. Schrieber
By definition of q, one or the other must halt. If B, halts first, then set d st2 = 4 + 4+1d4+1, d,) and go to stage s + 1. If A, halts first, then the procedure ends. At each stage s we have #r(d,+,)<&(d,+,) and so the procedure must eventually halt by finding an x such that d, = d,,, x. The usual argument then shows that
d,+l is a gcd of d,, and d,.
0
7.4. CoroIIary. If an r.p. Euclidean then it has a recursive gcd function. 7.5.
timma.
domain has a recursive Euclidean
Let R be an r.p. Euclidean
function,
domain, let Cc R be an r.e. set of distinct
primes, let P be a canonical set of primes extending C, and set S = R(P\ C). Then: (1) S is recursively presented. (2) Every normal Euclidean function b, for R induces a Euclidean for S. (3) If do is recursive, then & is recursive.
function &
S is recursively presented because S = Q(R, (R* U C)), and (R” U C) is r.e. Define & on So by +,(a/b) = &(a ,), where a = a, * gcd(a, b). It is a simple Proof.
matter
using the normality
of 4
to show that &
is well-defined.
Note that
+&lb) s+(a). It is clear from Corollary 7.4 that & is recursive if 4 is recursive. To see that & is Euclidean for S, suppose p/pl, d/d, E So and d/d, 1 p/pl. We may presume gcd(d, d,)- 1. Write 1$(pfdq)<4(d), some q E R. Then &(P/P~ + (dld,)(d,qlpJ)
= &((P + d&p,) s4(p+dq)
q
7.6. Theorem. There is an r.p. domain which has a recursive, finitely valued Euclidean function and a nonrecursive set of units. Proof.
Let
p :o + 2
be a one-to-one
enumeration
of the set P of positive
primes. Let Cc o be any n.r.e. set and set C = p[C]. Then S = Z(P\ C) is an r.p. domain which by Lemma 7.5 has a recursive Euclidean function. The recursive 0 function from w to S given by e + pJ1 shows that C s1 S”. Hence S” is n.r.e.
8. Monomial
maps
The construction of a Euclidean domain in which property E fails requires the ability to defeat each recursive function. The possibility of doing so is suggested by the existence of the minimum Euclidean function. What is needed is a method by which the Euclidean ranks of certain elements can be increased to arbitrarily high values. Monomial maps provide such a means. A monomial map is a map on
Recursive
properties of Euclidean
domains
69
a polynomial domain k[V] which takes each variable X to a positive power Y’ of another variable. Upon passing to the corresponding
monomial domain, the rank
of X/l changes, in effect, from 1 to t. A power map is a monomial map which maps distinct variables to powers of distinct variables. Under suitable conditions a monomial map can be approximated by power maps. Power maps preserve divisibility of monomials
and they
commute with the monomial content function. These properties will be useful in showing that the kernel of the monomial map to which the power maps belong is recursive. In this section we establish the basic properties of monomial and power maps. 8.1. De&&ion. Let P and V be sets of variables. We call a map y : k[P] -+ k[ V] a monomial map if y fixes k and there are functions h : P -+ V and t : P + 2’ for which yX = (AX)“, all X E P. We call A and t the defining functions of y. If A is one-to-one, then we call y a power map. It is clear that any monomial map y : k[P] * 8.2. Proposition. Let y : k[P] + k[V] X and t. Then (1) deg(ya) “deg(a), for a E F(P). (2) deg&r(p))
~deg(f,(p)),
for P E
k[ V] takes F(P) into F(V).
be a monomial map with defining functions
Wl\ker(y).
Proof. (1) is clear: since the values of the defining function t are assumed to be positive integers. (2) We calculate:
dedfdp)) = &d.fv~fp(p))+~4fmp(~)) 2 dd_f~rf&))
= deg(yfAp))
3 deg(f,(p)).
q
For a positive integer t denote by k’ the set {a* ) a E k}. 8.3.
Lemma.
Let P be a set of variables, t : P+
Z’
a function and p E k[P]O.
There is a valuation (a,, . . . , a,) of the variables (Yl, . . . , Y,) occurring in p for which p#O and aiEk’9, i=l,.. .,s.
Proof. Induct on the number s of variables occurring in p. If s = 0, then pf 0 is constant. For s+l, write Y for YS+l and write
p=peY”+.*.+po,
e>O,
p,#O,
PO, * * * 3 pe E W\WH. Applying the induction hypothesis, assign a value 4 E kiYc to each variable Yi occurring in pe so that pe # 0; assign zero to the remaining variables, except Y.
70
L. Schrieber
Then p becomes a nonzero polynomial in k[Y]. but kty is infinite. 0 8.4. Proposition. Let y : k[P] ---, k[V] and t. ‘Then: (1) y is one-to-one, (2) U]yp+XIp,
forXEPand
It has only finitely many roots,
be a power map with defining functions
A
pEkD?l,
(3) ya 1rb + a 1b, for a, b E P(P), (4) Y KW)I c GWI,
fvr = rfp and gvy = m.
(5) Proof.
(1) Given p E k[P]‘, to show -yp# 0 write p = p(X,, . . . ,X,). Then yp = PO.x,F, - . . , (AX,)“~). By Lemma 8.3 we may choose a,, . . . , a, E k for which p(aixl,. . . , a”) # 0. Assigning & to AXi, then yp# 0. (2) Assume AX divides yp. Write p = qX+ r, for some q E k[P] and rE
k[P\{X)]. Then yp = (yq)(U?Jti+ yr. Since AX divides -yp then AX divides yr. But MC does not occur in yr, because X does not occur in r and A is one-to-one. Hence yr = 0, so r = 0. Then X divides p. (3) Assume ya divides yb. Write a = X;ll. and nj may be zero). Then ya = (AXJmltX, . . . (xX,)“sas
and
**Q
and b = X;l-
yb = (~l)“~‘%
- *m
(some mi
. . . (~s)“sas.
Since A is one-to-one we may conclude that mitXi c n,tX, for each i = 1, . . . , s; hence mi s Q since tXi a 1. Consequently a divides b. (4) If p E G(P), then p is not divisible by any variable in P. To show that ?/pis not divisible by any variable in V, note that the only variables which could possibly divide yp are those variables which occur in yp. Every such variable is AX for some X E P. But X does not divide p, so AX does not divide yp by (2). (5) By Lemma 5.4. Cl
9. Renaming systems Throughout the remainder content function. 9.1.
Definition.
of this paper, by f we always mean the monomial
Let A c P c V be sets of variables,
Let A : A + V and t : A +
Z+. The map on k[P] determined by A and t is the map y : k[P] --j k[V] which fixes k and P\A and maps XE A to (AX)%. It is the same as the monomial map with extensions A and t of h and t as defining functions, where A is the identity and t equals 1 on P\ A. Note that if A is one-to-one and h[A] is disjoint from P, then A is one-to-one and so y is a power map.
71
Recursive properties of Euclidean domains
9.2. Definition. Let V be a set of variables.
We call (A, t, A) a renaming system
in Vif
and t:A+Z+.
For
h:A+V\A
AcV,
isone-to-one,
by yA, the map on k[Vl
any AOc A we denote
determined
by the
restrictions of X and t to AO. We denote yA simply as y. For A,,c A and PC V the restriction of yPnAo to k[P] is the map on k[P] determined
by the restrictions
of A and t to P rl AO. It is clear that yPnA, agrees
with yA, on k[P]. Hence if h[P n A,] is disjoint from P, then the restriction of y,+ to k[P] is a power map. 9.3.
Lemma. Let (A, t, A) be a renaming system in V. For A0 c A, c A: (1) ?*,?/a,= 3/A,- In particular, y?/a, = y.
(2) deg(fy,,p)==degvy&p),
for P E
Wl\k4yAI).
Proof. (1) We show yA,yAoX = yA,X for each variable X E TJ. If XEA~, then AXE V\A, so y,,hX=hX. Then YA,x4$=
If X$AO,
~Alwa*)=@X)*=YA,X.
then -y&X=X,
so ya,ybX=
(2) degCfyAlp) = dcgUyAlyhp)
yA,X.
sdcg(fy,p)
by Lemma 8.2.
9.4. Definition. Let (h, t, A) be a renaming {X E P n A 1hX E P} the internal set of P.
0
system in V. Given
P c V, call
It is plain that the internal set of P increases as P increases. If A,, is the internal set of P, then yA, maps k[P] into k[P\A,], and the internal set of P\A, is empty. If the internal set of P is empty, then the restriction of y to k[P] is a power map. 9.5.
Remark. If A, is the internal set of P, then combining Proposition
Lemma 9.3 we have the following commutative
8.4 with
diagram:
WI
%I \ W’\AoI be-to-one) A f
I
F(P\Ao) y,
kW\Al
I
f
WV-U
A similar diagram holds with g in place of f. In particular, for p E k[P] we have the useful formulas yfy&p = fyp and ygy&p = gyp.
72
L. Schrieber
9.6.Proposition. Let (A, t, A) be a renaming system in V and let A,
be the
internal set of P c V. Then for p E k[P]: (1) P E MyI (2)
P E
$7 Y&P = 0
y-l[G(V\A)I
ifi ~Y&P = 1.
Proof. Applying the diagram above, we compute: (1) poker iff ?/p=O, iff iff
YY~P=&
y&p=O.
(2) P E Tl[G(V\A)I
iff iff
YP E G(V\A), f’yp=l,
ifi iff
yf~&p = 1, fy&p=l.
Cl
10. Recursive renamingsystems Throughout presented.
the remainder
of this paper
we assume
k to be recursively
10.1.Definition. Denote by W the countable set {X,},,, of variables. We call (A, t, A) a recursive renaming system if it is a renaming system in W, h[A] is a recursive set, and A and t are partial recursive functions. It is clear that if A, c A is recursive, then rA, : k[ W] + k[ W \ A,] is a recursive function, determined uniformly from A,. In particular this is true if A0 is finite.
10.2.Lemma. Suppose A, B c W and A : A * B is a bijective partial recursive function. Assume B is recursive. Then given any finite set PC W, the internal set of P can be computed uniformly from P. Proof. Let {Ai}iEo be a recursive enumeration of all the finite subsets of A. We can then find i for which A[A,] = B n P, since A is onto. Since A is also one-to-one, it follows that Ai = A-‘[B n P]. The internal set of P is then P n Ai.
q
10.3.Lemma. If (A, t, A)
is a recursive ypl[ G( W \ A)] are recursive sets.
Proof.
renaming
system,
then
ker(r)
and
Given p E k[ W] we can compute the internal set A,, of the set of variables
occurring in p. Then, by Proposition 9.6, p E ker(y) iff r&p = 0, and p E q r-l[G( W\ A)] iff fy&p = 1, and both of these conditions are recursive.
Recursive
properiiesof Euclidean
73
domains
10.4. Definition. Let (A, t, A) be a renaming system R(h, t, A) the domain Q(k[W], y-l[G(W\A)], ker(y)).
in W. We denote
by
We see that y induces an isomorphism R(X, r, A) = k( W \A) given by [u/b] + -&yb, hence R(X, t, A) is a Euclidean domain. The isomorphism, which we also denote by y, shows that {[X,/l] ( e E W\ A} is a canonical, discrete set of primes for R(h, t, A). We summarize the results of this section. 10.5.Theorem. Let
(h, t, A)
be
a recursive
R(h, t, A). Then: (1) R is a recursively presented Euclidean
renaming
system
and
let R =
domain.
(2) & (Mql) = deg(fyp), for all [dsl E R”. (3) In particular, if XE W then: (a)
& ([x/11) = tX
if X E A,
(b)
&([X/ll)
if X# A.
= 1
(4) The map from W\A[A]
to R given by X-,[X/l]
is one-to-one.
proof. (1) follows from Lemma 10.3. (2) is by applying the isomorphism y between R and k(W\A), and Example 6.3. (3) follows directly from (2). For (4), suppose X and Y are in W\ A[A] and [X/l] = [Y/l]. Then yX = +yY. But y restricted to k[ W\ h[A]] is a power map, hence one-to-one, because the internal set of W\I\[A] is empty. So X= Y. Cl
11. The third case: -E&U 11.1.Lemma. Let r be a recursively presented Euclidean domain. Suppose w : w -+ R” is a recursive function
and &w(e)
recursive, finitely valued Euclidean
> @e(e) for all e EK.
Then
R has no
function.
Proof. Define a recursive function h : w + w to satisfy 4,,(e) = c#+w,all e in w. Suppose +,e : R” + o is any recursive function. Then &h(e) is total, so h(e) E K. men &w(We)) >h,dh(e)) = &NdeN. Thus it is not the case that 4e ~4~ Consequently 4= cannot be a Euclidean function for R.
0
11.2.Theorem. There is a recursively presented Euclidean domain R which has no recursive, finitely valued Euclidean
function.
Proof. Let B c o be any infinite, co-infinite recursive set. Set B = {X, 1b E B}. Let i: o + o \B be any one-to-one recursive function. Set A = i[Kj. Then A is r.e. and infinite. Set A = {X, ] a E A}.
L.
14
Schrieber
Let h : A + B be any partial recursive bijection. Define t : A ---, Z+ by tX& = &(e) + 1. Then (h, t, A) is a recursive renaming system, Euclidean domain. Define w : w + R” by w(e) = [X&/l]. &w(e)= BY Lemma
11.3.
tX;ce,>4e(e)
so R = R(h, t, A)
Then
for all e EK.
11.1 it follows that R has no recursive Euclidean
m-rem.
is an r.p.
function.
[7
Let R be the domain of the previous theorem. We then have:
(1) R” and Ral are recursive. (2) R,, is recursively isomorphic to K, for n 22. (3) 7’he set of primes in R is recursively isomorphic to w \K. Proof. Enumerate the set A as {Ye}_,, and set A, = {Yi 1i
if degCfy4,-,vc,,(p))
divergent,
2 n, for some e,
otherwise.
The function
4 is partial recursive, since the polynomial degree function is recursive and ya. nv(P) is a recursive function determined uniformly from e and p. If degCfyp)an, then 4(p) converges because ~~-,~(~)p = yp as soon as % n V(P) = A n V(P). If degCfyp)
diverges because then deg(fy%,,(,,p)
by Lemma 9.3. Thus R,, = {[p/q] 1p E domain(+)},
because
C n for all e,
&([p/q]) 2 n iff deg(fyp) 2 n iff
4(p) converges. Hence R,, is r.e. In particular, Ral is r.e. Since R* is always r.e. and R”= R” 6 Ral, we conclude that R* and Ral are both recursive. If n 32, then R,, $ K since R,, is r.e. Conversely to show that K s1 R,,, define a one-to-one
4dx)
=
recursive function v : w + o by &.(e)+n-1,
if eeK,
divergent,
otherwise
for all XEO. Consider the recursive function from o to R given by e + [XiuceJl]* Since i and v are one-to-one and i maps into w\B, Theorem 10.5 shows that the function is one-to-one. If e EK, then v(e) E K SO Xiuclv(e)E A. Then &([Xi”(,)/lI) = %,ce)= &,,(v(e)) + 15 n. Hence If e$K, then v(e)$K R-1.
[Xiu,e,/lI ER,,. SO
X,,,,$A.
Then 4,([Xiu,,/l])=
1. Hence
[Xiu(eJ1IE
15
Recursive properties of Euclidean domains
The function e + [Xioce,/l] thus shows that K s1 R,,. This same function
also shows that w\K
12. Further properties 12.1.
corollary.
of the domain R
The predicate &(x> < +R(y) is not decidable.
Proof. Suppose it is. Choose any variable X$ A. Then r([Xll]) = X/l, so the rank of [X/l]’ is exactly t. Given any [plp,]~ R”, decide successively &([p/pJ)< &([X/l]t) for t = 1,2,3, . . . . The first t for which it is true tells us that the rank of [p/p,] is t- 1.
Cl
12.2. Definition. A Euclidean function for a recursively presented domain is said to have a recursive division algorithm if it has a partial recursive division algorithm (Definition 7.1) and if the divisibility relation in the domain is decidable. We show that 12.3.
Lemma.
4R has a recursive division algorithm.
Let R be the domain of Theorem 11.2. For [d/d,], [p/pJ~ R”:
(1) Cd/d111[P/PJ $7 frd 1fry. (2) If P = W, A0 is the internal set of P and p, d E k[P], fyhd
then [d/d,] ) [p/p11 ijj
1fyhp.
Proof. (1) [d/&l 1[PIPJ
iff
r&r&
1YP~YPI,
iff frd 1fry. (2) By Remark frd
I frp
9.5 and Proposition iff
rfy&
iff fs$i
8.4:
I yfyhpy fybp.
17
12.4. Theorem. The following operations are effective, for [d/d,], [plp,]~ R”: (1) To decide whether [d/d,] ( [p/p,]. (2) If [d/&l
Proof.
1 [P/PJ,
to find
[s/sJE
R
for which
(1) Compute the internal set A0 of the set of variables occurring in p or in
76
L. Schrieber
d. It is then simply a question of deciding whether fy&d 1fy&p, which can be done by explicitly computing them and comparing exponents. (2) Compute the internal set A, of the set of variables occurring in p or in d. Set q = dlgy4p 9.5:
and q1 = pIgybd.
Note that frd 1 fyp. Then,
applying
Remark
4~ ([P/PII+ [d4l[qld)
= &~~~\/~YP/YPI)+ (rdrdd(r4 = d&(~p
* mphp,
- gd)
- gyd + w’ * gyp))
= hd..fCfyp + fr4) < dedfd
= 4+([d/dJ.
0
13. The fourth case:-E & -U To construct a Euclidean domain in which both of the conditions E and U fail, we apply the technique of Theorem 7.6 to the domain of Theorem 11.2. The property of prime-independence possessed by discrete domains allows the set of units to be introduced without collapsing the ranks of elements elevated by the renaming system. 13.1. Theorem. ‘There is a recursively presented Euclidean domain which has no recursive, finitely valued Euclidean function, and in which the set of units is not recursive. Proof. Let R be the proof of that Let Cc w be D = W\ (A U C).
the Euclidean domain of theorem. For any set E any n.r.e. set disjoint This gives a partition of
Theorem 11.2. We use the notation of c W, denote [E/l] = {[X/l] 1XE E}. from AUB. Set C = {X, 1c E C} and W into A, C and D; and B c D. Since
D is disjoint from A, [D/l] is a set of distinct primes in R and y maps [D/l] oneto-one onto D/l. We have seen that [C/l] U [D/l] is canonical and discrete for R ; hence the smaller set [D/l] is also discrete for R. Set S = R([D/l]). Writing S = Q(R, (R*U[C/l])) makes it clear that S is recursively presented, because C is r.e. and so (R* U [C/l]) is also r.e. Proposition 6.2 shows that S is Euclidean. Denote the element [X,/l]/[l/l] of S as x,. Then, applying the isomorphism y between R and k(W\A), we see that &(x,)
= deg,(yX,)
for all e E w.
We show C s,,, S* via the recursive function from o to S given by e + x,. If e E C, then yX, = X,, so deg,(yX,) = 0, since X, 6 D. If e$ C, then yX, is either a positive power of hX, E B c D (if X, E A) or X, itself (if X, ED). In either case deg,(yX,)>O. Thus e EC iff &(x,) = 0 iff x, E S*. Hence S” is not recursive.
Recursiue properties
To see that S has no recursive 11.2. Define
of Euclidean domains
Euclidean
function,
77
we proceed
as in Theorem
w : w + So by w(e) = x+_). If e EK, then:
&(w(e))
= degD(+G(,,) = deg, ((XXi(ej)*q(=l)
(since Xicej E A)
= txiCe)
(since
AXice, E B C 0)
=+e(e>+l. By Lemma
14. An open Church
11.1, S has no recursive
Euclidean
0
function.
question
and
Kleene
showed
how
infinite ordinals as values (see [5]). ordinal-valued recursive functions?
to define
recursive
Do
Theorems
Nerode
for
functions
11.2
and
which
take
13.1
hold
for
problem
and
for
Acknowledgements The
author
is indebted
to Anil
posing
the
supplying assistance and encouragement throughout the writing of this paper. He also thanks Jeff Remmel for drawing his attention to the fourth case of the main theorem and for suggesting a method of proof.
References [ll A. Seidenberg, Constructions in algebra, Trans. Amer. Math. Sot. 197 (1974) 273-313. [2] Z.I. Borevich and I.R. Shafarevich, Number Theory (Academic Press, New York, 1966). [3] A. Frohlich and J.C. Shepherdson, Effective procedures in field theory, Phil. Trans. Royal Sot. London (ser. A) 284 (1955) 407-432. [4] H.W. Lenstra, Jr., Lectures on Euclidean Rings (Bielefeld, 1974). [Sl H. Rogers, Theory of Recursive Functions and Effective Computability (McGraw-Hill, New York, 1967). [6] J.B. Tennenbaum, A Constructive Version of Hilbert’s Basis Theorem, Ph.D. Dissertation (1973), Univ. of Calif., San Diego. [7] G. Metakides and A. Nerode, Effective content of field theory, Ann. Math. Logic 17 (1979) 289-320. [81 M.O. Rabin, Computable algebra, general theory and theory of computable fields, Trans. Amer. Math. Sot. 95 (1960) 341-360. [9] P. Samuel, About Euclidean rings, J. Algebra 19 (1971) 282-301. Cl01 Th. Motzkin, On the Euclidean algorithm, Bull. Amer. Math. Sot. 55 (1949) 1142-1146.