THE COMPLEXITY OF PRIMES IN COMPUTABLE UFDS

Transcription

1 THE COMPLEXITY OF PRIMES IN COMPUTABLE UFDS DAMIR D. DZHAFAROV AND JOSEPH R. MILETI Abstract. In many smple ntegral domans, such as Z or Z[], there s a straghtforward procedure to determne f an element s prme by smply reducng to a drect check of fntely many potental dvsors. Despte the fact that such a nave approach does not mmedately translate to ntegral domans lke Z[x] or the rng of ntegers n an algebrac number feld, there stll exst computatonal procedures that work to determne the prme elements n these cases. In contrast, we wll show how to computably extend Z n such a way that we can control the ordnary nteger prmes n any Π 0 2 way, all whle mantanng unque factorzaton. As a corollary, we establsh the exstence of a computable UFD such that the set of prmes s Π 0 2-complete n every computable presentaton. 1. Introducton The power and versatlty of modern algebra arse from the abstract and axomatc approach t takes. However, wth the rse of computer algebra systems, t s mportant to fnd algorthms n order to perform computatons wthn these algebrac structures. Of course, n these settngs, one also cares about the effcency of these procedures. For example, although the prmes n Z are trvally computable, there s a great deal of nterest n how quckly we can determne whether an element s prme. In contrast, t s known that there are computable ntegral domans where t s mpossble even n prncple to determne the prmes computatonally (see below). In ths paper, we extend these examples to buld a computable UFD where the prmes are maxmally complcated n a very strong sense. We begn wth the followng defnton (see [13] for background on the formal defntons of computable sets and functon). Defnton 1.1. A computable rng s a rng whose underlyng set s a computable set A N, wth the property that + and are computable functons from A A to A. For example, t s easy to vew Z as a computable rng by usng the even natural numbers to code the postve elements n ascendng order and the odd natural numbers to code the negatve elements n descendng order. Of course, we can vew Z as a computable rng n a dfferent way by swtchng the roles of the evens and odds. Thus, a gven rng can have multple dstnct computable presentatons. Many other natural rngs can also be vewed as computable rngs. Snce we can code relatvely prme pars of natural numbers usng a sngle natural number, we can vew Q as a computable rng. Smlarly, snce we can code fnte sequences of ntegers as natural numbers, we can vew Z[x] as a computable rng as well. Generalzng Dzhafarov was partally supported by an NSF Postdoctoral Fellowshp. We thank Sean Sather- Wagstaff for pontng out Nagata a Crteron to us, and Keth Conrad for helpful dscussons. 1

2 2 DZHAFAROV AND MILETI ths, gven an arbtrary computable rng A, we can realze the polynomal rng A[x] as a computable rng n a natural way. In contrast, uncountable rngs can never be vewed as computable rngs, and there are some countable rngs that can not as well. For a general overvew of results about computable rngs and felds, see [14]. Computable felds have receved a great deal of attenton ([8], [10], [12]), and [11] provdes an excellent overvew of work n ths area. For computable rngs, several papers ([2], [4], [6], [7]) have studed the complexty of deals and radcals from the perspectve of computablty theory and reverse mathematcs. The followng algebrac defntons are standard. Defnton 1.2. Let A be an ntegral doman,.e. a commutatve rng wth 1 0 and wth no zero dvsors (so ab = 0 mples ether a = 0 or b = 0). Recall the followng defntons. (1) An element u A s a unt f there exsts w A wth uw = 1. We denote the set of unts by U(A). Notce that U(A) s a multplcatve group. (2) Gven a, b A, we say that a and b are assocates f there exsts u U(A) wth au = b. We denote the set of assocates of a by Assocates A (a). (3) An element p A s rreducble f t nonzero, not a unt, and has the property that whenever p = ab, ether a s a unt or b s a unt. An equvalent defnton s that p A s rreducble f t s nonzero, not a unt, and ts dvsors are precsely the unts and the assocates of p. (4) An element p A s prme f t nonzero, not a unt, and has the property that whenever p ab, ether p a or p b. We denotes the set of prmes of A by P rmes(a). (5) A s a unque factorzaton doman, or UFD, f t has the followng two propertes: For each a A such that a s nonzero and not a unt, there exst rreducble elements r 1, r 2,..., r n A wth a = r 1 r 2 r n. If r 1, r 2,..., r n, q 1, q 2,..., q m A are all rreducble and r 1 r 2 r n = q 1 q 2 q m, then n = m and there exsts a permutaton σ of {1, 2,..., n} such that r and q σ() are assocates for all. It s a smple fact that f A s an ntegral doman, then every prme element of A s rreducble. The converse fals n general, but s true n every UFD. In fact, we have the followng standard result. Theorem 1.3. Let A be an ntegral doman. The followng are equvalent: (1) A s a UFD. (2) Every element of A that s nonzero and not a unt s a product of rreducbles, and every rreducble element of A s prme. Of course, for most computable ntegral domans that arse n practce, the set of prmes form a computable set n any natural computable presentaton. For the rng Z, the set of prmes trvally form a computable set. Kronecker showed that the set of prmes n (any reasonable computable presentaton of) the UFD Z[x] s computable. Usng Gauss Lemma and the fact that every element of Q[x] s an assocate of an element of Z[x], t follows that the set of prmes n Q[x] s computable as well. Consder a number feld K wth [K : Q] = n and let O K be the set of algebrac ntegers n K. In general, O K s always a Dedeknd doman, but t may not be a

3 THE COMPLEXITY OF PRIMES IN COMPUTABLE UFDS 3 UFD. We may fx an ntegral bass of K over Q,.e. fx b 1, b 2,..., b n O K that form a bass for K over Q such that O K = {m 1 b 1 + m 2 b m n b n : m Z}. Now gven the fntely many values b b j, we can compute the multplcaton functon on K and hence on O K as well. Snce we can smply hard code n these values, t follows that any ntegral bass provdes a computable presentaton of the feld K (by workng wth underlyng set Q n ) and the rng O K (by workng wth underlyng set Z n ). We have the followng fact. Proposton 1.4. Let K be a number feld wth [K : Q] = n. If we fx an ntegral bass of K over Q, and represent elements of O K usng elements of Z n, then the set of prmes elements of O K s computable. Proof. Gven α K, the map ϕ α : K K defned by ϕ α (x) = α x s a Q-lnear map, and moreover we can unformly compute a matrx M α wth ratonal entres representng ths map because we need only express α b n terms of our bass. Furthermore, notce that f α O K, then ϕ α maps O K nto O K, and hence M α has nteger entres. From ths, we can conclude that the norm map N : O K Z defned by N(α) = det(ϕ α ) = det(m α ) s a computable functon. Snce an element α O K s a unt f and only f N(α) = ±1, t follows that U(O K ) s a computable set. Moreover, gven α, β K wth α 0 represented as elements of Q n, we can unformly compute β α as represented by an element of Qn by smply searchng through the effectvely countable set Q n untl we fnd γ K wth γ α = β. Now f α, β O K, we can effectvely determne f α β n O K by checkng f ths representaton of β α s n Zn. Therefore, the dvsblty relaton on O K s computable. Snce we can compute the norm of an element, and snce O K / α = N(α), we can compute the functon f : O K \{0} N defned by f(α) = O K / α. Now to determne f α s prme, we compute f(α) and then search untl we fnd f(α) many dstnct representatve of the quotent (ths s possble because the dvsblty relaton s computable). Wth these representatves, we can form the fnte multplcatve table of the quotent (agan usng the fact that the dvsblty relaton s computable). To determne f α s prme, we then check f the quotent has any zero dvsors, whch s now just a fnte check. Despte all of ths, there are computable ntegral domans such that the set of prmes s not computable. In fact, there s a computable feld F such that the set of prmes n F [x] s not computable (see [11, Lemma 3.4] or [14, Secton 3.2] for an example). There exst methods to measure the complexty of sets that are not computable, and we nvestgate the placement of such sets n the arthmetcal herarchy arsng from quantfyng over computable relatons. Defnton 1.5. Let Z N. We say that Z s a Σ 0 1 set f there exsts a computable R N 2 such that Z ( x)r(x, ). We say that Z s a Π 0 1 set f there exsts a computable R N 2 such that Z ( x)r(x, ).

4 4 DZHAFAROV AND MILETI We say that Z s a Π 0 2 set f there exsts a computable R N 3 such that Z ( x)( y)r(x, y, ). Snce t s possble to computably code fnte sequences of natural numbers wth a sngle natural number, the above defntons do not change f we allow fnte consecutve blocks of the same (exstental or unversal) quantfers. Although every computable set s Σ 0 1, there exsts a Σ 0 1 set that s not computable, such as the set of natural numbers codng programs that halt. Smlarly, the collecton of Σ 0 1 sets s a proper subset of the collecton of all Π 0 2 sets, and the collecton of Π 0 1 sets s a proper subset of the collecton of all Π 0 2 sets. See [13, Chapter 4] for more nformaton about the arthmetcal herarchy. Suppose that A s a computable ntegral doman. We then have that U(A) s a Σ 0 1 set because u U(A) ( w)[uw = 1], and the relaton uw = 1 s computable. The set of rreducbles of A s a Π 0 2 set because p s rreducble n A f and only f p 0 ( c)[pc 1] ( a)( b)[p = ab a U(A) b U(A)], and we already know that U(A) s a Σ 0 1 set. A smlar analyss shows that the set of prmes of A s a Π 0 2 set. Our man result s the followng, whch says that ths result s best possble n a very strong sense. Theorem 1.6. Let Q be a Π 0 2 set, and let p 0, p 1, p 2,... lst the usual prmes from N n ncreasng order. There exsts a computable UFD A such that: Z s a subrng of A. p s prme n A f and only f Q. Ths theorem dffers from the result that there s a computable feld F such that the set of prme elements n F [x] s not computable. One reason s that we are workng drectly wth the usual prmes rather than codng nto polynomals (such as x 2 p ), or creatng our own prmes to do the codng. As a result, our approach has a more number-theoretc flavor. Furthermore, f F s a computable feld, then U(F [x]) s a computable set n any reasonable computable presentaton of F [x], so the set of rreducbles (and hence prmes) of F [x] wll always be a Π 0 1 set by our above analyss, and hence could not be Π 0 2-complete. Moreover, we obtan the followng strong corollary that may not hold f we code complexty nto other prmes. Corollary 1.7. There exsts a computable UFD A such that the set of prmes of A s Π 0 2-complete n every computable presentaton of A, unformly n an ndex for the presentaton. Proof. Fx a Π 0 2-complete set Q (see [13], Theorem IV.3.2), and construct A usng ths Q as n Theorem 1.6. Now gven any computable presentaton of A, we can fnd the multplcatve dentty element of A by searchng untl we fnd a A such that a 2 = a and a + a a (notce that the multplcatve dentty s the only such element because A s an ntegral doman). Wth ths element n hand, we can fnd the representaton of each p n A by addng the multplcatve dentty to tself the requred number of tmes. Therefore, the set of prmes of A s Π 0 2-complete n every computable presentaton of A.

5 THE COMPLEXITY OF PRIMES IN COMPUTABLE UFDS 5 Snce we are workng wth the normal nteger prmes rather than creatng some new ones, we need to be much more careful because of the algebrac dependence relatonshps that exst between them. By adjustng the status of one prme,.e. keepng t prme or makng t not prme, t s certanly concevable that we could nterfere wth others. For example, suppose that A s a ntegral doman, that q P rmes(a), and that we want to break the prmeness/rreducblty of q,.e. we want to ntroduce a nontrval factorzaton of q. One dea s to ntroduce a square root of q,.e. to ntroduce a new element x wth x 2 = q. The natural way to do ths s to consder A[x]/ x 2 q, but ths s problematc for a few reasons. Wth ths approach, we mght destroy the prmeness/rreducblty of other elements n A, as t s well-known that f p, q N are dstnct odd prmes, then p s not prme n Z[ q] = Z[x]/ x 2 q f and only f q s a square modulo p. For example, n Z[ 7], we have that 3 s not prme because 3 (1 7)(1 + 7) but and Moreover, n Z[ q], rreducbles mght fal to be prme, and hence we may have lost the property of beng a UFD. Fnally, wth ths approach t s also mpossble to later destroy ths factorzaton as we can not make x a unt wthout makng q a unt. Another potental ssue arses f we do want to destroy a gven factorzaton by makng an element a unt, but we are not n a UFD and/or are workng wth rreducbles. For example, n [3], the followng example s gven: n Z[ 14] one has = ( )(5 2 14) where all of the above factors are rreducble. It follows that even though and 3 are not assocates n Z[ 14] (as the unts are ±1). Thus, n ths rng, f we later make 3 a unt, then we must make a unt as well. Wth all of these potental ssues n mnd, we now outlne the dea behnd the constructon. Start wth A 0 = Z. We want to turn the normal prmes p on and off based on a Π 0 2 set Q. Fx a computable R N 3 such that Q ( w)( z)r(w, z, ) So ntutvely f acts nfntely often (.e. f for each w n turn, we fnd a wtnessng z), then we want p to be prme n the end. If acts fntely often, we want p not to be prme. To work for, we assume fnte acton, and ntroduce a factorzaton p = x y for new elements x and y. If acts at a later stage, we want to destroy ths factorzaton. To do ths, we make y a unt. We wll show that ths keeps x prme, and snce p wll now be an assocate of x, we wll renstate the fact that p s prme. We then ntroduce another factorzaton p = x y for new x and y, and contnue, destroyng t f acts agan. We do ths forever, buldng a chan of ntegral domans Z = A 0 A 1 A Let A = n=0 A n. We buld ths rng n a computable fashon as follows. We thnk of the natural numbers as beng splt nto nfntely many nfnte columns through a computable parng functon. We start by puttng the ntegers n the frst column and call that A 0. Now each extenson wll add nfntely many elements to the rng, and to do ths at a gven stage we wll smply add these elements nto the next column and computably defne addton and multplcaton at ths pont both wthn ths column and between ths column and prevous ones. Eventually, we wll fll up all

6 6 DZHAFAROV AND MILETI of the columns n turn, and defne all of the operatons, resultng n a computable rng. Wth ths constructon, we wll need to keep track of several thngs. For example, when we make an element a unt, we wll localze our rng, and snce we have already constructed part of the rng so far we wll need to ensure that we can computably determne the new elements to add n order to form ths localzaton. As a result we wll need to ensure that we can computably keep track of the multples of the x and y that we ntroduce. Algebracally, we need to ensure that the rngs along the way are all Noetheran UFDs and that unrelated prmes are unaffected by these operatons. Fnally, we need to check that ths lmtng rng has the requred propertes snce a unon of UFDs need not be a UFD n general. 2. Turnng a Prme nto a Unt Let A be an ntegral doman and let q A be prme. Suppose that we want to embed A n another ntegral doman B such that q s a unt n B. Naturally, one consders the correspondng localzaton,.e. we take the multplcatve set S = {1, q, q 2,... } and let B = S 1 A. Thnkng of A as sttng nsde ts feld of fractons F by dentfyng a wth a 1, we have { } a B = q k : a A and k 0 { } a = A q k : a A and k 1. Now f A s a computable ntegral doman and we want to thnk about extendng to B n a computable fashon, then we need to know whch of the elements n the set on the rght are really new, along wth when they are dstnct from each other For example, we have that q2 q = q 1 s already an element of A, so we do not want to ntroduce t. Notce that every element of B\A can be wrtten n the form a where k 1 q k a and q a. To see ths, suppose that we are gven a general q B wth a A and m m 1. If q a, we can factor out a q from a and cancel terms to obtan a dfferent representaton of the same element wth a smaller power of q n the denomnator. We can now nduct (or take a mnmal power n the denomnator) to argue that ths element s represented n the above set. Thus, we have { } a B = A : a A, q a, and k 1. qk Moreover, t s not dffcult to show that the above representatons are unque (.e. that a / A when q a and k 1, and also that two elements of the rght set q k are equal exactly when the numerator and power of q are equal. For the latter, f a = b wth q a and q b, then q l a = q k b. Cancel common q s. If k = l, then q k q l a = b, and we are done. Otherwse, we have a q left over on sde, so ether q a or q b, a contradcton). As a result, f A s computable, and the set {a A : q a} s computable, then from A, q, and an ndex for ths set we can unformly computably buld B as an extenson of A. Snce we are gong to repeatedly apply ths constructon along wth a factorzaton constructon, we wll need to ensure that the set of multples of other prmes reman computable as well.

7 THE COMPLEXITY OF PRIMES IN COMPUTABLE UFDS 7 Proposton 2.1. We have U(B) = U(A) {uq k : k 1 and u U(A)} { } u : k 1 and u U(A). qk Proof. Snce q 1 q = 1, we have that q and 1 q are both elements of U(B). We trvally have that U(A) U(B) and also that U(B) s closed under multplcaton. It follows that every element of the sets on the rght s an element of U(B). We now show the reverse contanment. Let σ U(B) be arbtrary. Suppose frst that σ = a A. We have two cases. Suppose that there exsts b A wth σb = ab = 1. We then trvally have that σ = a U(A). Suppose nstead that there exsts b A wth q b and l 1 such that σ b q l = a 1 b q l = 1. We then have q l = ab. Snce q s prme and q b, t follows from Lemma 2.2 that q l a. Fx c A wth a = q l c. We then have q l = ab = q l cb, so cb = 1 and hence c U(A). Thus, σ = a = cq l where c U(A). Suppose now that that σ = a q k where a A wth q a and k 1. Suppose that there exsts b A wth σb = a b q k 1 = 1. We then have qk = ab. Snce q a and q s prme, we conclude from Lemma 2.2 that q k b. Fx c A wth b = cq k. We then have q k = ab = acq k, so ac = 1 and hence a U(A). Thus, σ = a wth a U(A). q k Suppose nstead that there exsts b A wth q b and l 1 such that b σ = a b = 1. We then have that ab = q k+l. Snce k, l 1, ths q l q k q l mples that q ab and hence ether q a or q b (snce q s prme), a contradcton. Ths completes the proof. Lemma 2.2. Let A be an ntegral doman, let p A be prme, and let k 1. If p k ab and p a, then p k b. Proof. By nducton on k. If k = 1, ths s mmedate from the defnton of prme. Suppose that the result s true for a fxed k 1. Suppose that p k+1 ab and p a. Snce k 1, we have p ab, so snce p s prme we know that p b. Wrte b = pc for some c A. We then have p k+1 apc, so p k ac. By nducton, we conclude that p k c. Snce b = pc, t follows that p k+1 b. Theorem 2.3. Let A be a computable Noetheran UFD and let q A be prme. Suppose that {a A : q a n A} s a computable set. Let S = {1, q, q 2,... } and let B = S 1 A as above. (1) We can buld B as a computable extenson of A unformly from A and an ndex for the set {a A : q a n A} of multples of q. (2) Let p P rmes(a)\assocates A (q). The multples of p n B are precsely the elements of the followng set: { } a {a A : p a n A} : a A, k 1, q a n A, and p a n A qk In partcular, there are no new elements of A that are multples of p n B. Furthermore, f we have a computable ndex for the set {a A : p a n

8 8 DZHAFAROV AND MILETI A}, then we can unformly computably obtan a computable ndex for the set {σ B : p σ n B}. (3) If p 1, p 2 P rmes(a) are not assocates n A, then they are not assocates n B. (4) P rmes(a)\assocates A (q) P rmes(b). (5) B s a Noetheran UFD. Proof. (1) Immedate from above. (2) It s easy to see that the elements n the gven sets are ndeed multples of p n B. Suppose then that σ B s arbtrary wth p σ n B. Suppose frst that σ = a A. We need to show that p a n B. We have two cases. Suppose that there exsts b A wth pb = σ = a. We then trvally have that p a n A. Suppose nstead there exsts b A wth q b and l 1 such that b p = a. We then have pb = aq l. Thus p aq l n A, and snce p s q l prme and p q (as p / Assocates A (q)), t follows that p a n A. Suppose nstead that that σ = a where a A wth q a and k 1. We q k need to show that p a n A. Suppose that there exsts b A wth pb = σ = a. We then have q k pbq k = a, so p a n A. Suppose nstead that there exsts b A wth q b and l 1 such that b p = σ = a. We then have pbq k = aq l. Thus p aq l n A, and q l q k snce p s prme and p q (as p / Assocates A (q)), t follows that p a n A. Ths completes the proof. (3) We prove the contrapostve. Suppose that p 1 and p 2 are assocates n B. Fx σ U(B) such that p 1 = σp 2. We know the unts of B from Proposton 2.1, so we handle the cases. If σ U(A), then clearly p 1 and p 2 are assocates n A. Suppose that σ = uq k wth u U(A). We then have p 1 = uq k p 2, so p 2 p 1 n A. Snce p 1 s prme n A, t s rreducble n A, so as p 2 s not a unt we can conclude that p 1 and p 2 are assocates n A. Suppose that σ = u q k where k 1 and u U(A). We then have p 1 = u p q k 2, so p 1 u 1 q k = p 2. Ths mples that p 1 p 2 n A. As n the prevous case, ths mples that p 1 and p 2 are assocates n A. (4) Let p P rmes(a)\assocates A (q). Frst notce that p / U(B) from Proposton 2.1 because p / U(A) and p s not an assocate of any q k (because f p q k, then p q as p s prme, and hence p s an assocate of q). Suppose that p 1 a q k b q l where we allow the possblty that k = 0 and/or l = 0. Fx c A and m 0 wth p 1 c q m = a q k b q l We then have pcq k+l = q m ab, so p q m ab n A. Now p s prme and p q (as p / Assocates A (q)), so ether p a n A or p b n A. If p a n A, then t s easy to see that p 1 a n B. Smlarly f p b. Therefore, p P rme(b). q k

9 THE COMPLEXITY OF PRIMES IN COMPUTABLE UFDS 9 (5) Ths s mmedate from the fact that the localzaton of a Noetheran rng s a Noetheran rng, and the localzaton of a UFD s a UFD. Notce that usng ths machnery we can prove the result (essentally appearng [1] and [14, Example 4.3.9]) that there exsts a computable PID A such that U(A) s Σ 0 1-complete n all computable presentatons. Fx a Σ 0 1-complete set Q. Start wth A 0 = Z and let p 0, p 1, p 2,... be a lstng of the usual prmes. As we go along, f we have A n and we ever see that e Q, then we perform our unt constructon to buld A n+1 extendng A n so that p e U(A n+1 ) whle mantanng prmeness of the p not equal to p e or to any elements we already made unts. Let A = A = n=0 A n and notce that Q f and only f p U(A ). Snce the fnal rng A s a localzaton of the PID A 0 = Z, t follows that A s a PID. 3. Introducng a Factorzaton In ths secton, we suppose that we have a computable Noetheran UFD A and an element q P rmes(a). We ntroduce a new factorzaton of q by gong to the rng B = A[x, y]/ xy q. The hope s that we only destroy the prmeness/rreducblty of q (and ts assocates), and we leave enough flexblty so that we can later make y a unt wthout makng x a unt (so that then q and x wll be assocates). Proposton 3.1. B s an ntegral doman. Proof. We clam that xy q s rreducble n A[x, y]. Use the lexcographc monomal orderng n A[x, y] wth x > y. Recall that, under ths orderng, the multdegree of n k=1 c kx k y j k A[x, y] s the lexcographcally largest of the pars ( 1, j 1 ),..., ( n, j n ). Notce that the mult-degree of xy q s (1, 1), so f t factors, the leadng terms must have mult-degree (1, 1) and (0, 0) or (1, 0) and (0, 1) (snce mult-degrees add upon multplcaton). The former mples that one of the factors s constant, and hencemust be a unt snce the leadng coeffcent of xy q s 1. Consder the latter. We have xy q = (ax + by + c)(dy + e) where a 0 and d 0. Comparng coeffcents of x on each sde, we conclude that ae = 0, so e = 0 (because A s an ntegral doman). Comparng constants, we conclude that q = ce, so q = 0, a contracton. Snce xy q s rreducble n A[x, y] and A[x, y] s a UFD (because A s a UFD), we conclude that xy q s prme n A[x, y]. Therefore, the quotent B = A[x, y]/ xy q s an ntegral doman. Proposton 3.2. Every element of B can be represented unquely n the form where each a A, b A, and c A. a m x m + + a 1 x + c + b 1 y + + b n y n Proof. Gven an arbtrary polynomal h(x, y) A[x, y], we can dvde by xy q (usng the fact that the leadng term s a unt) to obtan a remander where no monomal s dvsble by xy. In other words, n the quotent, reduce any monomal wth xy n t to q, and repeat untl there are no xy s. Ths proves exstence. For unqueness, the dfference of any polynomals of ths form s another polynomal of ths form, and hence has no monomal contanng both an x and a y. Any nonzero

10 10 DZHAFAROV AND MILETI multple of xy q must have a monomal dvsble by xy by lookng a leadng term under some monomal orderng (and agan usng the fact that A s an ntegral doman). Notce that n B we have xy = q. Thnkng of y = q x, we can alternatvely thnk about B n the followng way. Proposton 3.3. Consder the followng subrng of A(x): [ A x, q ] { = a m x m + + a 1 x + a 0 + a 1 q } x x + + a n qn x n : a A We have B = A[x, q x ]. Proof. Defne a rng homomorphsm ϕ: A[x, y] A(x) by fxng A pontwse, sendng x x, and sendng y q x. We clam that ker(ϕ) = xy q. Frst notce that xy q ker(ϕ), so we certanly have xy q ker(ϕ). Let f(x, y) ker(ϕ). Dvde by xy q to wrte f(x, y) = (xy q) g(x, y) + r(x, y) where r(x, y) has no monomal havng both an x and a y. Notce that ϕ(r(x, y)) = ϕ(f(x, y)) = 0. Wrtng r(x, y) = a m x m + + a 1 x + c + b 1 y + + b n y n we then have that a m x m + + a 1 x + c + b 1 q x + + b n qn x n = 0 Multplyng by x n and lookng at coeffcents, we conclude that each a = 0, each b = 0, and c = 0. Thus r(x, y) = 0, and hence f(x, y) xy q. Snce B = A[x, y]/ xy q, t follows that we get an nduced njectve homomorphsm ˆϕ: B A(x). Snce every element of B can be represented n the form a m x m + + a 1 x + c + b 1 y + + b n y n, we conclude that range(ϕ) = range( ˆϕ) = A[x, q x ], and hence B = A[x, q x ]. We wll use the dfferent ways of representng elements of the extenson B = A [ x, q x] nterchangeably dependng on whch s most convenent. Wth ths somorphsm n mnd, we defne the followng two functons. Defnton 3.4. Defne deg x : B\{0} Z as follows. Let σ B and consder the unque representaton of σ gven n Proposton 3.2. If there s a term contanng a power of x wth a nonzero coeffcent, then deg x (σ) s the largest such power of x. If there s no such power of x, but there s a nonzero constant term, then deg x (σ) = 0. If there s no such power of x and no constant term, let m be the least power of y wth a nonzero coeffcent, and defne deg x (σ) = m. We defne deg y : B\{0} Z smlarly. For example, we have deg x (y 2 + y 5 ) = 2 and deg y (y 2 + y 5 ) = 5. Proposton 3.5. Let σ, τ B\{0}. We then have deg x (στ) = deg x (σ) + deg x (τ) deg y (στ) = deg y (σ) + deg y (τ)

11 THE COMPLEXITY OF PRIMES IN COMPUTABLE UFDS 11 Proof. It s straghtforward to prove ths n the case when σ and τ are monomals,.e. of the form ax k, by l, or c 0 (notce that here we use the fact that A s an ntegral doman to conclude that the product s a nonzero monomal). For general σ and τ, we need only examne the leadng x-terms or y-terms. Proposton 3.6. We have deg x (σ) + deg y (σ) 0 for all σ B\{0}, wth equalty f and only f σ s a constant tmes a monomal. Proof. Let σ B\{0}. If the leadng x-term s x m, then deg x (σ) = m and deg y (σ) m, wth equalty f and only f x m s the leadng y-term as well. A smlar argument works f the leadng y-term s y n. Otherwse, we only have a constant, t whch case both deg x (σ) = 0 and deg y (σ) = 0. Proposton 3.7. Let σ, τ B. We then have that στ A n exactly the followng cases: σ = 0 or τ = 0. σ A and τ A. There exst a, b A and n N + wth σ = ax n and τ = by n, or there exsts a, b A and n N + wth σ = by n and τ = ax n. Proof. In each of these cases t s easy to see that στ A. Suppose conversely that στ A. We may assume that σ 0 and τ 0 or else we are done. We then have deg x (σ) + deg x (τ) = deg x (στ) = 0 so deg x (τ) = deg x (σ). Smlarly, we have deg y (τ) = deg y (σ). Addng these gves deg x (τ) + deg y (τ) = deg x (σ) deg y (σ) = (deg x (σ) + deg y (σ)) Usng Proposton 3.6, the only possblty s that deg x (τ)+deg y (τ) = 0 = deg x (σ)+ deg y (σ), and hence that both σ and τ are constants tmes monomals. The result now follows. Corollary 3.8. Let a A wth q a n A. If σ B and σ a n B, then σ A and σ a n A. In other words, the set of dvsors of a n B equals the set of dvsors of a n A. Proof. By Proposton 3.7, the only possble new dvsors of a are when a = bx n cyn wth n 1. However, ths mples that a = bc q n, so q a n A. Corollary 3.9. The unts of B are precsely the unts of A,.e. U(B) = U(A). Proof. Ths s mmedate because the set of unts s the set of dvsors of 1 A. Theorem Let A be a computable Noetheran UFD and let q A be prme. Let B = A[x, y]/ xy q as above. (1) We can buld B as a computable extenson of A unformly. (2) If p 1, p 2 P rmes(a) are not assocates n A, then they are not assocates n B.

12 12 DZHAFAROV AND MILETI (3) Let p P rmes(a)\assocates A (q) and let σ B. We have that p σ n B f and only every coeffcent of σ s dvsble by p n A. In partcular, there are no new elements of A that are multples of p n B. Furthermore, f we have a computable ndex for the set {a A : p a n A}, then we can unformly computably obtan a computable ndex for the set {σ B : p σ n B}. (4) P rmes(a)\assocates A (q) P rmes(b). (5) x σ n B f and only f the constant term and the coeffcents of each y j n σ are all dvsble by q n A. Therefore, f we have a computable ndex for the set {a A : q a n A}, then we can unformly computably obtan a computable ndex for the set {σ B : x σ n B}. (6) y σ n B f and only f the constant term and the coeffcents of each x n σ are all dvsble by q n A. Therefore, f we have a computable ndex for the set {a A : q a n A}, then we can unformly computably obtan a computable ndex for the set {σ B : y σ n B}. (7) x and y are prmes n B that are not assocates of each other n B. (8) x and y are not assocates n B wth any element of A, and hence not wth any element of P rmes(a). (9) B s a Noetheran UFD. Proof. (1) Immedate from Proposton 3.2. (2) Immedate from Corollary 3.9. (3) Ths follows from the fact that p (a m x m + + a 1 x + c + b 1 y + + b n y n ) = pa m x m + + pa 1 x + pc + pb 1 y + + pb n y n. (4) Let p P rmes(a)\assocates A (q). Notce that p s nonzero and s not a unt of B by Corollary 3.9. Let σ, τ B and suppose that p στ. Assume that p σ and p τ. We clearly have that both σ and τ are nonzero. Usng 3, we know that p dvdes every coeffcent of στ n A, but there are coeffcents of σ and τ that are not dvsble by p n A. Wrte and σ = a m x m + + a 1 x + a 0 + a 1 q x + + a n qn x n τ = b m x m + + b 1 x + b 0 + b 1 q x + + b n qn x n Let k and l be largest possble such that p a k n A and p b l n A. Look at the coeffcent of x k+l n στ. Ths coeffcent wll be a sum of terms, one of whch s a k b l q j for some j, whle other terms wll be dvsble by p n A. Snce p dvdes the resultng coeffcent, t follows that p a k b l q j n A. However, ths s a contradcton because p s prme n A but dvdes none of a k, b l, or q (the last because p s not an assocate of q n A). (5) Let σ B and wrte σ = a m x m + + a 1 x + c + b 1 y + + b n y n Suppose frst that q c and q b j n A for each j. Fx e A wth c = qe and fx d j A such that b j = qd j for all j. We then have x (a m x m a 1 + ey + d 1 y d n y n+1 ) = σ

13 THE COMPLEXITY OF PRIMES IN COMPUTABLE UFDS 13 Conversely, suppose that x σ, so that σ = x (a m x m + + a 1 x + c + b 1 y + + b n y n ) for some a, c, b j A. Then we have σ = a m x m a 1 x + cx + qb 1 + qb 2 y +... qb n y n 1 (6) Smlar to 5. (7) Notce that x s nonzero and s not a unt of B by Corollary 3.9. Let σ, τ B and suppose that x στ. Assume that x σ and x τ. We clearly have that both σ and τ are nonzero. Usng 5, we know that q dvdes the constant term and the coeffcents of each y j n στ n A. Wrte and σ = a m x m + + a 1 x + a 0 + b 1 y + + b n y n τ = c m x m + + c 1 x + c 0 + d 1 y + + d n y n By 5, we may let k and l be largest possble such that q a k n A and q c l n A. Look at the coeffcent of x k+l n στ. Ths coeffcent wll be a sum of terms, one of whch s a k c l, whle other terms wll be dvsble by q n A. Snce q dvdes the resultng coeffcent, t follows that q a k c l n A. However, ths s a contradcton because q s prme n A but dvdes nether of a k or c l n A. The proof that y s prme n B s smlar. The fact that x and y are not assocates n B follows from Corollary 3.9. (8) Immedate from Corollary 3.9. (9) We are assumng that A s a Noetheran UFD. Snce A s Noetheran, we know that A[x, y] s Noetheran by the Hlbert Bass Theorem. Snce B s a quotent of A[x, y], t follows that B s also Noetheran. We also know from 7 that x s prme n B = A[x, q x ]. To argue that B s a UFD, we use Nagata s Crteron (see [9, Theorem 20.2] or [5, Lemma 19.20]) whch says the followng. Theorem 3.11 (Nagata s Crteron). Let B be a Noetheran ntegral doman. Let Γ be a set of prme elements of B, and let S be the multplcatve set generated by Γ. If S 1 B s a UFD, then B s a UFD. Now x s prme n B = A[x, q x ] by 7. The localzaton of A[x, q x ] at x equals A[x, q x, 1 x ] = A[x, 1 x ], whch s the localzaton of A[x] at x. Snce A s a UFD, we know that A[x] s a UFD. Snce any localzaton of a UFD s a UFD, t follows that A[x, 1 x ] s a UFD. Snce B s a Noetheran ntegral doman, x s prme n B, and B localzed at x s a UFD, we may use Nagata s Crteron to conclude that B s a UFD. 4. Constructon and Verfcaton We now prove Theorem 1.6. Let Q be an arbtrary Π 0 2 set. Fx a computable R N 3 such that Q ( w)( z)r(w, z, ) Fx a bjectve computable parng functon, : N N N wth the property that, s <, t whenever s < t.

14 14 DZHAFAROV AND MILETI We work n stages, and begn by ntalzng wth A 0 = Z. We now start at stage 0. At a gven stage, we wll have ntroduced fntely many x (k) and y (k) for each, and we wll have marked a fnte ntal segment of N correspondng to those w N for whch we have found a wtnessng z and done an acton. Furthermore, f has been ntalzed and the frst unmarked w s k, then we wll have ntroduced x (l) and y (l) for each l k, but we wll not yet have ntroduced x (k+1) and y (k+1). Suppose that we are now at a stage, s and we have constructed through rng A n at ths stage. If s = 0, we do an ntalzaton for p by ntroducng a frst factorzaton. In other words, we ntroduce x (0) and y (0) and perform the factorzaton constructon on p to create the rng A n+1 (so we fll n one more column), and then move on to the next stage. Suppose that s 1, and let k be the frst unmarked w correspondng to. Check to see f there exsts z s such that R(w, z, ). If not, we do nothng and move to the next stage. If so, we mark k for, and we act for at ths stage, meanng that we do the followng. As mentoned above, we wll have ntroduced through x (k) and y (k). Frst, we perform the localzaton constructon to make y (k) a unt n order to create the rng A n+1. Next, we ntroduce x (k+1) and y (k+1) and perform the factorzaton constructon wth these on p to create the rng A n+2. Thus, we fll n two more columns n successon, and then move on to the next stage. Fnally, let A = n=0 A n. Theorem 4.1. Suppose that we are at the begnnng of a gven stage and we have constructed through A n. For each that has been ntalzed, let x (k) and y (k) be the last elements ntroduced for (so k s the frst unmarked w for ). Suppose that has not yet been ntalzed. We have the followng: p s prme n A n. The set {a A n : p a n A n } s computable and we can unformly fnd a computable ndex for t. For any unntalzed j, we have that p s not an assocate of p j n A n. For any ntalzed j, we have that p s not an assocate of ether x (kj) j or y (kj) j n A n. Suppose that has been ntalzed. We have the followng: x (k) and y (k) are prme n A n, and are not assocates n A n. The sets {a A n : x (k) a n A n } and {a A n : y (k) a n A n } are computable and we can unformly fnd computable ndces for them. For any unntalzed j, we have that x (k) and y (k) are not assocates of p j n A n. For any ntalzed j, we have that x (k) s not an assocate of ether x (kj) j or y (kj) j n A n, and y (k) s not an assocate of ether x (kj) j or y (kj) j n A n. Suppose that we act for at ths stage. We then have y (k) U(A n+1 ), that x (k) s prme n A n+1, and that p s prme n A n+1.

15 THE COMPLEXITY OF PRIMES IN COMPUTABLE UFDS 15 Proof. The proof s mmedate by usng nducton on the stages along wth Theorem 2.3 and Theorem Defnton 4.2. Let, k N and suppose that we ntroduce x (k) constructon. We call x (k) y (k+1) for. and y (k) Proposton 4.3. We have the followng. and y (k) termnal for f we never ntroduce x (k+1) (1) Suppose that we ntroduce x (k) and y (k) n A m. If x (k) and y (k) for, then they are non-assocate prmes n A n for each n m. (2) If r A m s prme n A m and s not an assocate of any p, x (k) n our and are termnal, or y (k) (whether termnal or nontermnal) n A m, then r remans prme n A n for each n m. Proof. Agan, ths follows by nducton usng Theorem 2.3 and Theorem Proposton 4.4. Let a A, so a A m for some m N. The followng are equvalent: (1) a U(A ). (2) a U(A n ) for all suffcently large n m. (3) a U(A n ) for some n m. Proof. If a U(A ), then fxng b A wth ab = 1, we have that a U(A n ) for any n large enough such that a, b A n. If a U(A n ) for some n m, then fxng b A n wth ab = 1, we have a, b A, so a U(A ). Proposton 4.5. Let r A, so r A m for some m N. If there are nfntely many n m such that r s prme n A n, then r s prme n A. Proof. Suppose that there are nfntely many n m such that r s prme n A n. Fx a, b A and suppose that r ab n A. Fx c A wth rc = ab. Go to a pont where each of r, c, a, b exst, and then fx an n beyond that such that r s prme n A n. We then have r ab n A n, so as r s prme n A n, ether r a n A n or r b n A n. Therefore, ether r a n A or r b n A. Fnally, notce that r s nonzero and not a unt n A because nfntely often t s not a unt n A n (as nfntely often t s prme n A n ). Corollary 4.6. We have the followng. (1) If x (k) are ntroduced and are termnal for, then they are non- and y (k) assocate prmes n A. (2) If x (k) and y (k) are ntroduced and are nontermnal for, then y (k) U(A ), and x (k) s an assocate of p n A. (3) If r A m s prme n A m and s not an assocate of any p, x (k), or y (k) n A m (whether termnal or nontermnal), then r remans prme n A. Proof. Immedate from Theorem 4.1, Proposton 4.3, Proposton 4.4, and Proposton 4.5. Corollary 4.7. p s prme n A f and only f Q.

16 16 DZHAFAROV AND MILETI Proof. Suppose frst that Q. We then act for nfntely often, and hence p s prme n nfntely many A n by Theorem 4.1. Thus, p s prme n A by Proposton 4.5. Suppose now that / Q. We then act for fntely often, so we may fx k such that x (k) and y (k) are termnal for. By Corollary 4.6, each of x (k) and y (k) are prme n A. Snce p = x (k) y (k), t follows that p s not rreducble n A, and hence not prme n A. Lemma 4.8. Let m N. Let r A and suppose that r s prme n A m. We then have that ether r U(A ), r s prme s A, or r s the product of two prmes n A. Proof. We handle the varous cases. If there exsts Q such that r s an assocate of p n A m, then r s prme n A by Corollary 4.7. Suppose that there exsts / Q such that r s an assocate of p n A m. We then act for fntely often, so we may fx k such that x (k) termnal for. By Corollary 4.6, each of x (k) then have that p = x (k) Snce ux (k) prmes n A. and y (k) and y (k) and y (k) are are prme n A. We y (k), so r = ux (k) y (k) for some unt u U(A ). are prme n A, we see that r s the product of two If there exsts, k N such that r s an assocate of a termnal x (k) n A m, then r s prme n A by Corollary 4.6. If there exsts, k N such that r s an assocate of a termnal y (k) n A m, then r s prme n A by Corollary 4.6. If there exsts Q and k N such that r s an assocate of a nontermnal x (k) n A m, then r s an assocate of p n A by Corollary 4.6 and hence r s prme n A by Corollary 4.7. If there exsts / Q and k N such that r s an assocate of a nontermnal x (k) n A m, then r s an assocate of p n A by Corollary 4.6, and hence r s a product of two prmes n A from above. If there exsts, k N such that r s an assocate of a nontermnal y (k) n A m, then r U(A ) by Corollary 4.6. If r s not an assocate of any p, x (k), or y (k) n A m, then r s prme n A by Corollary 4.6. Theorem 4.9. A s a UFD. Proof. We prove that every nonzero nonunt element of A s a product of rreducbles and that every rreducble s prme, whch suffces by Theorem 1.3. Let a A be nonzero and not a unt. Fx n wth a A n and notce that a s not a unt n A n. Snce A n s a UFD, we may wrte a = r 1 r 2 r l where each r s rreducble and hence prme n A n. By Lemma 4.8, each r j s ether a unt n A, s prme n A, or s the product of two prmes n A. It s not possble that all r j are unts n A, because ths would mply that a s a unt n A. Thus, a s a product of prmes n A (snce we can absorb the unts n one of the prmes). Snce prmes are rreducble, we conclude that a s a product of rreducbles n A.

17 THE COMPLEXITY OF PRIMES IN COMPUTABLE UFDS 17 We now show that every rreducble element of A s prme. Let a A be rreducble. Fx n wth a A n. Notce that a s nonzero and not a unt n A n because otherwse t would be zero or a unt n A. Snce A n s a UFD, we may wrte a = r 1 r 2 r l where each r j s rreducble and hence prme n A n. By Lemma 4.8, each r j s ether a unt n A, s prme n A, or s the product of two prmes n A. It s not possble that all r j are unts n A, because ths would mply that a s a unt n A. If some r j s a product of two prmes n A, then a s not rreducble n A, a contradcton. Also, f two of the r j are prme n A, then A s not rreducble n A, a contradcton. Thus, exactly one of the r s prme n A and the rest are unts. It follows that A s a prme tmes some unts n A, so a s prme n A. Ths completes the proof of Theorem 1.6. References [1] Walter Baur. Rekursve Algebren mt Kettenbedngungen. Z. Math. Logk Grundlagen Math., 20:37 46, [2] Chrs J. Conds. On the complexty of radcals n noncommutatve rngs. J. Algebra, 322(10): , [3] Keth Conrad. Factorng n quadratc felds, June [4] Rodney G. Downey, Steffen Lempp, and Joseph R. Mlet. Ideals n computable rngs. J. Algebra, 314(2): , [5] Davd Esenbud. Commutatve algebra, volume 150 of Graduate Texts n Mathematcs. Sprnger-Verlag, New York, Wth a vew toward algebrac geometry. [6] Harvey M. Fredman, Stephen G. Smpson, and Rck L. Smth. Countable algebra and set exstence axoms. Ann. Pure Appl. Logc, 25(2): , [7] Harvey M. Fredman, Stephen G. Smpson, and Rck L. Smth. Addendum to: Countable algebra and set exstence axoms [Ann. Pure Appl. Logc 25 (1983), no. 2, ; MR (85:03157)]. Ann. Pure Appl. Logc, 28(3): , [8] A. Fröhlch and J. C. Shepherdson. Effectve procedures n feld theory. Phlos. Trans. Roy. Soc. London. Ser. A., 248: , [9] Hdeyuk Matsumura. Commutatve rng theory, volume 8 of Cambrdge Studes n Advanced Mathematcs. Cambrdge Unversty Press, Cambrdge, Translated from the Japanese by M. Red. [10] G. Metakdes and A. Nerode. Effectve content of feld theory. Ann. Math. Logc, 17(3): , [11] Russell Mller. Computable felds and Galos theory. Notces Amer. Math. Soc., 55(7): , [12] Mchael O. Rabn. Computable algebra, general theory and theory of computable felds. Trans. Amer. Math. Soc., 95: , [13] Robert I. Soare. Recursvely enumerable sets and degrees. Perspectves n Mathematcal Logc. Sprnger-Verlag, Berln, A study of computable functons and computably generated sets. [14] V. Stoltenberg-Hansen and J. V. Tucker. Computable rngs and felds. In Handbook of computablty theory, volume 140 of Stud. Logc Found. Math., pages North-Holland, Amsterdam, Department of Mathematcs, Department of Mathematcs, Unversty of Connectcut, Storrs, Connectcut U.S.A. E-mal address: damr@math.uconn.edu Department of Mathematcs and Statstcs, Grnnell College, Grnnell, Iowa U.S.A. E-mal address: mletjo@grnnell.edu