CYCLOTOMIC EQUATIONS AND SQUARE PROPERTIES IN RINGS

If R is a ring, the structure of the projective special linear group PSL2(R) is used to investigate the existence of sum of square properties holding in R. Rngs which satisfy Fermat's two-square theorem are called sum of squares rings and have been studied previously. The present study considers a related property called square property one. It is shown that this holds in an infinite class of rings which includes the integers, polynomial rings over many fields and Z n where p is a prime P such that -3 is not a square rood p. Finally, it is shown that the class of sume of squares rings and the class satisfying square property one are non-coincidental.


INTRODUCTION
Fermat's classical two-square theorem gives the relationship between those integers n which are sums of two squares and those integers n for which the cyclotomic x 2 equation + ! 0can be solved modulo n.Specifically -i is a quadratic residue mod n if and only if n is expressible as the sum of two relatively prime squares.In [i] a proof of this was given which involved the group theoretical structure of the modular group PSL 2(Z).Fermat's theorem is then, in a sense, independent of number theory in that the structure of PSL2(Z) can be deduced by purely analytic (Fuchsian Croup) ,thods.(Lehner [2]) TT.is dea was used by Fine [%] to show that Fermat's result holds in an infinite class of rings.Such rings were termed sum of squares rins.In this note we first use a technique similar to 3] to investigate the structure of" those integers n for which the cyclotomic equation x2+ x + I 0 has solutions mod n.A square result smmilar to Fermat's theorem is obtained.It is then shown that this square property holds in infinitely many sum of squares rings but that the two square properties are independent.Finally, some questions that were raised in 3] are answered.Before beginning, we note that the technique of 3] has been extended by Kern-Isberner and Rosenberger 4] to consider those integers which can be 2 d,v2 expressed in the form n x + where d is a fixed positive integer.In a different direction, the general situation for which the equation x2+ dx + 1 0 has solutions rood n was considered by Fine in [5].A collection of square results was obtained 2.
We first prove the following theorem concerning the integers Z.It is well know (Lehner 2], New-man 6]) that M Z2* Z 3 that is group theoretically M is a free product of a cyclic group of order 2 and a cyclic group of order 3.
If A is the map z' -i/z and B is the map z' -i/z+l then M has the presen- tation <A, B: i> [6] Since in a free product anF element of finite order must be conjugate to an element of finite order in one of the factors 7] it follows tat any element of order 3 in M must be conjugate to either B: z' -i/z+l or to B -I z (-z-l)/z.
If U: z' (az+b)/(cz+d) is an element of M then conjugating B and B -I we obtain U IBU: z' (dz-b) -i (az+b) -c.+a (z-J -) c-qT" (ab+cd+bc)z + (b2+ d2+ bd) 2+ -(a2+ c ac)z (be + ab + cd) or z (dz-b -z-l)(az+b] U-BU: ----g--zz+a)(--q c-qT" (ad+ab+cd)z + (b2+ d2+ bd) has determinant +! and is thus in M. Further it has trace +i.Elements of M with trace 1 have order 3, [6] so the map in (2.3) has order 3. Therefore, this map must have either form (2.1) or form (2o2). Thus, n b2+ d2+ bd for some b, de Z. Further, x + i 0 has a solution modulo n.In an identical manner, by dealing with traces of -i we get the reufLt concerning the equation 2 x x + i 0 mod n.By quadratic formula, x + x + i 0 and x + i 0 have solutions rood p (where p is an odd prime) if and only if -3 is a square mod p.Thus as a corollary we have COROLLARY" If p is an odd prime then -3 is a quadratic residue rood p if and 2

2+
only if p a + b ab for some relatively prime integers a, b The structure of the modular group can be used to effectively classify all the trace classes.This was done in 5].From this is is obtained that for each d 0 there exist finitely many quadratic forms (depending on d) such that x2+ dx + i 0 has solutions mod n if and only if n is represented by one of these forms.Further, there exists an effective procedure to write down each of these forms for each d.In tlis paper we take a different tract and consider those rings for which theorem i is val i d.

3.
Recall from 33 that a sum of squares ring is a commutative ring R with an identity (not a field) with -i not a square in R which satisfies the following two square properties: SSi: If re R and -i is a quadratic residue mod(r) then r (u2+ If r + with (u, v) i then -i is a quadratic residue mod (r) For SS2 a ZCD ring was not required (u, v) !indicating only that u and v have no co.ondivisors.In 3] it was shown that there are infinitely many sum of squares rings.Specifically, the following classes of rings were proven to be sum of squares rings.
B. FINE i) Z n where p is a prime, p 3 nod 4 and n > i P 2) The polynomial rings F x] where F is a field with -i not a square in F and where PSLo(F) has only one conjugacy class in trace zero.In particular a) Z Ix] where p is a prime congruent to 3 nod 4 P b) K Ix] where K is an ordered field permitting square roots of all positive elements.Examples here are Ix] and A [x] where is the real field and A the suhfie!l of algebraic numbers 3) General Euclidean domains D with trivial units, of characteristic # 2 and with a subadditive norm function satisfying 0 # N(b) <-N(a) implies N(a+kb) < N(a) for some ks Do (The integers Z provide an example of this last type of ring.) We now consider rings which satisfy the results of Theorem io We say that a commutative ring with an identity (not a field) satisfies square property one abbreviated S?I if the ring R satisfies + SPIa" if rci and x2+ x + !0 has solutions mod (r) then r (u2+ v2+ uv) for some ring elements u, v.
As in the case of sume of squares ring (u, v) I indicates that u and v have no cornon dvisors.A GCD ring is not required.We obtain: THEOP[ 2: The following classes of rings all satisfy square property one.a) n wher n>! an'] is a trine such +/-hat -3 is not a square od b) F Ix] where F is a field of characteristic #2, with -3 not a square in F and every matric of trace i in PSL2(F)is conjugate within PSL2(F)to either-+(0 -01 )I or + (-ii ) c) Euclidean domains D of char # 2 with trivial units and a sub-additive norm function satisfying 0 # N(b) -< N(a) implies N(a+kb) N(a) for some kDo PROOF" All of the above rings have the property that the GCD of two elements is expressible as a linear combination of these elements.Thus if (u, v) I there exist a, b with au + bv i. Employing the idential formal method as in the second part of the proof of Theorem i, it is seen that these rings all satisfy SPIb.
Since each of these rings has only trivial idempotents, the center of their special + SL2(k) is i with I the identity matrix.Thus PSL2(R) SL2(R)/-+ I linear grou}s for any of the above rings.It follows that any element of PSL2(R) can be considered as + or a matrix in SL2(R).Modeled on the first part of Theorem i, it is seen that these rings w=i! satisfy SPIa if every matrix of trace i is conjugate within We will show in turn that this is true for PSL 2(I{) either to -(i i -i 0 each of the above classes.a) If -3 is not a square in Z then -3 is not a square in Z n for all n > io p P !nerefore, there exists no solutions to x2+ x + I 0 in Z n Since n > i, Z n is P P not a field By a result of D L. McQuillan [8] if APSL2(Z n) and tr(A) i then A p 0 -i).Thus Z n satisfies square property one.
is conjugate to + (I i p b) if F is a field with char F # 2 and -3 not a square to F, we show that every matrix of trace 1 in PSL2(F [X])every matrix of trace one is conjugate to either _+ (0-1) + i-i) the result follows as above "l l" or-(1 0 Suppose TPSL2(F[x]) with T -+ (-f Let S be the set of all conjugates of h +l T in PSL2F [x]) and suppose VS with V + (-u v w v+l and with the degree of u minimal among all the conjugates of T. If v 0 or W 0 then -u(u+l) 1. Then u2+ u + 1 0 has a solution in F (u must be in F since the only units in F x] are in F) contrac- dicting -3 not being a square in F. Therefore, v # 0 and w # O.If ueF then since 2 -u u _(wT) + i, it follows that deg v + des w 2 deg u, and thus v and w are also in F.
Assume dec u->l.Since des v + deg w 2 deg u then deg v _< des u or des w _< deg u.By the division algorithm a polynomial q can be determined so that det (u + qv) < des u.
1 q) gives a matrix + (u+qv * Conjugating the matrix V by (0 1 , ,).This matrix is in S being a conjugate of V But this contradicts the minimality of des u among the elements of S since des (u+qv) < des u.Therefore deg u < 1 and so ugF From the argument giv=n tefore *hen v and must also be in F and thus VePSL2(F).Therefore, every matrix in PSL2(F [x]) of trace one in PSL2(F).As a corollary, we obtain COROLLARY 2 i) If -3 is not a square mod p, then Z [x] satisfies square property one.P 2) If F is an ordered field where every positive element has a square root, then F x] satisfies square property one.In particular if R is the real field and A the subfield of algebraic numbers when both R Ix] and A [x] satisfy square property one.
The proofs of the two statements in the corollary consist in showing that over there fields the conjugacy conditions on matrices of trace one hold.In fact, over 0 l) 3] 8] these fields every matrix of trace one is conjugate to _+ (-i 1 3.Each of the rings discussed in the previous section, except for the conditions on primes, are sum of squares rings.There are trivial examples of sum of squares rings which do not satisfy square property one for example Z 3 [x] which is a sum of squares ring but does not satisfy square property one since -3 is a square in Z 3. In this section, we give a non-trivial example of a ring which satisfies square property one but which is not a sum of squares ring.The example will lead us to pose a question about smm of squares rings.
Let I denote the ring of integers in the quadratic imaginary number field 2 Then ;e get with a b c d integers and ad-bc i cz + d

2 -
(a2+ c + ac)z-(be + ab + cd) where multiplicatcn Zs done va .matrixmultiplication by U (2.2) Therefore, any element of . of order 3 must have form (2.1) or (2.2). 2 Now let a > 0, n Z. Suppose the equation x + x + i 0 has a solution modulo n.Then there exists integers m and k with 2 m +m+ i nk Therefore, the linear fractional transformation -mz + n z kz + (m+) with (b, d) io Since b and d are relatively prime, there exist integers a, c with ad bc i.Then there exists a map U: z' (az+b)/(cz+d) in M. Conjugating the map B -I by this U give us form (2) that is z' (ad+cd+ab)z + (b2+d2+bd) -mz + n -(a +c +ac)z-(bc+ab+cd) conjugation preserves determinants, we have - c) Finally in a Euclidean domain the stated conditions on the subadditive norm functions are exactly what is necessary to allow the proof of theorm 1 to go through.