Piotr Blass. Jerey Lang

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1 Ulam Quarterly Volume 2, Number 1, 1993 Piotr Blass Deartment of Mathematics Palm Beach Atlantic College West Palm Beach, FL Joseh Blass Deartment of Mathematics Bowling Green State University Bowling Green, OH Jerey Lang Deartment of Mathematics University of Kansas Lawrence, KS The simlest tye of integral extension of a unique factorization domain B is obtained by adjoining to B an n-th root of one of its elements; that is, by forming the ring B n = B[z]=(z n 0 g), where n 2 Z + and g 2 B. When n = 1, B n = B. When n 2 it is natural to ask for what g 2 B will unique factorization be reserved? The case where B is a olynomial ring over an algebraically closed eld k immediately comes to mind. In the one variable case we have a denitive answer: B n is factorial if and only if g 2 B = k[x] is linear; for if deg(g) 2 then g factors comletely in B as a roduct of linear factors and in B n g will have two distinct factorizations. With more variables the issue becomes more cloudy. This aer concentrates on the two variable case so that our rincial question is 0.1. Question. For what g 2 k[x; y], will B n = k[x; y; z]=(z n 0 g) be a unique factorization domain? The one variable case suggests that the answer may be that if g is irreducible in k[x; y] then B n will be factorial, but the following examle shows that this is not enough Examle. Assume char(k) = 6= 0 and let g = xy + x +1 + y +1. Then B is not factorial since (z + xy) = (x + y +1 )(y + x +1 ) is a nonunique factorization in B. 11

2 12 Piotr Blass, Joseh Blass and Jerey Lang There are similar examles in characteristic 0. Perhas g in (0.2) is not irreducible enough; what if g is in some sense very irreducible? A generic g is about as irreducible as we can think of. We say that g = ij x i y j 2 k[x; y] of degree d is generic if the set of coecients f ij : 0 i + j dg is algebraically indeendent over the rime subeld of k. Thus we obtain the following conjecture Conjecture. If g 2 k[x; y] is generic and deg(g) 4, then for all n 2 Z +, B n is a unique factorization domain. This conjecture is artly motivated by the classical result of Max Noether [4], which was extended to all characteristics by Deligne [2]: a generic surface of degree 4 in rojective 3-sace has innite cyclic divisor class grou (see [5,. 130{131] for a denition of divisor class grou). When R is a Krull domain, R is factorial if and only if the divisor class grou of Sec R is 0. For each n 2 Z + let n = Sec B n A 3 k and denote the divisor class grou of n by Cl( n ). Then (0.3) becomes 0.4. Conjecture. If g is generic and deg(g) 4, then Cl( n ) = 0 for all n 2 Z + Note that Cl(n) = 0 if and only if every curve on n is a comlete intersection. The following results summarize the rogress we've made so far on (0.4) when char(k) = > 2. Here and throughout the rest of the aer g 2 k[x; y] is generic of degree Theorem. If m 2 Z + such that gcd(; m) = 1 and Cl( m ) = 0, then Cl( r m) = 0 for all r 2 Z Corollary. If n = s for some s 2 Z +, then Cl( n ) = Corollary. Cl( n ) = 0 for all n if and only if Cl( n ) = 0 for all n relatively rime to. (0.7) is an obvious corollary of (0.5) and (0.6) is a consequence of (0.5) and the fact that B 1 = k[x; y]. In the next section we outline the roof of (0.5). We simly mention here that when = 2, Cl( 2 s) = Z=2Z for all s 2 Z + (see [7,. 359{360]). This leads us to reformulate (0.4) Conjecture. If g is generic of degree 4, then Cl( n ) = 0 for all n if char(k) 6= Radical Descent If A is a Krull domain of characteristic 6= 0 and 1 : A! A is a derivation with kernel B, then B is a Krull domain and A is integral over

3 Some Questions on Unique Factorization 13 B. If the degree of the eld of fractions of A over the eld of fractions of B is 6= 0 and 1(A) is not contained in any height one rime of A then (1.1) (a) there is a grou homomorhism : Cl(Sec B)! Cl(Sec A) such that ker is isomorhic to the quotient L=L 0 of the additive grous L = ft 01 1t : t and t 01 1t 2 Ag and L 0 = fu 01 1u : u 2 A 3 g; (b) there is a b 2 B such that 1 = b1; (c) an element x of A belongs to L if and only if (See [8,. 62{64].) 1 01 x 0 bx + x = 0: 1.2. Proosition. Let D be the derivation on k(x; y) dened by D g 0 Then (a) ker D \ k[x; y] = k[x ; y ; g] = B ; (b) Cl(Sec B ) is isomorhic to L 0 = ft 01 Dt : t and t 01 Dt 2 k[x; y]g; (c) there exists a 0 2 k[x ; y ; g] such that D = a 0 D and deg(a 0 ) ( 0 1)(deg(g) 0 2). ([6,. 616{622]) 1.3. Discussion. Let S = fq 2 k 2 : g x (Q) = g y (Q) = 0g. The cardinality of S is (deg(g) 0 1) 2 when deg(g) 6= 0 (mod ) and (deg(g)) 2 0 3deg(g) + 3 otherwise ([6,. 287{288]). A simle alication of Bezout's theorem yields that if t 2 k[x; y], then the number of Q 2 S such that t(q) = 0 can not exceed deg(t)(deg(g) 0 1). In articular, if t(q) = 0 for all Q 2 S then deg(t) > deg(g) 0 2 or t = Notation. The derivation D in (1.2) extends to a derivation on k(x; y; z) with D(z) = 0. Since D(z n 0 g) = 0, D induces a derivation on B n which we denote by D n. Let L n = fu 01 D n u : u and u 01 D n u 2 B n g and L 0 n = fu01 D n u : u 2 Bn 3g. Then B n is isomorhic to ker D n \ B n and by (1.1) there is a grou homomorhism n : Cl( n )! Cl( n ) with ker ( n ) = L n =L 0 n Proosition. Let t = n01 i=0 t i z i 2 B n, where t i 2 k[x; y], 0 i n01. For each i let J(i) = fj : 0 j < n and j i (mod n)g. Then t 2 L n if and only if for each i, 0 i < n, D 01 t i 0 a 0 t i = 0 t j g(j0i)=n ; where a 0 is as in (1.2(c)). j2j(i) Proof. By (1.1), t 2 L n if and only if Dn 01 t 0 a 0t = 0t, which holds if 0 1 and only if D 01 t i 0 a 0 t i z i = 0 t i zi. Since 1; z; : :: ; z n01, is a i i

4 14 Piotr Blass, Joseh Blass and Jerey Lang basis for the eld of fractions of B n over k(x; y) and since z n = g in B n, we obtain the conclusion by comaring owers of z on both sides of the above equation Lemma. Let t = n01 i=0 t 2 L n, then deg(t i ) deg(g) 0 2 for each i. t i z i 2 B n, where t i 2 k[x; y], 0 i < n. Proof. We consider the case deg(g) 6= 0 (mod ); the other case is similar. Let r be such that deg(t r ) deg(t i ) for each i. We have r = nq + s for q; s 2 Z with q 0, 0 s < n. By (1.5), D 01 t s 0 a 0 t s = 0t r gq. By (1.2), deg(a 0 ) ( 0 1)(deg(g) 0 2). A simle induction shows that Thus deg 0 D 01 t s 1 deg(ts ) + ( 0 1)(deg(g) 0 2): deg(t r ) deg 0 D 01 t s 0 a 0 t s 1 deg(ts ) + ( 0 1)(deg(g) 0 2) deg(t r ) + ( 0 1)(deg(g) 0 2): Hence deg(t r ) deg(g) Theorem. Let H = g xx g yy 0 gxy 2, the hessian of g. For each Q 2 S let H(Q) denote a xed root of the equation T 2 = H(Q). Then a 0 (Q) = 01 H(Q) for each Q 2 S. ([3, Theorem 1.5]) If 1.8. Lemma. Suose n = m, where m; n; 2 Z +. Then the comosition B m = 0! k[x; y; z ]=(z n 0 g),! B n mas L m isomorhically onto L n. Proof. Let t = n01 i=0 t i z i 2 L n where t i 2 k[x; y]. By (1.5), gcd(i; ) = 1 imlies D 01 t i = a 0 t i. Then for each Q 2 S, 0 = 0 D 01 t i 1 (Q) = a 0 (Q)t i (Q). For a generic g, H(Q) 6= 0 for all Q 2 S. By (1.3), (1.6) and (1.7), t i = 0 whenever gcd(i; ) = 1. Thus t 2 k[x; y; z ]=(z n 0 g) = B m. Therefore the isomorhism B m! k[x; y; z ]=(z n 0 g) mas L m onto L n Discussion. Assume m 2 Z + and gcd(; m) = 1. We need to study the action of G = Gal(k; F ( ij )) on S (recall g = ij x i y j ). Let N denote the cardinality of S. Let S m = f(; ; ) 2 k 3 : (; ) 2 S and m = g(; )g. Since g is generic g(q) 6= 0 for all Q 2 S. Then S m consists of mn oints which we can list as Q ij, 1 i N, 1 j m, in such a way that if Q ij = (; ; ), then (; ) = Q i. Let! be a rimitive m-th root of unity in k and the k(x; y)-automorhism on k(x; y; z) dened by (z) =!z. Then induces an automorhism on B m and let T : B m! k[x; y] denote the trace ma. Since each

5 Some Questions on Unique Factorization 15 Q ij 2 m we may dene t(q ij ) for t 2 B m by evaluating any reimage of t in k[x; y; z] at Q ij. Observe that if i is xed and t(q ij ) = 0 for each j, then [T (t)](q ij ) = 0 for each j, which yields [T (t)](q i ) = 0. Now t = t r z r for unique t r 2 k[x; y], 0 r < m. If s is a non negative integer less than m, then t(q ij ) = 0 for each j imlies (z m0s t)(q ij ) = 0 r for each j, which we just saw imlies 0 = [T (z m0s t)](q i ) = (mz m t s ) (Q i ) = mg(q i )t s (Q i ); and hence t s (Q i ) = 0 for each s. We summarize this in a lemma Lemma. Assume gcd(; m) = 1 and t = m01 r=0 xed i, t(q ij ) = 0 for each j, then t r (Q i ) = 0 for each r. t r z r 2 B m. If for a Lemma. Assume gcd(; m) = 1. For each t 2 L m and Q ij 2 S m there exists r ij 2 F such that t(q ij ) = r ij H(Q L i ). Furthermore, the ma : L m! F 1 H(Q ij ) dened by (t) = (t(q ij )) is an injection of i;j additive grous. Proof. Given t 2 L m, Dm 01 t 0 a 0t = 0t by (1.1). Evaluate both sides of this equality at Q ij we obtain a 0 (Q i )t(q ij ) = t (Q ij ). By (1.7) we obtain the rst assertion. The second one follows by (1.3), (1.6) and (1.10) Theorem. Let G = Gal(k; F ( ij )). Then G acts on S as the full symmetric grou. Furthermore, for each air Q 0 ; Q 00 2 S, there exists a 2 G such that acts as the identity on S and ( H(Q)) = ([1,. 297] and [3,. 354]) 0 H(Q); Q = Q 0 ; Q 00 H(Q); otherwise: Lemma. Assume gcd(; m) = 1. Then L m = 0. Proof. Let t 2 L m and suose t 6= 0. Assume (t) = r ij H(Q i ). Given 2 G, (t) 2 L m and the action of on t is comatible with the action of on (t). By (1.12), there are 0 ; 00 2 G such that 0 ( H(Q)) = 00 ( H(Q)) = 0 H(Qi )); i = 1; 2 H(Q i ); otherwise; 0 H(Q i )); i = 1; 3 H(Qi ); otherwise:

6 16 Piotr Blass, Joseh Blass and Jerey Lang Then ^t = t 0 0 (t) 0 00 (t) (t) 2 L m and has the roerty that ^t(q ij ) = 0 for all i 2 and 1 j m. Note also that ^t 6= 0 since the m01 rst coordinate of (^t) = 4r11 H(Q 1 ). ^t = t s z s, where t s 2 k[x; y], 0 s < m. By (1.10) t s (Q i ) = 0 for each s and i 2. If deg(g) 6= 0 (mod ), S has (deg(g) 0 1) 2 oints. By (1.3) and (1.6) we get each t s = 0, a contradiction. The case where divides deg(g) is slightly more comlicated but very similar and we omit the details Theorem. If gcd(; m) = 1, then Cl( r m) injects into Cl( r01 m) for all r 2 Z +. In articular, Cl( m ) = 0 imlies Cl( r m) = 0 for all r 2 Z +. Proof. Use (1.4), (1.8) and (1.13). s=0 References 1. Blass, P. and Lang, J., Zariski Surfaces and Dierential Equations in Characteristic > 0, Marcel Decker, Monograhs in Pure and Alied Math., 106, Deligne, P. and Katz, N., Groues de Monodromie en Geometrie Algebrique, Lecture Notes in Mathematics, 340, Sringer-Verlag, Grant, A. and Lang, J., Alications of the fundamental grou and urely insearable descent to the study of curves on Zariski surfaces, J. Alg., 132:2 (1990). 4. Grothendieck, A., SGAI, Lecture Notes in Mathematics, 224, Sringer- Verlag, New York, Hartshorne, R., Algebraic Geometry, Sringer-Verlag, New York, Lang, J., The divisor class grou of the surface z m = G(x; y) over elds of characteristic > 0, J. Alg., 84 (1983). 7. Lang, J., Generic Zariski surfaces, Comositio Mathematica, 73 (1990). 8. Samuel, P., Lectures on unique factorization domains, Tata Lectures Notes, Please note: This coyright notice has been revised and varies slightly from the original statement. This ublication and its contents are ccoyright Ulam Quarterly. Permission is hereby granted to individuals to freely make coies of the Journal and its contents for noncommercial use only, within the fair use rovisions of the USA coyright law. For any use beyond this, lease contact Dr. Piotr Blass, Editor-in-Chief of the Ulam Quarterly. This notication must accomany all distribution Ulam Quarterly as well as any ortion of its contents.