Proc. Indian Acad. Sci. (Math. Sci.), Vol. 98, Nos 2 and 3, December 1988, pp. 109-116. 9Printed in India. Generalized epimorphism theorem S M BHATWADEKAR School of Mathematics, Tata Institute of Fundamental Research, Homi Bhabha Road, Bombay 400 005, India MS received 12 January 1988; revised 24 March 1988 Abstract. Let R[X, Y] be a polynomial ring in two variables over a commutative ring R and let F~R[X, Y] such that R[X, Y]/(F) = R[Z] (a polynomial ring in one variable). In this set-up we prove that RfX, Y] = R[F,G] for some G~R[X, Y] if either R contains a field of characteristic zero or R is a seminormal domain of characteristic zero. Keywortts. Epimorphism theorem; polynomial ring; seminormal domain; characteristic zero. 1. Introduction Let k be a field of characteristic zero. Let k[X, Y] be a polynomial ring in two variables over k and Fek[X, Y] such that k[X, Y]/(F)= k[Z] (a polynomial ring in one variable). In this set-up the famous epimorphism theorem of Abhyankar and Moh ([2], Theorem 1.2) says that k[X, Y] = k[F, G] for some G~k[X, Y]. Russell and Sathaye had obtained the following analog of the epimorphism theorem ([6], Theorem 2.6.2): IfR is a locally factorial Krull domain of characteristic zero and Fe R[X, Y] such that R[X, Y]/(F) = R[Z], then R[X, Y] = R[F, G]. Therefore one asks the following natural question: Is the foregoing result valid for an arbitrary commutative domain R of characteristic zero? In this paper we answer this question affirmatively under the assumption that R is seminormal. We prove: Theorem A. Let R be a seminormal commutative domain of characteristic zero. Let I be an ideal of R[ X, Y] such that R[ X, Y-J/I = R[ Z]. Then I is a principal ideal say generated by F and R[X, Y] = R[F, G] for some Gr Y]. Moreover we give an example (Example 3.8) to show that I need not be principal if R is not seminormal. When R contains a field of characteristic zero we prove the following (weaker) epimorphism theorem: Theorem B. Let R be a commutative ring containing afield of characteristic zero. Let FeR[X, Y] such that R[X, Y]/(F) = R[Z]. Then R[X, Y] = R[F, G] for some G~R[X, Y]. 109 110 S M Bhatwadekar 2. Preliminaries Throughout this paper all rings will be commutative. In this section we set up notations and state some results for later use. R will denote a commutative ring. Rt"l: polynomial ring in n variables over R. R": free R-module of rank n. For a finitely generated R-algebra A, f2A/g: universal module of R-differentials of A. For a prime ideal q0 of R, k(~): R,O/~OR~. DEFINITION A reduced ring R is said to be seminormal if it satisfies the condition: for b, cER with b3= c 2, there is an a~R with a 2 = b, a 3 = c. Lemma (2.1). Let R be a noetherian ring and let s~R be a non-zero divisor. Let M be a finitely generated R-module. lf M~ is a projective R,-module of rank d and M/sM is R/sR- projective of rank d then M is R-projective of rank d. Proof. Without loss of generality we can assume that R is local. Since M/sM is R/s-projective and R is local there exists a surjective R-linear map fl: R d --* M (d = rank M/sM). Let N = ker ft. Since M~ is R~-projective of rank d and fl is surjective we get N~ = 0. But s is a non-zero-divisor of R and N c R ~, therefore N~ = 0 ~ N = 0 and fl is an isomorphism. Lemma (2.2). Let R be a noetherian ring and I be an ideal of R t"J such that Rt'l/I ~ R t"- tl as R-algebras. Then for an ideal (fi of R, I c~ fb R t~l = (ill. Moreover if I is a principal ideal of R t~j say generated by F, then (i) F is a non-zero-divisor of R [~1. (ii) F is a algebraically independent over R, i.e. R[F] ~ R TM. (iii) R[F]nff)R E~l =(fiR[F] for any ideal ff~ of R. Proof. Since for any non-negative integer l, R tn is a free R-module, the exact sequence 0 --. I ~ R t"j & R t~- ~j ---,0; ~: R-algebra homomorphism of R-modules gives rise to the exact sequence 0 ~ l(~R/ffi ---* Rt"J(~R/ffJ .~ ,R L -'l(~R/fb ---* 0 R R R proving that the canonical map l/grI--,I + (_gRtn~/@RI~1 of R/@t~Lmodules is an isomorphism. Hence I n (fir t"l = 051. Now we assume that I = (F). (i) It is easy to see that Fr c"1 for any maximal ideal ~ of R. This shows that F is a non-zero-divisor of R t'l. (ii) Suppose c o + elf + ... + c,F" = 0 where ci~RVi, 0 <~ i <~ r. Then 0 = ~(c o + elF + ... + c,F') = c o i.e. F(cl + c:F + ... + c~F'- 1) = 0. Therefore, as by (i) F is a non-zero- Generalized epimorphism theorem 111 divisor, cl + c2F + ... c,F'-1 = 0 showing that cl = 0. Repeating this argument we see that ci = 0Vi, 0 ~< i ~< r. (iii) Let ff be the image of F in R/15 t"l (=Rt"l/15Rt"J). Then obviously R/15t,~/(ff) ~ R/15t,- n. Therefore by (ii) ff is algebraically independent over R/15 and hence R [F] c~ 15R t"J = 15R [F]. Lemma (2.3). Let R be a noetherian ring and let 15 be the nilradical of R. Let I be an ideal of R Enl such that Rtnl/l ~ R En- xl as R-algebras. If 1/15I is a projective R/15t~Lmodule of (constant) rank I then I is a projective R[nLmodule of (constant) rank I. Proof Since 15 is nilpotent, the canonical map Pic(R t"l) ~ Pic(R/15 t"l) is an isomorph- ism. Therefore there exists a projective R~"Lmodule L of constant rank 1 such that L/15L ~ 1/151. Hence there exists a Rt"Llinear map 0: L ~ I such that the induced map ~: L/15L ~ 1/151 is an isomorphism. We claim that r is an isomorphism. Surjectivity of~: Since ~ is an isomorphism, we have I = ~//(L) + 151. But 15 is nilpotent and hence I = ~(L). Injectivity of 0: Let M =" Jr 0. Then we get the following exact sequence of R t"L modules: O~M~L&I~O. As in Lemma 2.2 we see that I is a projective R-module. Therefore the above exact sequence gives rise to the following exact seouence: 0 ~ M/15M -~ L/15L ~ 1/15I ~ O. But ~ff is an isomorphism. Therefore M/15M = 0 i.e. M = 15M. The nilpotency of 15 shows that M = 0. Thus ~, is an isomorphism. 3. Main theorems In this section we prove Theorem A and Theorem B which are quoted in the introduction. For the proof of these theorems we need some lemmas and a proposition. Lemma 3.1 is well known but for the lack of a proper reference we give a proof. Lemma 3.1. Let R be a noetherian ring and S be a noetherian R-algebra. Let nER be such that S~ is a fiat R~-algebra and S/zt S is a fiat R/Tz R-algebra. Moreover assume that Torlg(S, R/zcR)= O. Then S is a fiat R-algebra. Proof. Let M and N be finitely generated R-modules and let f: M ~ N be a R-linear injective map. Then we want to show that the map f | Is: M | ~ N | is injective. Let K = ker(f@ Is). 112 S M Bhatwadekar Since S,, is R~-tlat we have K,~ = 0. Let T = 1 + 7rR and T' = 1 + nS. Then since Tor~r(Sr,, RT/TZRT)= Tor~(S, R/TrR)| r, = 0 and Rr/lrR T = R/TzR, Sr,/rrS r, = S/rrS, by ([1], Theorem 3.2, p. 91) S r, is flat over R r and hence K T. =0. Thus K r, = 0, K, = 0. Therefore K = 0 showing that S is a fiat R-algebra. Lemma 3.2. Let R be a noetherian ring offinite Krull dimension (denoted by dim R). Let F be an element of R[ X, Y] such that R[ X, Y]/(F) = R[ Z] as R-algebras. Then R[ X, Y] is a fiat g[F]-algebra. Proof Let 65 be the nilradical of R. Since R[F] (F being algebraically independent over R) and R[X, Y] are flat over R, for every module M over R we have Torf m (R[X, Y], M| In particular for every ideal J of R we have To~m(R[X, Y], R[F]/JR[F]) = 0. Therefore, since (5 is nilpotent, by ([1], Theorem 3.2, p. 91) R[X, Y] is flat over R[F] ifR[X, Y]/(SR[X, Y] (= R/(5[X, Y]) is flat over R[F]/15R[F]. So it is enough to prove the result when R is a reduced ring. We prove the result by induction on dim R.