An Approach to the Jacobian Conjecture in Two Variables An Approach to the Jacobian Conjecture in Two Variables Ryan Bruner, Mathematics Steven J. Rosenberg, Ph.D. Department of Mathematics and Computer Science ABSTRACT We will make strides toward proving the Jacobian Conjecture in two variables when the degrees of the polynomials defining the endomorphism are relatively prime. We will use techniques that are less elementary than are strictly necessary to prove this case in hope of generalizing the proof to the case when these degrees are not relatively prime. 51 An Approach to the Jacobian Conjecture in Two Variables 1 Introduction The Jacobian Conjecture is a major outstanding conjecture in algebra from 1939 when it was first stated by mathematician Ott-Heinrich Keller. The Jacobian Conjecture says if you form a ring endomorphism s on a polynomial ring in n variables over a field with characteristic zero, then the ring endomorphism is bijective if and only if the determinant of the jacobian of s is a nonzero element of the given field. To date despite multiple published attempted proofs the conjecture remains unproven. 2 Notation and Definitions Commutative Ring with 1 [Ei] A commutative ring with 1 is an additive abelian group R with an operation (a;b) 7! ab referred to as multiplication such that 8a;b;c 2 R: (ab)c = a(bc) ab = ba a(b + c) = ab + ac 1a = a1 = a In our work we will use ring to mean a commutative ring with 1. Polynomial Ring [La] A polynomial ring k[x] is a set of all possible polynomials with coefficients from the integral domain k and the place holder or variable x. n i m j Given the polynomials f (x) = ∑i=0 aix and g(x) = ∑ j=0 b jx we define the sum to be max(m;n) k f (x) + g(x) = ∑ (ak + bk)x k=0 we define the product to be à ! m+n k f (x)g(x) = ∑ ∑ aib j x k=0 i+ j=k A polynomial ring in n variables we can write as k[x1;:::;xn] := (k[x1;:::;xn¡1])[xn]. Ring Homomorphism [La] Let A;B be rings, and let f : A ! B be a function. We say that f is a ring homomorphism if f (1A) = 1B and 8a;b 2 A, f (a + b) = f (a) + f (b) and f (ab) = f (a) f (b): 52 An Approach to the Jacobian Conjecture in Two Variables Jacobian [Ab] Let F;G be polynomials in variables x and y, then J(F;G), the jacobian of F;G relative to x, y, is defined by J(F;G) = FxGy ¡ GxFy The Jacobian Conjecture in two variables Let k be a field with characteristic zero. Let f : k[x;y] ! k[x;y] be a ring endomorphism, fixing k pointwise, and f = f(x), g = f(y). Then f is bijective if and only if jJ( f ;g)j 2 k£. Algebraic vs. Transcendental Suppose R and S are rings and that S is a subring of R. Then we say that R extends S. Let a 2 R. Then: Definition: a is called algebraic over S if 9 f 2 S[x] ¡ f0g s.t. f (a) = 0. Definition: a is called integral over S if 9 f 2 S[x] ¡ f0g s.t. f(a) = 0 and f is monic. Definition: a is said to be transcendental over S if a is not algebraic. Further: Definition: If 8a 2 R, a is algebraic over S, then R is called an algebraic extension of S. Definition: If 8a 2 R, a is integral over S, then R is called an integral extension of S. Transcendence Degree Let k be a field. Let K be an extension field of k. Definition: Let S be a subset of K. Let T = k[X1;:::;Xn] be a polynomial ring in variables X1;:::;Xn. We say S is algebraically independent over k when for every finite sequence of distinct elements x1;:::;xn 2 S, If f 2 T, and f (x1;:::;xn) = 0, then f is the zero polynomial. Definition: Let S be a subset of K. If S is algebraically independent over k, and if the cardinality of S is the greatest cardinality of all algebraically independent subsets of K over k, then the cardinality of S is the transcendence degree of K over k.[La] Laurent Series Let k be a field with characteristic 0. A Laurent series is written as a power series which may include terms of negative degree. A Laurent series with a placeholder x about a given point a 2 k (its center) is defined as: ¥ j ∑a j(x ¡ a) j 53 An Approach to the Jacobian Conjecture in Two Variables where j 2 Z, a j 2 k. Notation: k((x)) denotes the ring of Laurent series about 0 in the variable x. Notation: k[[x]] denotes the ring of power series about 0 in the variable x. 3 Results Proposition 6 [Ro] Let k be a field of characteristic 0. Let F;G 2 k[[y]][x] be monic in x with degx(F) = ¡1=n 1=n n > 0 and degx(G) = m > 0. Let r 2 k[[y]]((x )) satisfy F(r;y) = x, where r has leading term x . ¥ ¡ j=n Write G(r;y) = ∑ j=¡m c j(y)x with c j(y) 2 k[[y]]. Then J(F;G) = 1 if and only if c j(y) 2 k for all j : ¡m · j · n ¡ 2 and cn¡1(y) = y=n + c for some c 2 k: Example 1 Let k be a field of characteristic 0. Let A0;B2;B1;B0 be algebraically independent over k. Let R = k[A0;B2;B1;B0]. We start our example using the setup from prop 6 of [Ro] by constructing F;G 2 R[x] 2 3 2 to be monic in x. We let F = x +A0 and let G = x +B2x +B1x+B0 so that degx(F) = 2 and degx(G) = 3. 1=2 1=2 ¡1 1=2 We now want to construct r so that it satisfies F(r) = x, thus r = (x ¡ A0) = x (1 ¡ A0x ) . Next we wish to expand this expression for r as a Laurent series about 0 in x¡1=2 using the Binomial Theorem. à ! ¥ µ ¶ 1=2 ¡1 1=2 1=2 1=2 ¡1 j x (1 ¡ x A0) = x ¢ ∑ (¡A0x ) j=0 j ¡1 1=2 In order to find G(r) we will use big-O notation with the series (1¡x A0) by truncating it after the third term in the series and adding O(x¡3). à ! ¥ µ ¶ µ 2 ¶ 1=2 1=2 ¡1 j 1=2 (¡A0) ¡1 (¡A0) ¡2 ¡3 x ∑ (¡A0x ) = x 1 + x ¡ x + O(x ) j=0 j 2 8 If we now construct G(r) we get: 1=2 3 1=2 2 1=2 G(r) = ((x ¡ A0) ) + B2((x ¡ A0) ) + B1(x ¡ A0) + B0 1=2 1=2 = (x ¡ A0)(x ¡ A0) + B2(x ¡ A0) + B1(x ¡ A0) + B0 1=2 = (x ¡ A0 + B1)(x ¡ A0) + B2x ¡ A0B2 + B0 1=2 ¡1 1=2 = (x ¡ A0 + B1)x (1 ¡ A0x ) + B2x ¡ A0B2 + B0 µ ¶ ³ ´ (¡A ) (¡A2) = x3=2 ¡ A x1=2 + B x1=2 1 + 0 x¡1 ¡ 0 (x¡1)2 + O(x¡3) + B x + (B ¡ A B ) 0 1 2 8 2 0 0 2 µ ¶ µ ¶ µ ¶ ¡3A 3A2 ¡ 4B A A3 ¡ B A2 = x3=2 +B x2=2 + 0 + B x1=2 +(B ¡B A )x0=2 + 0 1 0 x¡1=2 + 0 1 0 x¡3=2 +O(x¡3=2) 2 2 1 0 2 0 8 8 and we can now see the desired coefficients of G(r) are: µ ¶ µ ¶ ¡3A 3A2 ¡ 4B A C = B ; C = 0 + B ; C = (B ¡ B A ); C = 0 1 0 1 2 2 2 1 3 0 2 0 4 8 If we construct the ring S = k[C1;C2;C3;C4] we now wish to show that A0 is integral over S. 54 An Approach to the Jacobian Conjecture in Two Variables Lemma 1. Let k be a field with characteristic 0, and let R = k[a0;b0;b1;b2] and let S = k[c1;c2;c3;c4]: ³ ´ ³ 2 ´ ¡3a0 3a0¡4b1a0 If c1 = b2; c2 = 2 + b1 ; c3 = (b0 ¡ b2a0);and c4 = 8 , then a0 is integral over S. ³ ´ ³ 2 ´ ¡3a0 3a0¡4b1a0 PROOF. Assume: c1 = b2, c2 = 2 + b1 , c3 = (b0 ¡ b2a0), and c4 = 8 . ³ ´ ¡3a0 3 We know c2 = 2 + b1 so solving for b1 we get b1 = 2 a0 + c2: ³ 2 ´ 2 3 3a0¡4b1a0 3a0¡4( 2 a0+c2)a0 Using this in c4 = 8 we get by substitution c4 = 8 : Simplifying and rewriting this equation in terms of a0 we have 4 8 a2 + c a + c = 0: 0 3 2 0 3 4 4 8 2 4 8 Since 1; 3 c2; 3 c4 2 S;9h(x) = x + 3 c2x + 3 c4 2 S[x] ¡ f0g which is monic and satisfies h(a0) = 0: 2 3 Lemma 2. Let k be a field of characteristic 0. Let a0;b0;b1;b2 2 k[y]. Let f = x + a0 and let g = x + 2 b2x + b1x + b0. If J( f ;g) = 1, then a0 is constant in k[y]. PROOF. Assume J( f ;g) = 1 and assume for a contradiction a0 is not constant in k[y].