On Integrality and Irreducible Polynomials In a Certain Ring Extension On Integrality and Irreducible Polynomials In a Certain Ring Extension Michael Ross, Mathematics Dr. Steve Rosenberg, Department of Math and Computer Science ABSTRACT In this paper, we make incremental progress on Dr. Rosenberg’s approach to proving the Jacobian conjecture in two variables. Accordingly, the results of this paper were reached by working on the ring Pn i Pm i extension that is constructed in the following manner. Let f = i=0 aix and let g = i=0 bix with 1 m−i −1 P n an; bm = 1 and an−1 = 0. Then write g ◦ f = i=0 cix . Let k be a field of characteristic zero and let R = k[a0; :::; an−2; b0; :::; bn−1] and S = k[c0; :::; cm+n−1]. Herein, we prove a necessary and sufficient condition under which α 2 R is integral over S, give a closed form expression for Irr(a0; S; x) for n = 2, and provide conjectures for deg(Irr(a0; S; x)) for general n. 1 Introduction 1.1 A Jacobian History n n Conjecture 1. (Jacobian Conjecture) Let F := (F1; :::; Fn): C ! C be a polynomial mapping. If ∗ J(F ) 2 C , then F is invertible. The Jacobian Conjecture was originally stated in 1939 by O.H. Keller, for n = 2, with integer coeffi- cients. In 1962, a result proven by Bialynicki-Birula and Rosenlicht gave that if F was injective, then F was surjective, and F ’s inverse was also a polynomial map. (i.e. F is an automorphism.)[Essen] This paved the way to the slightly stronger form of the Jacobian Conjecture: Conjecture 2. (Jacobian Conjecture, Automorphism Version) Let k be a field of characteristic zero, and n n ∗ let F := (F1; :::; Fn): k ! k be a polynomial mapping. If J(F ) 2 k , then F is a polynomial automorphism. Because of their result, this conjecture follows from proving that J(F ) 2 k∗ ) F is injective. Then in the late 1960s, the conjecture was named by Shreeram Shankar Abhyankar, due to the central point of the problem being the Jacobian matrix. [Essen] Then, further partial results were proven. For example, T.T. Moh proved that the conjecture was true for n = 2 when deg(F ) ≤ 100, and S. Wang proved it was true for all n, when deg(F ) ≤ 2. Here, deg(F ) means max(deg(Fi)).[Bass] Further, mathematicians such as Arno van den Essen and Wenhua Zhao have proven that the Jacobian Conjecture is equivalent to or follows from cases of other conjectures. This can occasionally be an effective technique, as it is a similar approach to that which was used to prove Fermat’s infamous Last Theorem, a problem that remained open for nearly 360 years, despite innumerable attempts to resolve it. 1.2 Approach and Motivation In this work, we are concerned only with the two variable case of the Jacobian Conjecture. Let k be a field of characteristic zero, let A0; :::; An−2;B0; :::; Bm−1 be algebraically independent over k, and let 1 On Integrality and Irreducible Polynomials In a Certain Ring Extension Pn i Pm i R = k[A0; :::; An−2;B0; :::Bm−1]. Let f = i=0 Aix and let g = i=0 Bix with An;Bm = 1 and −1 P1 m−i An−1 = 0. Then write g ◦ f = i=0 Cix n , and let S = k[C0; :::Cm+n−1]. Then by Proposition 20 and Corollary 21 of Rosenberg’s Observations, R is integral over S iff gcd(m; n) = 1, and if R is integral over S, then the Jacobian Conjecture is true for polynomials of degree m, n. To put this result in terms of the geometric Jacobian Conjecture as listed above, Rosenberg proved that the Conjecture was true for n = 2, when gcd(degx(F1); degx(F2)) = 1. Since integrality proved to be a key concept in Rosenberg’s results, it thus became clear that an ex- amination of the conditions for integrality, and in particular a look at the integral closure of S in R when gcd(m; n) 6= 1, could prove to produce some insightful and useful results. Thus, two approaches were taken in this paper, in an attempt to do so. The first, examining conditions on integrality from a theoretical perspective; and the second, examining Irr(A0; S; x) via a computer program, with the hope of spotting a significant pattern. 2 Definitions and Notation Definition 1. Let S ≤ R be an extension of commutative rings with 1, and let α 2 R. Then α is called integral over S in case 9 f 2 S[x] − f0g such that f is monic, and f(α) = 0. Further, if every α 2 R is integral over S, we say that R is integral over S. Definition 2. Let R be a commutative ring with 1. An R-module M is an ordered triple (M; +; ·) with the following properities: 1. (M; +) is an abelian group. 2. · : R × M ! M. 3. 8a; b 2 R; and 8m 2 M; (a · b) · m = a · (b · m) 4. 8a; b 2 R and 8m; n 2 M, a · (m + n) = a · m + a · n and (a + b) · m = a · m + b · m: 5. 8m 2 M; 1 · m = m: Definition 3. Let R be a commutative ring with 1. An R-algebra is a ring S, and a ring homomorphism σ : R ! S. Notation 1. Let F be a field, and let A; B ⊆ F . Then 1. A[x] := ff 2 F [x] j the coefficients of f belong to Ag . 2. A1[x] := ff 2 A[x] j f is monicg. 3. F [x]A := ff 2 F [x] j all roots of f in F are in Ag. 1 1 4. F [x]A := F [x] \ F [x]A. The following are from Eisenbud’s Commutative Algebra. Definition 4. Given a field k and a subset I ⊆ k[x1; :::; xn] we define a corresponding algebraic subset of kn to be n Z(I) = f(a1; :::; an) 2 k jf(a1; :::; an) = 0 for all f 2 Ig : Z(X) is called the zero-set of X. 2 On Integrality and Irreducible Polynomials In a Certain Ring Extension Definition 5. Given a field k and any set X ⊆ kn, we define I(X) = ff 2 k[x1; :::; xn]jf(a1; :::; an) = 0 for all (a1; :::; an) 2 Xg : I(X) is called the ideal of X, and fittingly is an ideal. Definition 6. If R is a ring and I ⊆ R is an ideal, then the set p I := ff 2 Rjf m 2 I for some positive integer mg p is called the radical of I. If I = I then I is called a radical ideal. The following is well known, and will not be proven here. p Proposition 1. I is an ideal of R. 3 Results 3.1 Theoretical These first two results are some very basic topological lemmas, which are used further on in the paper. The first result shows that for monic f of degree n, if the set of possible coefficients f is allowed to have is bounded, then the set of possible roots f can have is bounded. The second result shows a sort of converse: if the set of possible roots f is allowed to have is bounded, then the set possible coefficients it can have is bounded. + Lemma 1. Let n 2 Z , let A ⊆ C, suppose A is bounded by some m, and let 1 R = fr 2 C j f(r) = 0 for some f 2 A [x] with deg(f) = ng: Then R is bounded by max(1; nm). Proof. Suppose A is bounded by some m, and let r 2 R. Then f(r) = 0 for some f 2 A1[x], where n−1 n X i f = x + aix i=0 with ai 2 A, and n−1 n X i f(r) = r + air = 0: i=0 Case 1: jrj < 1. Case 2: jrj ≥ 1. Then, 3 On Integrality and Irreducible Polynomials In a Certain Ring Extension n−1 n X i r = − air i=0 n−1 X i+1−n r = − air i=0 n−1 X i+1−n jrj = air i=0 n−1 X i+1−n ≤ jaij r i=0 n−1 X i+1−n ≤ jaj r (where a = max(ja0j ; ja1j ; :::; jan−1j)) i=0 n−1 X 0 ≤ jaj r (since jrj ≥ 1) i=0 = n jaj j1j = n jaj ≤ nm: Thus, R is bounded by max(1; nm). + Lemma 2. Let n 2 Z , let R ⊆ C be bounded, and let 1 A = fa 2 C j a is a coefficient of f; for some f 2 C [x]R and deg(f) = ng: Then A is bounded. 1 Proof. Let R ⊆ C be bounded by some m, and let f 2 C [x]R with deg(f) = n. Then 9ρi 2 R, so that we Qn Pn j can write f = i=1(x − ρi) = j=0 ajx with aj 2 C. So, 0 1 ! j B X Y C aj = (−1) @ ρi A [n] i2I I2( j ) ! X Y jajj = ρi [n] i2I I2( j ) X Y ≤ ρi (Triangle Inequality) [n] i2I I2( j ) 4 On Integrality and Irreducible Polynomials In a Certain Ring Extension X Y = jρij [n] i2I I2( j ) X Y ≤ m (Since ρi 2 R) [n] i2I I2( j ) X = mj [n] I2( j ) n j = (j )m : Thus, A is bounded by some M dependent only on n and m.