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 ∞ 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 ∈ 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 ) ∈ 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 ) ∈ k , then F is a polynomial automorphism.

Because of their result, this conjecture follows from proving that J(F ) ∈ 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 P∞ 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 ∈ R. Then α is called integral over S in case ∃ f ∈ S[x] − {0} such that f is monic, and f(α) = 0. Further, if every α ∈ 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. ∀a, b ∈ R, and ∀m ∈ M, (a · b) · m = a · (b · m) 4. ∀a, b ∈ R and ∀m, n ∈ M, a · (m + n) = a · m + a · n and (a + b) · m = a · m + b · m.

5. ∀m ∈ 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] := {f ∈ F [x] | the coefficients of f belong to A} . 2. A1[x] := {f ∈ A[x] | f is monic}.

3. F [x]A := {f ∈ F [x] | all roots of f in F are in A}.

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) = {(a1, ..., an) ∈ k |f(a1, ..., an) = 0 for all f ∈ I} . 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) = {f ∈ k[x1, ..., xn]|f(a1, ..., an) = 0 for all (a1, ..., an) ∈ X} .

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 √ I := {f ∈ R|f m ∈ I for some positive integer m} √ 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. √ 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 ∈ Z , let A ⊆ C, suppose A is bounded by some m, and let

1 R = {r ∈ C | f(r) = 0 for some f ∈ A [x] with deg(f) = n}.

Then R is bounded by max(1, nm).

Proof. Suppose A is bounded by some m, and let r ∈ R. Then f(r) = 0 for some f ∈ A1[x], where

n−1 n X i f = x + aix i=0 with ai ∈ A, and n−1 n X i f(r) = r + air = 0. i=0

Case 1: |r| < 1. Case 2: |r| ≥ 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 |r| = air i=0 n−1 X i+1−n ≤ |ai| r i=0 n−1 X i+1−n ≤ |a| r (where a = max(|a0| , |a1| , ..., |an−1|)) i=0 n−1 X 0 ≤ |a| r (since |r| ≥ 1) i=0 = n |a| |1| = n |a| ≤ nm.

Thus, R is bounded by max(1, nm).

+ Lemma 2. Let n ∈ Z , let R ⊆ C be bounded, and let

1 A = {a ∈ C | a is a coefficient of f, for some f ∈ C [x]R and deg(f) = n}.

Then A is bounded.

1 Proof. Let R ⊆ C be bounded by some m, and let f ∈ C [x]R with deg(f) = n. Then ∃ρi ∈ R, so that we Qn Pn j can write f = i=1(x − ρi) = j=0 ajx with aj ∈ C. So,

  ! j  X Y  aj = (−1)  ρi  [n] i∈I I∈( j )

!

X Y |aj| = ρi [n] i∈I I∈( j )

X Y ≤ ρi (Triangle Inequality) [n] i∈I I∈( j )

4 On Integrality and Irreducible Polynomials In a Certain Ring Extension

X Y = |ρi| [n] i∈I I∈( j ) X Y ≤ m (Since ρi ∈ R) [n] i∈I I∈( j ) X = mj [n] I∈( j ) n j = (j )m .

Thus, A is bounded by some M dependent only on n and m.

This next lemma is used in the proof of Lemma 4.

k+1 Lemma 3. Let S = C[x1, ..., xk] be a polynomial ring, and let f ∈ S[x]. Let A ⊆ Bδ(r) ⊂ C , where A k+1 is dense in Bδ(r), a ball of radius δ > 0 about r ∈ C . If ∀a ∈ A, f(a) = 0, then f is identically zero.

Proof. Since f is continuous, ∀b ∈ Bδ(r), f(b) = 0. Since f = 0 everywhere in Bδ, ∃p ∈ Bδ s.t. fs(p) = 0  α1 αk α for all s ∈ x1 ...xk x |α1, ..., α ∈ N , where fs means the partial derivative of f with respect to x1, α1 times, etc. Since the coefficient for every monomial in the Taylor series expansion of f about p is 0, and since the coefficients of the Taylor series of a polynomial at a point are equal to the coefficients of the polynomial at that point, the coefficients of f are zero, and so f is identically zero.

Hilbert’s Nullstellensatz is a well known theorem, which serves as the primary link between algebra and geometry, linking algebraic objects to corresponding geometric ones. It is stated as follows in Eisenbud’s Commutative Algebra with a View Toward Algebraic Geometry, wherein it is proved several times, in several variants.

Theorem 1. (Nullstellensatz). [Eis:1.6:Theorem 1.6] Let k be an algebraically closed field. If I ⊂ k[x , ..., x ] is an ideal, then 1 n √ I(Z(I)) = I.

Thus, the correspondences I 7→ Z(I) and X 7→ I(X) induce a bijection between the collection of algebraic n n subsets of A (k) = k and radical ideals of k[x1, ..., xn].

One of the key results in the present paper, the following lemma serves as a simplified version of the Integrality Theorem that follows it, and proved to be a crucial tool in proving that result.

Lemma 4. Let S = C[x1, ..., xk], let α ∈ S (An algebraic closure of the field of fractions of S), let f = Irr(α, S, x), let n = deg(f), let

n k+1 o Z = Z({f}) = (c1, ..., ck, a) ∈ C |f(c)(a) = 0 ,

π k and let Z → C be projection onto the first k components. Then the following are equivalent:

1. α is integral over S.

k −1 2. ∀X ⊆ C s.t. X is bounded, π (X) is bounded.

5 On Integrality and Irreducible Polynomials In a Certain Ring Extension

n Pn−1 i Proof. (1 ⇒ 2) Suppose α is integral over S. Then write f = x + i=0 aix with ai ∈ S. Without k loss of generality, let X ⊂ C be a ball of radius m about the origin. Now, each ai is a polynomial in k the xi’s, and is thus continuous as a function from C to C. So, ∀ai, ai(X) is bounded by some mi. Let S M = max(mi). Then ai(X) ⊆ BM =: A, a ball of radius M about the origin. Then by Lemma 1, 1 R := {r ∈ C | f(r) = 0 for some f ∈ A [x] with deg(f) = n} is bounded by Mˆ := max(1, nM). Then   1 −1 2 ˆ 2 2 π (X) ⊆ (X × R), which is bounded by m + M .(X) k −1 Pn i (2 ⇒ 1) Suppose that ∀X ⊆ C s.t. X is bounded, π (X) is bounded. Now, f = i=0 aix , where k ai ∈ S. Assume for a contradiction that ∃r ∈ C s.t. an(r) = 0 and ai(r) 6= 0 for some i 6= n. By Lemma 3, k ∃rs, a non-terminating sequence s.t. rs → r and an(rs) 6= 0 for all s. Since ∀j, aj : C → C is continuous, as rs → r, an(rs) → an(r) and ai(s) → ai(r). Let X be an open neighborhood of positive radius d around the point r, and note that X is bounded. Choose N s.t. ∀s > N, rs ∈ X, and from now on, assume s > N. Then, ∀s, ∃bn(rs), such that an(rs) ∗ (bn)(rs) = 1. Then fs := bn(rs) ∗ f(rs) is monic. However, since an(rs) → an(r) = 0, |bn(rs)| → ∞, so |bn(rs)ai(rs)| → ∞. Then A := {coefficients of fs|s > N} is unbounded, and so by the converse of Lemma 2, R = {β ∈ C|fs(β) = 0} is unbounded. Therefore, −1 π (X) = {rs × {β}|s > N and β ∈ R} is unbounded, which is a contradiction. Thus, ∀r ∈ C such that an(r) = 0, ai(r) = 0 for all i. So, r ∈ Z((an)) ⇒ ∀i, r ∈ Z((ai)). So for all i, Z((an)) ⊆ Z((ai)), so I(Z((ai))) ⊆ I(Z((an))) for all i, which by Hilbert’s Nullstellensatz implies p p m m ∗ ∀i, ai ∈ (ai) ⊆ (an). Thus for m >> 0, ai ∈ (an), and so an|ai in S, for all i. If an ∈ S , then f ∗ can be made monic, and so α is integral over S. If an ∈/ S , then ∃ a prime p of S such that p|am in S, but then ∀i, p|ai, and so p|f, implying f was not irreducible, which is a contradiction. Thus, α is integral over S.

Theorem 2. Let R = C[An−2, ..., A0,Bm−1, ..., B0] and S = C[C1, ..., Ck] where k = m+n−1, Ci ∈ R, k and suppose R is algebraic over S. Let α ∈ R and let r = (r1, ..., rk) ∈ C . ∀H ∈ R, H(r) := H k k+1 evaluated at (An−1 = r1, ..., B0 = rk). Let τ : C → C be given by τ(r) = (C1(r), ..., Ck(r), α(r)). k k k Let π : τ(C ) → C be the projection onto the first k components. Then α is integral over S iff ∀X ⊆ C such that X is bounded, π−1(X) is bounded.

k Proof. Since α is algebraic over S, we can let f := Irr(α, S, x), and let π˜ : Z(f) → C be projection onto the first k components. k −1 (⇒) Suppose α is integral over S. Let X ⊂ C be bounded. Then by Lemma 4, π˜ (X) is bounded by k −1 some ball B of positive radius. By Lemma 5, τ(C ) ⊆ Z(f), so π =π ˜|τ(Ck), and thus π (X) ⊆ B. k −1 (⇐) Suppose that ∀X ⊆ C s.t. X is bounded, π (X) is bounded. Without loss of generality, let X be k −1 a ball of finite radius around the origin in C , and let p ∈ π˜ (X). Then π˜(p) =: q ∈ X. Since X is open, −1 we can choose positive m so that N := Bm(q) ⊆ X. Then, since π˜ is continuous, π˜ (N) is open, and so −1 k ∃B ⊆ π˜ (N) s.t. B is a ball of positive radius around p. Then, by Lemma 5, ∃(zn) ⊆ τ(C ) s.t. zn → p. −1 −1 −1 Then for n >> 0, π(zn) ∈ X, and so zn ∈ π (X). Thus p ∈ π (X), and so π˜ (X) is bounded. Then by Lemma 4, α is integral over S.

k Lemma 5. If everything is as in the statement of Theorem 2, and f := Irr(α, S, x), then τ(C ) ⊆ Z(f) k and τ(C ) is dense in Z(f).

k k Proof. Let t ∈ τ(C ). Then ∃r ∈ C s.t. τ(r) = t. Then t = (C1, ...Ck, α)(r). Thus, f(t) = f(r)(α). But f(α) = 0 identically, so f(r)(α) = 0, and thus t ∈ Z(f).(X) k From there, Chevalley’s Theorem gives us that τ(C ) is dense in Z(f).

6 On Integrality and Irreducible Polynomials In a Certain Ring Extension

3.2 Experimental

Using the PARI script listed in Appendix A, we were able to compute Irr(A0, S, x) for a given m and n, representing the degrees of g and f, respectively. The following conjectures were reached through observation of patterns in the output of the script. Unfortunately, hardware limitations have provided a roadblock in pursuing these lines of inquiry. The n = 2 family was calculable only up to the m = 13 case, the n = 3 family up to the m = 9 case, and the n = 4 family could only be computed up to m = 4. After these points, the program would run for a great deal of time before succumbing to an overflow error. Accordingly, the later results are neither as strong, nor as conclusive as the first result.

Theorem 3. If deg(F ) = 2, Then for deg(G) = m > 0,

dm/2e 2i−1
X Irr(A , S, x) = i C xi. 0 22i−1 m+1−2i i=0

−1 P∞ m−i Proof. We know that g ◦ f = i=0 Cix 2 . Thus:

∞ −1 X m−i g ◦ f ◦ f = g = Cif 2 i=0 ∞ X 2 m−i = Ci(x + A0) 2 i=0 ∞ ∞  m−i  ! X X 2 m−i −k k = C 2 (x ) 2 A i k 0 i=0 k=0 ∞ ∞ X X  m−i  = C 2 xm−i−2kAk i k 0 i=0 k=0  m−j  m b 2 c 2k+j X X   = xj  C 2 Ak  m−j−2k k 0 j=−∞ k=0

Since g is a polynomial in x, ∀j < 0, the coefficent of xj = 0. Thus, for j = −1,

j m−(−1) k 2 X  2k−1  0 = C 2 Ak m−(−1)−2k k 0 k=0 m d 2 e 2i−1
X = C k Ai m+1−2i 22i−1 0 i=0

m 2i−1 Pd 2 e ( k ) i by the following lemma. Thus, h := i=0 Cm+1−2i 22i−1 x is a polynomial for which h(A0) = 0. Then Eisenstein’s criterion implies that h is irreducible in S.

Lemma 6.  2k−1  2k−1
2 = k k 22k−1

7 On Integrality and Irreducible Polynomials In a Certain Ring Extension

Proof.  2k−1  (k − 1/2)(k − 3/2) ··· (1/2) 2k 2 = · k k! 2k (2k − 1)(2k − 3) ··· (1) 2k−1(k − 1)! = · 2kk! 2k−1(k − 1)! (2k − 1)! = 22k−1k!(k − 1)! 2k−1
= k 22k−1

Conjecture 3. If deg(F ) = 3, then for deg(G) = m > 0, ( m(m+1) 2·3 if 3|n, deg(Irr(A0, S, x)) = (m+1)(m+2) 2·3 otherwise.

Conjecture 4. If deg(F ) = n, Then for deg(G) = m > 0, deg(Irr(A0, S, x)) can be expressed as a quasi-polynomial dependent on m, with its cases dependent on (m mod (n)), and it can be written in a form similar to that in Conjecture 2; namely, a product of (n − 1) consecutive integers near m, over the product n(n − 1). Note. It can be seen via Theorem 3 that Conjecture 4 holds for the n = 2 case, where ( m 2∗1 if 2|n, deg(Irr(A0, S, x)) = (m+1) 2∗1 otherwise.

4 Future Research

Future works would continue to examine Irr(A0, S, x), as well as make further progress towards un- derstanding the integral closure of S in R, utilizing the main integrality theorem (Theorem 2).

5 References

1. Eisenbud, D.: Commutatitve Algebra with a View Toward Algebraic Geometry. Springer, New York (2004) 2. Moh, T.T.: Jacobian Conjecture. Purdue University. http://www.math.purdue.edu/ ttm/jacobian.html. Accessed 31 June 2011 3. van den Essen, A.R.P.:Polynomial automorphisms and the jacobian conjecture. Societ´ e´ Mathematique´ de France. http://emis.mi.ras.ru/journals/SC/1997/2/pdf/smf sem-cong 2 55-81.pdf (1995) Accessed 31 June 2011 4. Bass, H., Connell, E.H., Wright, D.: The jacobian conjecture, reduction of degree and formal expan- sion of the inverse. American Mathematical Society. Vol 7, Num2. 287-330 (1982) 5. Rosenberg, S.:Obvious Observations on Algebra. Unpublished. 6. Bruner, R.:An approach to the jacobian conjecture in two variables. The McNair Scholars Journal of the University of Wisconsin – Superior. Vol 11 51-58 (2010)

8 On Integrality and Irreducible Polynomials In a Certain Ring Extension

A PARI script

What follows is the PARI script used to generate Irr(A0, S, x), for a given m and n. The indented lines are continuations of the lines above them, modified solely to fit on the page. allocatemem(900000000) As=[a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12] Bs=[b0,b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11,b12,b13] m=7 \* The degree of G *\ n=3 \* The degree of F *\ g=xˆm f=xˆn for(i=0,m-1,g=g+Bs[i+1]*xˆi) print("g=",g) \* Building G *\ for(i=0,n-2,f=f+As[i+1]*xˆi) print("f=",f) \* Building F *\ ff=0 for(i=0, n, ff=ff+polcoeff(f,i,x)*xˆ(n-i)) print("ff=") print(ff) ffr=Ser(ffˆ(1/n),x) ffr=ffr/x ffr=Ser(ffr,x) print("g=") print(g) finv=x diff=xˆ(-1)-ffr for(i=0,m+n+1,a=polcoeff(diff,i,x); finv=finv+a*xˆ(-i);diff=diff-a*ffrˆ(-i)) finv=subst(finv,x,xˆ(-1)) finv=Ser(finv,x) gcompfinv = subst(g, x, finv) C=vector(m+n+1) Cs=[c1,c2,c3,c4,c5,c6,c7,c8,c9,c10, c11,c12,c13,c14,c15,c16,c17,c18,c19,c20] for(i=1,m+n+1,C[i]=polcoeff(gcompfinv,i-m,x)) Corig=C for(i=1,m+n+1,C[i]=Cs[i]-C[i]) j=0 substVec=vector(n+m-1) for(i=0,m-1,substVec[i+1]=Bs[m-i]) for(i=0,n-2,substVec[m+i+1]=As[n-1-i]) print("substVec:", substVec) print("Here") goal=a0 for(s=1,m+n-2,for(i=s+1,m+n-1,C[i]=polresultant(C[i],C[s], substVec[s]);print("i=",i));print(s)) check=C[m+n-1] for(i=1,m+n-1,check=subst(check,Cs[i],Corig[i])) check

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