Undecidability of C(T 0,T 1 )

Undecidability of C(T 0,T 1 ) David A. Pierce 1997; recompiled, April 4, 2017 Mathematics Dept Mimar Sinan Fine Arts University, Istanbul david.pierce@msgsu.edu.tr http://mat.msgsu.edu.tr/~dpierce/ We show how to interpret (with parameters) the ring of rational integers in the field of rational functions in two indeterminates over C. The argument is based on the article [3] of Kim and Roush, but with more details filled in. The first lemma concerns a (multiplicative) valuation v on a field K with associated place ϕ. Recall what this means: There is a ring O K such that x K O = x 1 O; we let U comprise the units of O, so that m = O U is the unique maximal ideal of O; then the place ϕ is the obvious map K O/m { }, with the arithmetic operations involving defined when possible to make ϕ behave like a 1

homomorphism; and the valuation v is the homomorphism K K /U, where the value group K /U is ordered so that xu yu xy 1 O. Then v(x + y) max{v(x),v(y)}, and if v(x) v(y), then v(x+y) = max{v(x),v(y)}. We shall say that x K is odd with respect to v if there is no y K such that v(x) = v(y) 2. Lemma 1. Let a,b K be such that v(a) = 1 and b is odd with respect to v. If the equation ax 2 +by 2 = 1 has a solution from K, then ϕ(a) is a square in the residue field O/m. Proof. For any x,y K, we have v(ax 2 ) = v(a)v(x) 2 = v(x) 2 v(b)v(y) 2 = v(by 2 ) since b is odd. Now suppose ax 2 +by 2 = 1; then 1 = v(1) = v(ax 2 +by 2 ) = max{v(ax 2 ),v(by 2 )}, v(by 2 ) 1, v(by 2 ) < 1 since b is odd. Hence by 2 m, so ϕ(by 2 ) = 0. Therefore 1 = ϕ(1) = ϕ(ax 2 +by 2 ) = ϕ(ax 2 ), 1 = ϕ(a)ϕ(x) 2 ( ) since ϕ(a) 0; hence ϕ(x), so ϕ(a) = 1 2 1 ϕ(x) 2 = ϕ. x We shall use the case s = 2 of the following: 2

Lemma 2 (Tsen Lang Theorem). Let k be an algebraically closed field, and let K be a function field of transcendence degree m over k. Let S = {f j (X 0,...,X n 1 ) : j < r} be a system of r forms of degrees s in n variables over K, where n > s m r. Then S represents 0, that is, its elements have a common nontrivial zero. Proof. We first establish the theorem in the case where K is the pure transcendental extension k(t 0,...,T m 1 ) of k. We proceed by induction on m. If m = 0, then K = k, and the claim is a consequence of the Projective Dimension Theorem ([2, I.7.2, p. 48]). That is, each form in S defines a (possibly reducible) variety of codimension 1 in P n over k; by the Theorem, the intersection of the varieties contains a variety of dimension n 1 r 0. An alternative argument is in [4, ch. II, p. 43]. By this argument, there is a place ϕ on k( X) that is finite on k[f 0 ( X),...,f r 1 ( X)], but is infinite at some X i (which is transcendental over k[f 0 ( X),...,f r 1 ( X)]; there is such an X i, since n > r). The place ϕ being given, let i be such that X i has the greatest value under the valuation determined by ϕ. Then ϕ(1/x i ) = 0. Write y j = X j /X i ; then ϕ(y j ) is finite for all j < n, and ϕ(y i ) = ϕ(1) = 1. For all j < r, if d is the degree of f j, then by the homogeneity of f j we have 0 = ϕ(f j ( X)/X d i) = ϕ(f j (ȳ)) = f j (ϕ(ȳ)); so ϕ(ȳ) is a common nontrivial zero of the f j. Now suppose that the theorem holds for m = l. Let S comprise forms f j (X 0,...,X n 1 ) of degrees s over k(t 0,...,T l ) for j < r; we want to show that they represent 0. The coefficients in thef j are rational functions oft l overk(t 0,...,T l 1 ). We may clear fractions and assume that these coefficients are 3

polynomials in T l over k(t 0,...,T l 1 ) of degrees p for some p. Now replace each X i in each f j with Y i,0 +Y i,1 T l + +Y i,t T t l, the size of t to be determined presently. Each f j thus becomes a form of degree s in the n(t + 1) variables Y 0,0,...,Y n 1,t with coefficients from k(t 0,...,T l 1 )[T l ]. Regrouping terms as necessary, we can understand each f j as a polynomial in T l of degree st + p whose coefficients are forms of degrees s in the n(t + 1) variables Y i,j over k(t 0,...,T l 1 ). If the collection of these forms represents 0, then so does S. The number of such forms is r(st+p+1); by inductive hypothesis, they represent 0 nontrivially if n(t+1) >s l r(st+p+1) =s l+1 r(t+1)+s l r(p+1 s), equivalently, (n s l+1 r)(t+1) > s l r(p+1 s). This inequality holds, for t large enough, provided n > s l+1 r. Thus the theorem is established for K = k(t 0,...,T m 1 ) when m = l+1. Finally, suppose K is a finite extension of k(t 0,...,T m 1 ). Say that K has the basis {w 1,...,w t } as a vector space over k(t 0,...,T m 1 ). If we replace each variable X i in S with Y i,1 w 1 + +Y i,t w t, then we can write each form in S as a linear form in w 1,...,w t whose coefficients are forms of degree s in the nt variables Y 0,1,...,Y n 1,t over k(t 0,...,T m 1 ). These rt forms represent 0 nontrivially if nt > s m rt, 4

equivalently, n > s m r. Thus the general theorem is established. Lemma 3. Let the field K be a finite Galois extension of F. In F there is an interpretation of K together with the elements of its Galois group over F. Proof. Suppose [K : F] = n, and let f : F n K be an F-vector space isomorphism. It is enough to show that there is a polynomial map g : F 2n F n over F, and for each σ Gal(K/F) there is a polynomial map h : F n F n over F, such that for all ā, b F n we have f(ā) f( b) = f(g(ā, b)), σ(f(ā)) = f(h(ā)). Let {e i } i be the natural basis of F n, so that ā F n is i a ie i. Suppose f(e i ) f(e j ) = f(c ij ) and σ(f(e i )) = f(d i ). Then we have f(ā) f( b) = i,j a i b j f(e i )f(e j ) = a i b j f(c ij ) i,j ( ) =f a i b j c ij, and also σ(f(ā)) = i a iσ(f(e i )) = i a if(d i ) = f( i a id i ). i,j Theorem 4. The ring (Z,+, ) is interpretable in the field (C(T 0,T 1 ),+, ) (with parameters). 5

Proof. We establish a sequence of structures: 1. (C(T 0,T 1 ),+, ), 2. (C(T 0,T 1 ),+,,σ), 3. (C(T 0,U 0,T 1,U 1 ),+,, σ,c(t 0,U 0,T 1 )), 4. (Z Z,+,,,Z 0 ), 5. (Z,+, ), each interpreting the next. The most work goes into interpreting the structure (4) in (3). Let us first note what these structures are. We get the structure (2) from (1) by adding the C-automorphism σ that takes T 0 to T 1 and T 1 to T 0. In (3), the elements U i satisfy U 2 i = T 3 i +at i +b for certain complex numbers a and b as follows: Let E be the affine curve over C given by Y 2 = X 3 + ax + b. We choose a and b so that this curve is nonsingular; hence its projective closure Ē is an elliptic curve. We also require Ē to have no complex multiplication. We extend σ to the C-automorphism σ of C(T 0,U 0,T 1,U 1 ) that switches U 0 and U 1. When naming the structures (1) (3) henceforth, we may not explicitly name the field operations. In the structure (4), the binary operation is given by (x,y) (z,w) (z,w) = (y,x), and is a certain binary operation such that (n,1) (m,s) m = ns. 6

We now have four interpretations to give. Interpretation of (2) in (1): We do not define the automorphism σ in the field C(T 0,T 1 ). Rather, we define the fieldwith-automorphism (C(T 0,T 1 ),σ) in a field C(S 0,S 1 ) that is isomorphic to C(T 0,T 1 ). Let S 0 and S 1 be the elementary symmetric functions of T 0 and T 1, that is, S 0 = T 0 +T 1, S 1 = T 0 T 1, so that T 0 and T 1 are the roots of the polynomial X 2 S 0 X +S 1 C(S 0,S 1 )[X]. Hence the field C(T 0,T 1 ) is a finite Galois extension of C(S 0,S 1 ). The nontrivial element of the Galois group is the automorphism σ taking T 0 to T 1 and T 1 to T 0. By Lemma 3, the structure (C(T 0,T 1 ),σ) is interpretable in C(S 0,S 1 ), but the latter field is also isomorphic to C(T 0,T 1 ). Interpretation of (3) in (2): We interpret the field C(T 0,U 0,T 1,U 1 ) in the field C(T 0,T 1 ) by means of the coordinate map Moreover, we have g : C(T 0,T 1 ) 4 C(T 0,U 0,T 1,U 1 ) (a,b,c,d) a+u 0 b+u 1 c+u 0 U 1 d. σ(g(a,b,c,d)) = g(σ(a),σ(c),σ(b),σ(d)). Thus (C(T 0,U 0,T 1,U 1 ), σ) is interpretable in (C(T 0,T 1 ),σ). Finally, we have g(a,b,c,d) C(T 0,U 0,T 1 ) c = d = 0, 7

so in fact the structure (C(T 0,U 0,T 1,U 1 ), σ,c(t 0,U 0,T 1 )) is interpretable in (C(T 0,T 1 ),σ). Interpretation of (5) in (4): We interpret the group (Z,+) in (Z Z,+,Z 0 ) trivially by means of the coordinate map h : Z 0 Z (m,0) m. Then to interpret multiplication, we note: h(n,0) h(s,0) = h(m,0) m = ns (Z Z,+, ) = (n,1) (m,s) (Z Z,+,,,Z 0 ) = x[x (s,0) (n,0)+(0,1) (m,0)+x]. So the whole structure (Z Z,+,,,Z 0 ) interprets Z as a ring. Interpretation of (4) in (3): First, let us identify the transcendentals T i with the rational maps E E C (x 0,y 0,x 1,y 1 ) x i. Then we can identify the U i with the maps (x 0,y 0,x 1,y 1 ) y i. Hence we can consider C(T 0,U 0,T 1,U 1 ) to be the field C(E E) of all rational functions on E E over C. The elliptic curve Ē is given by the homogeneous equation Y 2 Z = X 3 +axz 2 +bz 3. Hence it comprises points P of projective 2-space such that: 1. P = (x : y : 1), where (x,y) E; or: 8

2. P = (0 : 1 : 0). Generally we shall treat Ē as literally being E with an extra point, the point at infinity. We know that Ē is an algebraic group, that is, there is a group operation + on Ē given by polynomials. The neutral element 0 of the group is arbitrary, but we shall take it to be (0 : 1 : 0). If K is a field extending C, then by Ē(K) we mean the set of solutions from K of the homogeneous equation defining Ē. Then of course Ē(K) is an algebraic group as well. In particular, if C(E) is the field of rational functions on E over C, then we can identify Ē(C(E)) with the group of rational maps E Ē. (The group operation is pointwise addition.) Such a map uniquely determines a rational map Ē Ē. Let f : Ē Ē be a rational map. Then the map f f(0) takes 0 to 0. We know that such a map is necessarily an endomorphism of Ē. Because we assume that Ē has no complex multiplication, its endomorphisms are just the multiples of the identity. Thus every rational map Ē Ē is the sum of a constant map and an endomorphism, and in particular Ē(C(E)) = id Ē. Ē Likewise, the group of rational maps E E Ē is Ē(C(E E)) = π 0 π 1 Ē, where π 0 and π 1 are the projections Ē Ē Ē. Then the π i are determined by the projections (T i,u i ) : E E E, and we shall henceforth write π i for (T i,u i ). The affine curve E(C(E E)) lacks only the neutral element of Ē(C(E E)); so every element of E(C(E E)) can be 9

written as mπ 0 +nπ 1 +P for (m,n) Z Z and P Ē, where not all three of m, n and P are zero. Let us say that two distinct elements of E(C(E E)) are equivalent if their difference lies in E, and let us denote the equivalence class of mπ 0 +nπ 1 +P by [m,n]. Then the collection of these equivalence classes is a group G = Z Z; moreover, we can interpret G in C(E E), first by defining the affine curve E(C(E E)) in the obvious way, then noting that E = E(C) is also definable in C(E E) because C is definable (say by the formula y y 4 = 1+x 4 ). Now let H be the subgroup {[m,0] : m Z} of G, and let and be defined for G as for Z Z. We have to interpret H, and in the field-expansion (C(E E), σ,c(t 0,U 0,T 1 )). To interpret H in the field C(E E) with distinguished subfield C(T 0,U 0,T 1 ), we recall that any f [m,n] G is a certain ordered pair from E(C(E E)) C(E E) 2. We show that its coordinates are in C(T 0,U 0,T 1 ) if and only if n = 0. The sufficiency of n = 0 is clear. For the necessity, note that T 1 (P, Q) = T 1 (P,Q) for all (P,Q) E E. Since obviously the same is true with T 0 and also U 0 in place of T 1, if the coordinates of f are from C(T 0,U 0,T 1 ), then f(p, Q) = f(p,q), mp nq = mp +nq, nq = nq, and so n = 0. We interpret in (C(E E), σ) by noting that (x,y) [m,n] ( σ(x), σ(y)) [n,m]. 10

It now remains to interpret. Fix n Z and Q E, and let τ be the rational map (π 0,π 1 nπ 0 Q) : E E E E. As τ has the inverse (π 0,π 1 + nπ 0 + Q), so τ induces an isomorphism τ : C(E E) C(E E) f f τ. Let T be the projection E C, (x, y) x. We have τ (T (mπ 0 +sπ 1 +P)) = T (mπ 0 +sπ 1 +P) (π 0,π 1 nπ 0 Q) = T (mπ 0 +s[π 1 nπ 0 Q]+P) = T ([m ns]π 0 +sπ 1 +P sq). Hence the sentences and x y[t (mπ 0 +sπ 1 +P)x 2 +T (nπ 0 +π 1 +Q)y 2 = 1 xy 0] x y[t ([m ns]π 0 +sπ 1 +P sq)x 2 +T (π 1 )y 2 = 1 xy 0] are equivalent in C(E E). The same is true if we replace each π i with cπ i for some c Z. Presently we shall establish: 11

Claim. There is c > 0 such that, for all k,l Z, we have k = 0 if and only if C(E E) = x y[t (kπ 0 +lπ 1 +R)x 2 +T (cπ 1 )y 2 xy 0] for some R Ē. We shall use the Claim in case k = c[m ns] and l = cs. Once the claim is established, we can define in (C(E E), σ,c(t 0,U 0,T 1 )) by setting [n,p] [m,s] if and only if C(E E) = x y[(t f)x 2 +(T g)y 2 = 1 xy 0] for some f c[m,s], g c[n,p]. To establish the Claim, we first note that the forward direction is a consequence of the Tsen Lang Theorem (Lemma 2), since T (lπ 1 +P) and T (cπ 1 ) are from C(T 1,U 1 ). For the reverse direction, by Lemma 1, it suffices to show that for some c > 0 and all R Ē, if k 0, then there is a valuation v on C(E E) with associated place ϕ such that T (cπ 1 ) is odd, but v(t (kπ 0 +lπ 1 +R)) = 1 and ϕ(t (kπ 0 +lπ 1 +R)) is a nonsquare in the residue field. Now, any D = (d 0,d 1 ) E determines a place ϕ : C(E E) C(T 0,U 0 ) { }, namely the function of replacing T 1 and U 1 with d 0 and d 1 respectively. In particular, ϕ(t (pπ 0 +qπ 1 +R)) = T (pπ 0 +qd+r). If p 0, then ϕ(t (pπ 0 +qπ 1 +R)) / {0, }, so if v is the valuation associated with ϕ, then v(t (pπ 0 +qπ 1 +R)) = 1. Moreover, suppose (f 2,g) E(C(E)) = E(C(T 0,U 0 )), where f C(E). Let C be the curve given by Y 2 = X 6 +ax 2 +b. Then (f,g) is a rational map E C; but since the genus of C exceeds 1 (see the appendix to [3]), this map 12

must be constant, by the Hurwitz Theorem (see [2]). Now, (ϕ(t (pπ 0 + qπ 1 + R)),g) E(C(T 0,U 0 )) for some g, so if p 0, then ϕ(t (pπ 0 +qπ 1 +R)) is not a square in C(T 0,U 0 ). This result has a generalization. Let F be a finite extension of C(E), so that F is the function field of some curve C 1. If (f 2,g) E(C(E)) for some f F, then (f,g) is a rational map C 1 C. By the Hurwitz Theorem, the degree of this map is bounded. Note that T (cπ 1 ) is in C(T 1 ) for all positive integers c: for, let C 2 be the curve over C(T 1 ) given by the Denef equation Y 2 (T 3 1 +at 1 +b) = X 3 +ax +b described in [1, 3]. Then the projective closure of C 2 is an elliptic curve, and we have a rational map f : C 2 E(C(T 1,U 1 )) (x,y) (x,yu 1 ). Then (T 1,1) C 2, and f(c(t 1,1)) = c(t 1,U 1 ) = cπ 1, so T (cπ 1 ) = (first coordinate of f(c(t 1,1)), an element of C(T 1 ). Let r i for i < 3 be the roots of X 3 + ax + b. If c is large enough, then T (cπ 1 ) is not a square in the extension C( T 1 r 0, T 1 r 1, T 1 r 2 ) of C(T 1 ). Let d 0 be distinct from each r i. Then T (cπ 1 ) is odd with respect to the valuation v of C(T 1 ) associated with the place T 1 d 0. Extend this place to C(T 1,U 1 ) by sending U 1 to d 1 = d 3 0 +ad 0 +b = (d0 r 0 )(d 0 r 1 )(d 0 r 2 ). Then no square root of v(t (cπ 1 )) is introduced; T (cπ 1 ) is still odd. 13

References [1] J. Denef. The diophantine problem for polynomial rings and fields of rational functions. Transactions of the American Mathematical Society, 242:391 399, August 1978. [2] Robin Hartshorne. Algebraic Geometry. Springer, New York, 1974. [3] K. H. Kim and F. W. Roush. Diophantine undecidability of C(t 1,t 2 ). Journal of Algebra, 150:35 44, 1992. [4] Serge Lang. Introduction to Algebraic Geometry. Addison Wesley, Reading, Mass., 1972. 14