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THE AUTOMORPHISM GROUP OF THE COMBINATORIAL GEOMETRY OF AN ALGEBRAICALLY CLOSED FIELD DAVID M. EVANS AND EHUD HRUSHOVSKI

1. Introduction Throughout this paper L will be an algebraically closed field and K an algebraically closed subfield such that the transcendence rank of L over K is at least 3. In [4] we considered the pregeometry G(L/K) = (L,clK), where, for Y a subset of L, the closure clK( Y) is the algebraic closure in L of the subfield generated by K and Y. In this paper we shall be concerned mainly with the geometry G(L/K) associated to this: so that the point set P ofG(L/K) is the set of algebraically closed subfields of L which contain K and are of transcendence rank 1 over K, and the closure clK(Z) of Z £ P is the set of points which are contained in clK(JJ Z) (see [4, 1.1] for more details). One of the aims of [4] was to compare these geometries with the more familiar projective geometries (where the pregeometry is given by linear closure in a vector space and the corresponding geometry is the projective geometry). In this paper we pursue this theme further and obtain results on the automorphism group of G(L/K), and the recovery of L from G(L/K) (so these are analogues of two classical results from projective geometry: the Fundamental Theorem and the Coordinatisation Procedure). Let Aut(L){K} be the group of automorphisms of L which fix AT setwise. It is clear that any ge Aut(£){K) is an automorphism of the pregeometry G(L/K), and therefore induces an automorphism %(g) of the geometry G(L/K) (given by x(g)(c\K(x)) = c\K(gx) for all xeL\K). The map x- Aut(L){K}-^ Aut(G(L/K)) is a homomorphism. Daniel Lascar asked whether % is surjective. Our main result is a partial answer to this. THEOREM A. Suppose that L is an algebraically closedfieldand K an algebraically closed subfield such that the transcendence rank of L over K is at least 5. Then any automorphism of the geometry G(L/K) is induced by afield automorphism of L which fixes K setwise.

The proof of Theorem A makes use of the group configuration results of the second author. As well as making use of the 'rank one' configuration (the partial quadrangle) used in [4], we require some of the 'higher rank' configurations (see 3.1). This accounts for the restriction on the transcedence rank of L over K in the hypotheses of Theorem A. We expect that the conclusion of Theorem A holds more generally, assuming only that the transcendence rank of L over K is at least 3 (the question makes little sense without such an assumption, as the geometry is then just a single line with a collection of points on it). Received 24 August 1992; revised 7 March 1994. 1991 Mathematics Subject Classification 03C60. The second author was supported by the NSF. J. London Math. Soc. (2) 52 (1995) 209-225

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With additional restrictions on K, it is possible to avoid using the 'higher rank' group configurations, and indeed to compute the automorphism group it is only necessary to consider closed sets of small rank. We make this precise as follows. For ne N, let i?(«) be the first-order language with equality which has, for each / < «, a single /-ary relation symbol. Consider the points ofG(L/K) as the domain of an S£{ri)structure Gn(L/K), by interpreting the /-ary relation symbol as the set of /-tuples of independent points from G(L/K). Clearly any automorphism of G(L/K) induces an automorphism of Gn(L/K), but the converse to this is a priori far from clear. We can prove the following. THEOREM B. (1) Suppose that either the characteristic of K is 0, or K is algebraic {or both). Then any automorphism ofG3(L/K) is induced by a field automorphism of L whichfixesK setwise. (2) In any case, the image ofx'- Aut(X){K}-* Aut (G3(L/K)) is a subgroup of index at most \K\ in Aut (G3(L/K)).

Again, we expect that (1) of this result is true without the additional restrictions on K. The obstruction to removing these restrictions, as far as this paper is concerned, is explained in Remark 4.6. The method of proof of Theorem A shows that the field L can be recovered in a uniform way from the geometry G(L/K). More precisely, we prove the following theorem. THEOREM C. Suppose that the transcendence rank of'L over K is at least 5. Then the field L is uniformly interpretable in Gb(L/K). using one imaginary parameter.

Here 'uniform' means that the same interpretation works for all G5(L/K), and that any parameter from a certain 0-definable set gives the same result. As an immediate consequence of this we obtain the following, which can be looked upon as a clarification of Remark 4.1 of [4] (we withdraw the statement made there that a suitable interpretation can be made if the transcendence rank of L over K is 4, as our current methods cannot prove this). THEOREM D . Suppose that the transcendence rank of L over K is at least 5. If G(L/K) and G(L'/K') are isomorphic geometries, then the field extensions L.K and L ' : K' are isomorphic.

In [4,4.1], we suggested the following alternative strategy for proving Theorem D, assuming only that the transcendence rank of L over K is at least 3. By considering the coordinatising objects of (maximal) projective planes in G{L/K), we can read off the characteristic of K and the isomorphism type of O(K), the skew-field of quotients of the ring of algebraic endomorphisms of the additive group of K[4, Corollary 3.3.2]. Now clearly once we know K, then L is determined by its transcendence rank over K, and this is the rank of the geometry G(L/K). So to prove Theorem D, it will suffice to show for algebraically closed A'and K' that if 0{K) and O(K') are isomorphic, then K and K' have the same characteristic and transcendence rank over their prime

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subfields (and considerations of cardinality show that it is enough to do this assuming that K is countable). This has now been done by Kitty Holland (see [5]), so Theorem D holds, assuming that the transcendence rank of L over K is at least 3. Daniel Lascar (personal communication) has proved, using Theorem A and results of his on the small index property (see [7, 8]), the following interesting result. THEOREM (Lascar, unpublished). Suppose that L is an algebraically closed field and K a countable algebraically closed subfield. Assume either that the transcendence rank of L over K is countably infinite, or that the Continuum Hypothesis holds and the transcendence rank of L over K is 2N°. Then Aut(L){K] is complete (that is, any automorphism is inner).

We shall assume that the reader is reasonably familiar with our paper [4]. We shall use the notation and terminology of that paper, except that we distinguish between the pregeometry and the geometry (as already remarked). We shall denote clK simply by cl; 'independence' etc., refers to independence in G(L/K). We use r() to denote rank in both the geometry and pregeometry. If A e P we say that a e A is generic if the rank of a is 1. A line in G(L/K) is a closed subset of P of rank 2. Throughout we shall denote the characteristic of L (and K) by q, and we shall always assume that the transcendence rank of L over AT is at least 3. The Galois group of L over K will be denoted by Gal (L/K). We shall use frequently the following consequence of results in [4]. (1.1) THEOREM. Let x1,x2,y1,y2eL\K be such that cl(xx) = cl(x2), c l O ^ ^ c\(y2), andc\(x1.y1) = c\(x2.y2). Suppose further that x1 andyx are independent. Then there exist a,beK and a rational number r such that x2 = a.x[ andy2 = b.y\. Proof. This follows from [4, 2.2.2 and 3.1(ii)] (actually, it is necessary to check that the same r will work for both x2 and y2 simultaneously, but this can be seen from the proof of 2.2.2). We shall make extensive use of the following straightforward lemma. An algebraic proof of this can be found in [4], but at the suggestion of the referee, we give a model theoretic proof which is valid in any strongly minimal set with infinitely many algebraic elements. The definition of a partial quadrangle is as in [4, Section 2.1], and at the start of Section 4 of this paper. (1.2) LEMMA [4, Lemma 2.1.1]. Let (A,B,C,X, Y,Z) be a partial quadrangle in G(L/K) and a,x,y be generic tuples in A,X,Y, respectively. Then there exist generic tuples a',x' in B,Z, respectively, such that a*x~y and a"x"y lie in the same Gsi\(L/K)-orbit. Proof

Let beB, ZEZ and ceC be generic elements. We know that L is a

strongly minimal set when considered as a structure in the first-order language of fields with a constant added for each element of K. By definability of Morley rank, there is a formula 0 (in this language) such that (a) 0(a, x, y, b, z, c) holds in L; and if (a',x',y',b/,z',c/) holds in L, then (b) x{a'b'/cf) ^ 1 and x{x'z'/c') ^ 1;

(c) and if additionally, r(a'x'y') = r(axy), then tp(a'x"y") = tp(axy).

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The one-variable formula (with parameters) 3a/x/4 be given by

2. = {(cl (x), cl (y), cl (x+y), cl (x .y~l): x, y independent generic elements of L) = {(cl (x), cl (x. z), cl (x. z + x), cl (z)): x, z independent generic elements of L). (2.2) LEMMA. Suppose that we have the configuration of points and lines in G(L/K) as in Figure 1, where (X, Y, W,Z)e£.

B

Then there exist xeX and ye Y with x+ye W and x. ye T. Proof. Let x, y be such that (cl (x), cl (y), cl (x+y), cl (xjr1)) = (X, Y, Z, W). By [4,2.1.1] (which is our 1.2), there exist generic x'eB and / e C such that x'y'~l = xy'1. By 1.1 (with A in place of K), there exist a, a' e A and a rational r such that x' = axr and / = a'yr. It follows that r = 1 and a'a^eK. So c\(BY) C\d(XC) - c\(axy), whence T= c\(xy). (2.3) DEFINITION. Denote by Si' the set of 4-tuples (X, Y, W, T) appearing in configurations (A, B, C, D, X, Y, W, Z, T) as in 2.2.

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Note that 2! = {(cl (x), cl (y), cl (x+y), cl (x.y): x, y independent generic elements of L). (2.4) PROPOSITION. Let (X, P, Q, R, A) be a 5-tuple of distinct collinear points with (X, Q,R,A)€& and (X, A,P, Q),(X, A,P,R)e£'. Then there exist generic xeX and aeA such that x + aeP, x.aeQ and x + a.xeR. Proof. We can find generic x,x',x"eX and a, a', a" EA such that x' + a', x" + a"eP, a.x, a'.x'sQ and a.x + x, a".x"eR (see Figure 2). X X

X x"

x -ta

ax ax

P

Q

ax+x R

x"+a"

x"a" FIG.

a a A a"

2

By 1.1 there exist a.,P,y,SeKand rationals r,s such that x' = a.x r , a! = ft.ar and x" = y.xs, a" = 5.(a+ \)s. If the characteristic q of K is non-zero, then we assume, without loss of generality, that q does not divide the numerator or the denominator in s or r. Now, as u — K* by gt = a(g)t. As h(gt) = cc(h)(x(g)t, it follows that a is a homomorphism. Also, h(gs) = s+P(h) + , , , , < / » / = • It is necessary to check that this is well defined and that the map ft': f\/ = -> L\{0} given by M'(((x'}>(*})/ = ) = M(-x:/»//( , < x » / = (using 2.11(1) to show that this is definable). If x'+y'ed(a) but is non-zero, then take z independent from d(a,x,x') and consider x'z,y'z,xz instead of x', y' and x. If x'+y' = 0, we need a new class which will be denoted by 0/ =, and we define the sum to be this. One now shows that n' extends to a bijection //: (/l/ = ) U{0/ = }->L which is an additive homomorphism. This completes the interpretation of L. For uniformity, it only remains to note that Aut (Gb(L/KJ) is transitive on the classes // = , so the interpreted field is independent of choice of parameter. Proof of Theorem D. It will suffice to show that we can recover from G(L/K) the transcendence rank of K over its prime subfield. To do this, take a closed subset of

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rank 5 in G(L/K) and consider the geometry restricted to it. This is isomorphic to G(LJK), where K^ Lo^ L and the transcendence rank of Lo over K is 5. By Theorem C we can recover Lo, and in particular its characteristic and its absolute transcendence rank. This is enough to obtain the absolute transcendence rank of K. 4. Proof of Theorem B Recall [4, 2.1] that a 6-tuple of distinct points (A,B, C,X, Y,Z) in G{L/K) is a partial quadrangle if (A,B, C), (A,X, Y), (B,Z, Y), (C,Z,X) are the only triples of collinear points from {A,B, C, X, Y,Z). (4.1) DEFINITION. Let y be the set of triples of distinct collinear points which lie in some partial quadrangle in G(L/K). Let £ = Aut (G3(L/K)). Note that 9~ is a Z-invariant subset of P3. Moreover, by [4, 2.1.2] a triple (A,B,C) is in 9~ if and only if there exists a connected 1-dimensional algebraic group (G, *) in L (defined over K) and generic elements aeA, beB of G such that a*beC. Obviously $~ is invariant under permuting coordinates. (4.2) DEFINITION. For T^T^F, write Tx ~ T2 if and only if TuT2 appear together in a partial quadrangle. Let — be the transitive extension of ~ . (4.3) LEMMA. (1) The relation — is a yL-invariant equivalence relation on 3~. (2) If(A,B,C), {A',B',C')e2T, then (A,B,C)—(A',B',C') if and only if there exists a connected \-dimensional algebraic group (G, *) in L and generic elements a, a', b,b' of G lying in A, A',B,B', respectively, such that a*beC and a' *b''eC. Proof. (1) Clear. (2) If (A,B, C) ~ (A',B',C) then the existence of the group and its generic elements as in the statement follows from [4, 2.1.2]. So we obtain the property for {A,B,C)—{A',B\C') using [4, 2.1.1]. Conversely, suppose that there exist generic elements of (G, *) as in (2). We may clearly assume that the lines c\(A,B) and cl^',/?') are distinct. Without loss, assume that A' $ cl {A, B) and A $ cl (A\ B'). Then {A, B, C, cl (a * a'), A', cl (a' * b'1)) is a partial quadrangle, and so is (A',B', C,c\(a * a'), A,c\(a * b'~l). Thus (A,B, C) ~ (A,c\(a*a'),A')~(A',B',C'). REMARK. With a little more care (for example, using the construction in the second part of the proof of 4.3(2)) one can show that the equivalence relation — is O-definable in GZ{L/K).

(4.4) DEFINITION. (1) Suppose that x,ysL are independent and generic. Set STm = {d{x),d{y),c\{x.y))/and ZTa = (cl(x),cl(^),cl(x+y))/-. (2) Let Zo denote the stabiliser in I of the equivalence class 5~m. (4.5) LEMMA. (1) The index o/S 0 in S is at most \K\; (2) We have \m{x) ^^. Proof. (1) There are only \K\ possibilities for the group G in (4.3)(2).

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(2) This is clear from (4.3)(2). (4.6) REMARKS. (1) By considering projective planes [4, Theorem 3.3.1] one sees that 2Ta is E-invariant. (2) We shall show that im/ = Lo. So the question which remains is whether or not there exist (cl(jc),cl(_y),cl(*.;>)), {c\(a),d{b),c\{a*b))e3T which lie in the same £orbit, where (G,*) is a connected 1-dimensional algebraic group, a,beG are generic elements and (G, *) is not (isogenous to) the multiplicative group of L. By again considering projective planes in G(L/K) as in [4, Theorem 3.3.1] one sees that (G, *) must be an elliptic curve with Horn (G, G) = Z. If K is algebraic and of non-zero characteristic, then by a result of Deuring, there are no such curves [10]. If the characteristic is zero, then Theorem 4.7 gives a geometric characterisation of &~m. So in these cases I = I o . In fact, we expect this to be true assuming only that the transcendence rank of L over AT is at least 3, but at present we see no way of proving this. (4.7) THEOREM. Suppose that we have the configuration of points and lines in GZ{L/K) as shown in Figure 8, and that (X, Y,W)e3Ta and (X, Y,Z)e^m.

B

FIG. 8

Then there exist generic elements a,x,y of A,X, Y, respectively, such that a.xeB,

a.yeC,

x+yeW,

a.(x+y)eD

and

x.y~lsZ.

Thus(X,Y,W,Z)e± If the characteristic of K is zero, then we can dispense with the assumption that (X, Y, Z) e 2Tm and assume instead that {X, Y, Z) £ 2Ta. Proof. Let x,y e L be such that (cl (x), cl (y), cl (x+y)) = (X, Y, W). Case 1. The characteristic of K is zero. By [4, Lemma 2.1.1] there exist x',y' such that {c\(x'),c\(y'),c\{x'+y')) = (B, C,D). By [4, Theorem 2.2.2] and the structure of Hom(L,L) there exist a,b,csA such that x' = a.x + b and y' = a.y + c. One can now use derivations, and the technique of [1] to show quite easily that a$K (otherwise (X, Y,Z)e^a) and there exist a,/?, y, 0 then the derivative (over K) of z'v with respect to z is non-zero, and z is separable over K(z'). Note also that yp.z'pe W, and that we can assume that x',y' are separable over K(x,y) and at least one of them has non-zero derivative with respect to either JC or y. By the symmetry of the situation (and the fact that something is separable over K{z) if and only if it is separable over K{z~x)) we may assume that (d/dy)(y') ^ 0. Let F= K(x,y,x',y',z'p). Note that F is separably algebraic over K(x,y) and so x,y is a separating transcendence basis for F over K. Let* the differential pairing (cf. [6, p. 269] and the proof of 2.4 here). There exist x1 eX and yteY such that dx' = xxdx and dy' = y1 dy (just compute the derivatives of x' and y' with respect to x and y). Remember that we can assume that yx # 0. Furthermore, note that dz = (dx — zdy)/y. Let d(z') = z1dz (so Case a = 0,z1^0. So dz' = z1dz = (zjy)(dx — zdy), and y,z' is a separating transcendence basis for F over K. Clearly d(x'+y') ^ 0, and as y.z' and x'+y' are mutually algebraic over K, there exists f eFsuch that td(x' +y') = d(y.z') and, in fact, / e W. Thus comparing coefficients of dx and dy in the above equation gives tx1 = z15 tyx = (z'-zx.z).

(1) (2)

Note that / # 0 (from (1)). Subcase t e AT\{0}. Then xx e K, zx e K and y^ e K, whence z' — zx. z e K. So we can write z' = yz+p for some y,f5eK. Replacing x by yx and jy by fly gives what we require. Subcase t$K. Note that xx =£ 0 and so neither xx nor zx is in K (by (1)). From (1), (2) we get x1O>r1 = z 1 .(z / -z 1 .z)-\ (3) and comparing this with x.y'1 = z and using 1.1, we get that there exists a rational number r and 0. Let z" = z'v. Remember that in this case, z2 = (d/dz)(z/p) ^ 0. Furthermore, x' +y' and yv. z" are mutually algebraic (over K). Then as before, there exist s, te W, not both zero, such that s.(x1dx+y1dy) = t.d(yp.z") = t.yp.z2.dz = t.yp~1.z2.(dx-zdy). So s.x^t.yv-'.z,, (14) s.yx = -t.yp-l.zz.z.

(15)

In particular, s ^ 0 and t ^ 0. Furthermore, from (14) and (15) we have >^1 = - x 1 . z . So yl.y = —x1.xeXC\

(16)

Y = K, which again contradicts the exercise below.

EXERCISE. Suppose that v is transcendental over K and w is separably algebraic over K(v) and has non-zero derivative (d/dv)(w) = wv Then w.wx^K.

We are now in a position to complete the proof of Theorem B. Proof of Theorem B. Theorems 4.7 and 2.12 show that im(^) = !„• The proof then follows from Lemma 4.5 and Remarks 4.6. References 1. C. J. ASH and J. W. ROSENTHAL, 'Intersections of algebraically closed fields', Ann. Pure Appl. Logic, 30(1986) 103-119. 2. ELISABETH BOUSCAREN, 'The group configuration—after E. Hrushovski', The model theory of groups (eds A. Nesin and A. Pillay; Notre Dame Press, Notre Dame, 1989). 3. GREGORY CHERLIN, 'Groups of small Morley rank', Ann. Math. Logic, 17 (1979) 1-28. 4. DAVID M. EVANS and EHUD HRUSHOVSKI, 'Projective planes in algebraically closed fields', Proc. London Math. Soc. (3) 62 (1991) 1-24. 5. K. L. HOLLAND, 'On finding fields from their algebraic closure geometries', Proc. Amer. Math. Soc. 116(1992) 1135-1142. 6. S. LANG, Algebra (Addison-Wesley, Reading, 1965).

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7. DANIEL LASCAR, 'Autour de la propriete du petit indice', Proc. London Math. Soc. (3) 62 (1991) 25-53. 8. DANIEL LASCAR, ' Les automorphismes d'un ensemble fortement minimal', / . Symbolic Logic 57(1992) 238-251. 9. BRUNO POIZAT, Groupes stables (Nur al-Mantiq wal-Ma'rifah, Villeurbanne, 1987). 10. J. H. SILVERMAN, The arithmetic of elliptic curves, Graduate Texts in Mathematics 106 (Springer, Berlin, 1986).

School of Mathematics University of East Anglia Norwich NR4 7TJ E-mail: [email protected]

Department of Mathematics MIT Cambridge Massachusetts 02139 USA. and Department of Mathematics The Hebrew University Jerusalem Israel.