GOOD MODELS FOR CUBIC SURFACES. 1. Introduction

GOOD MODELS FOR CUBIC SURFACES ANDREAS-STEPHAN ELSENHANS Abstract. This article describes an algorithm for finding a model of a hyersurface with small coefficients. It is shown that the aroach works in arbitrary dimension and degree. In the secial case of a cubic surface it is comletely exlicit. 1. Introduction In [EJ] J. Jahnel and the author constructed cubic surfaces by using exlicit Galois descent and the hexahedral form. This leads to equations with big coefficients. The aim of this note is to exlain a way to find an isomorhic surface with small coefficients. This isomorhism is an isomorhism of Q-schemes. In general, the corresonding Z-schemes are not isomorhic. The construction of a good equation consists of two arts. In the first art one has to imrove the model for the scheme over Z for each rime of bad reduction. In an otimal situation, one can remove a bad rime comletely. In the second art, one has to look at the infinite rime. This is classical known as reduction theory. It means to modify the embedding of the surface by oerating with Sl 4 (Z). 2. The finite laces In [K] J. Kollar discusses the roblem of choosing good models in a very general way. We follow this aroach but we focus on the arithmetic situation. Let V be the hyersurface given by a homogeneous olynomial f(x 0,..., x n ) = 0 in the rojective sace. Assuming that V is semi-stable leads to a least one nonzero invariant I(f). Modifying f locally at the rime changes this invariant by a ower of. We use this to control the finiteness of the algorithm. More recisely, we construct a sequence of olynomials f n with integral coefficients defining isomorhic schemes over Q such that v (I(f n )) decreases. As I is a olynomial in the coefficients of f, this rocess terminates after finitely many stes. 1 The author was artially suorted by the Deutsche Forschungsgemeinschaft (DFG). 1

2 ANDREAS-STEPHAN ELSENHANS Remark 1. Recall from invariant theory that a smooth hyersurface of degree at least 3 is stable [MFK, Ch. IV, Proosition 4.2]. In this case one could choose the discriminant as the non-zero invariant. The main tool for the descrition of the algorithm are weights. Definition 2. Let f(x 0,..., x n ) be a olynomial with integral coefficients. Denote by mult f(x 0,..., x n ) the smallest -adic valuation of a coefficient of f. Remark 3. Let f be a homogeneous olynomial of degree d with mult f(x 0,..., x n ) = 0. Then mult f(x 0,..., x n ) = d. Let M Mat((n + 1) (n + 1), Z) Gl n+1 (Q) be given such that det(m) is a ower of. Then we can use M to modify the model. We get g = e f(mx) with e := mult (f(mx)). The new equation g(x 0,..., x n ) = 0 is better if v (I(g)) < v (I(f)). This is equivalent to e > v (det(m)) deg(f) n+1 Remark 4. Observe the following ambiguity. Suose that the columns of M and M define the same Z-lattice. Then the resulting new models are equal (i.e. isomorhic as Z-schemes). Using the elementary divisor theorem, one can factor the matrix M into a roduct of M 1 Gl n+1 (Z) and a diagonal matrix M 2, whose entries are owers of. The exonents of these diagonal entries are called a weight system. Without loss of generality the entries of M 2 are sorted. Remark 5. One does not have to look at arbitrary weight systems here are some obvious restrictions: As the entries of M 2 are sorted the weights are sorted too. Relacing a weight system (w 0,..., w n ) by a translated weight system (c + w 0,..., c + w n ) does not affect the model. So we can assume 0 = w 0 w 1 w n. All this can be interreted in the language of affine Bruhat-Tits buildings. Recall 6. The Bruhat-Tits building. The affine Bruhat-Tits building of tye à n+1 for Q is a n-dimensional simlicial comlex. The vertices are classes of (n + 1)-dimensional lattices in Q n+1. The class of a lattice L Q n+1 is [L] := {cl c Q }. Assume that the lattices L 0,..., L k satisfy L 0 L 1 L k L 0. Then their classes form a k-simlex. Let b 0,..., b n be a basis of Q n+1. Then the classes of all lattices of the form Z e 0 b 0 Z e n b n for e 0,..., e n Z form a sub-comlex called an aartment.

GOOD MODELS FOR CUBIC SURFACES 3 The aartment corresonding to the standard basis is called the standard aartment. Any two simlices are contained in one aartment. For a detailed descrition of buildings see, e.g., [B]. Remark 7. The factorization of the matrix M described above has the following interretation. The matrix M 1 chooses the aartment containing the old and the new lattice. The matrix M 2 describes the osition of the new lattice in the aartment. Doing the reduction with a concrete equation consists of two stes. First one has to choose an aartment and then one has to choose weights, i.e., a lattice in the aartment. If we assume that we have already chosen the right aartment then the matrix M is diagonal. Observation 8. Let f be a form of degree d and n + 1 variables. The weight system (0, w 1,..., w n ) imroves our equation with resect to the standard aartment if and only if f(x 0, w 1 x 1,..., w n x n ) 0 (mod k ) for k = d (w n+1 1 + + w n ) + 1. Writing f = a i1...i d x i1 x id we get the equivalent statement 0 i 1 i d n k w i 1 w id ai1...i d for each d-tule (i 1,..., i d ) such that k w i1 w id is ositive. From this the following finiteness result is easily derived: Proosition 9. Let n and d be fixed. Then a finite set of weight systems W exists such that the following holds. A form f of degree d in n + 1 variables can be imroved if and only if it can be imroved by using one of the weight systems of W. Proof. Without loss of generality the imrovement takes lace in the standard aartment. A form of degree d and n + 1 variables has m := ( ) n+d d coefficients. We identify the sace of all forms with Z m. A form can be imroved with a given weight system w if and only if the divisibility conditions listed above are satisfied. This condition can be reformulated in terms of the -adic valuation of the coefficients. More recisely a form can be imroved with the weight system w if and only if v(a i ) e i (w). The e i (w) are non-negative integers deending on w and not on. We have to show that the infinite union of all cones C(e 1 (w),..., e m (w)) := {a N m a i e i (w)} has a finite sub-covering.

4 ANDREAS-STEPHAN ELSENHANS We show this urely geometric statement by induction on the dimension. Assume the the dimension is 1. Then the statement reduces to the fact that each nonemty subset of N has a smallest element. Assume that the statement is roven in dimension m 1. We have to show that it is true in dimension m. We start with the cones C(e), e E. We roject onto the first m 1 coordinates. By induction we find finitely many cones that cover the rojection. We take arbitrary reimages C(e 1 ),..., C(e l ) of the cones that cover the rojection. The union of the cones chosen in dimension m covers most of the union of all cones. All missing oints have the roerty that the last coordinate is smaller than M := max{(e j ) m : j = 1,..., l}. As everything takes lace in N m the missing oints are located in finitely many layers with last coordinate between 1 and M. Insecting one layer we find the same kind of roblem in dimension m 1. By induction hyothesis each layer leads to finitely many additional cones. Remark 10. This finiteness result roves the existence of an algorithm that constructs an otimal model for a given lace. The algorithm uses the fact that the Bruhat-Tits-building is locally finite. This ensures that only finitely many lattices exist which corresond to a given weight system. So one gets the existence of a finite list of lattices that imrove an arbitrary equation if it can be imroved. Testing all these lattices either leads to a better equation or to the roof that no better equation exists. Reeating this until no better model can be found on gets an algorithm that chooses a otimal one. For semi-stable varieties this algorithm stos after finitely many stes. Now we turn to a more concrete descrition of the necessary weight systems in the case of cubic surfaces. A roosition stated at the end of J. Kollar s aer claims the following. Proosition 11. Let f(x 0,..., x 3 ) = 0 be a model of a cubic surface. Assume that this model can be imroved. Then it can be imroved with one of the following five weight systems (0, 0, 0, 1), (0, 0, 1, 1), (0, 1, 1, 1), (0, 1, 2, 2), (0, 2, 2, 3). Proof. The roof has the following strategy: Assuming that a given weight system with resect to the standard aartment leads to a better model, we show that one of the given five weight systems imroves the model, too. The (0, 0, 0, 1)-case The weight system (0, 0, 0, 1) imroves the cubic equation f(x 0, x 1, x 2, x 3 ) = 0 if

GOOD MODELS FOR CUBIC SURFACES 5 and only if all coefficients of the olynomial f(x 0, x 1, x 2, x 3 ) are divisible by. Equivalently the coefficients of x 3 0, x 2 0x 1, x 2 0x 2, x 0 x 2 1, x 0 x 1 x 2, x 0 x 2 2, x 3 1, x 2 1x 3, x 1 x 2 2, x 3 2 in the original olynomial are divisible by. All other weight systems which imly these divisibilities can be relaced by the weight system (0, 0, 0, 1). Claim: Assume that the sorted weight system (0, w 1, w 2, w 3 ) imroves our equation and satisfies the extra condition w 1 + w 3 3w 2 0. Then the weight system (0, 0, 0, 1) imroves the equation, too. Check: We have to show that all monomials without x 3 have coefficients that are divisible by. This is equivalent to k w i1 w i2 w i3 1 for i 1, i 2, i 3 {0, 1, 2}. Substituting k = 3(w1 +w 2 +w 3 ) + 1 the asserted inequality becomes 3(w 4 1 + w 2 + w 3 ) w i1 w i2 w i3 0. As the weights are sorted it is enough to look at the case i 1 = i 2 = i 3 = 2. We get 3(w 4 1 + w 3 ) 9w 4 2 0, which was assumed to hold. 4 The (0, 1, 2, 2)-case Every sorted weight system (0, w 1, w 2, w 3 ) which satisfies w 1 + w 3 3w 2 0, w 1 1, and 3w 1 w 2 w 3 0 can be relaced by (0, 1, 2, 2). We have to show that the coefficient of x i 0x 3 i 1 is divisible by 1+i, that the coefficients of x 2 0x 2 and x 2 0x 3 is divisible by 2, and finally that the coefficients of x 0 x 1 x 2 and x 0 x 1 x 3 are divisible by. As w 1 is assumed to be at least 1, the first statement reduces to k 3w 1 1. The second requirement reduces to k w 3 2 and the last condition is entailed by k w 1 w 3 1. Using w 1 1 we have to show k 3w 1 1 and k w 1 w 3 1. The formula k = 3(w1 +w 2 +w 3 ) + 1 leads to the inequalities 3(w 2+w 3 ) 9w 4 4 4 1 0 and 3w 4 2 1w 4 1 1w 4 3 0, which are assumed to be true. The (0, 2, 2, 3)-case Every sorted weight system (0, w 1, w 2, w 3 ) which satisfies w 3 > w 2, 3w 1 w 2 w 3 0, and w 3 3 can be relaced by (0, 2, 2, 3). We have to show the following divisibilities Monomial Coefficient is divisible by x 3 0 6 x 2 0x 1, x 2 0x 2 4 x 2 0x 3 3 x 0 x 2 1, x 0 x 1 x 2, x 0 x 2 2 2 x 0 x 1 x 3, x 0 x 2 x 3

6 ANDREAS-STEPHAN ELSENHANS These are entailed by the inequalities k 6 k w 1, k w 2 4 k w 3 3 k 2w 1, k w 1 w 2, k 2w 2 2 k w 1 w 3, k w 2 w 3 1. Using w 3 > w 2 w 1 and the integrality of the w i we reduce this system of inequalities to k 6 k w 3 3 k w 2 w 3 1. Note that our assumtions imly w 2 2 and w 3 3. Hence it remains to show k w 2 w 3 1. Substituting k = 3(w1 +w 2 +w 3 ) + 1 leads to the inequality 3 w 4 1 1w 4 2 1w 4 3 0 which was our assumtion. The (0, 1, 1, 1)-case Every sorted weight system (0, w 1, w 2, w 3 ) which satisfies 3w 1 + 3w 2 5w 3 0 can be relaced by (0, 1, 1, 1). We have to show the following divisibilities Monomial Coefficient is divisible by x 3 0 3 x 2 0x 1, x 2 0x 2, x 2 0x 3 2 x 0 x 2 1, x 0 x 1 x 2, x 0 x 1 x 3, x 0 x 2 2, x 0 x 2 x 3, x 0 x 2 3 The resulting system of inequalities can be simlified to k 2w 3 1 by using 1 w 3 w 2 w 1. The equality k = 3(w1 +w 2 +w 3 ) + 1 leads to the assumed 4 inequality. The (0, 0, 1, 1)-case Every sorted weight system (0, w 1, w 2, w 3 ) which satisfies 5w 1 + 3w 2 w 3 0 can be relaced by (0, 0, 1, 1). We have to show the following divisibilities Monomial Coefficient is divisible by x 3 0, x 2 0x 1, x 0 x 2 1, x 3 1 2 x 2 0x 2, x 2 0x 3, x 0 x 1 x 2, x 0 x 1 x 3, x 2 1x 2, x 2 1x 3 The resulting system of inequalities reduces to k 3w 1 2 and k 2w 1 w 3 1. Our assumtion contradicts w 1 = w 3. So we can use w 3 w 1 + 1. Only the inequality k 2w 1 w 3 1 is necessary. 4

Using k = 3(w1 +w 2 +w 3 ) 4 GOOD MODELS FOR CUBIC SURFACES 7 +1 we get 3 4 w 2 5 4 w 1 1 4 w 3 0. This is our assumtion. The general case We have to show that a weight system (0, w 1, w 2, w 3 ) with 0 w 1 w 2 w 3 can be relaced by one of the five given by J. Kollar. If w 1 = 0 then 3w 2 w 3 is either 0 or < 0. So we are in the (0, 0, 1, 1) or the (0, 0, 0, 1) case. If w 3 = 1 then we are in one of the three cases (0, 0, 0, 1), (0, 0, 1, 1), (0, 1, 1, 1). So we can assume w 3 2. If w 1 1 and w 3 = 2 then we are in one of the cases (0, 1, 1, 2), (0, 1, 2, 2), or (0, 2, 2, 2). They are covered by the (0, 0, 0, 1), (0, 1, 2, 2), and (0, 1, 1, 1) cases. Now we can assume w 1 1 and w 3 3. We test w 1 + w 3 3w 2. If this is not negative we are in the (0, 0, 0, 1)-case. Otherwise we test 3w 1 w 2 w 3. If this is not ositive we are in the (0, 1, 2, 2) case. It remains to treat the case w 1 1, w 3 3, and 3w 1 w 2 w 3 > 0. This is the (0, 2, 2, 3)-case if w 3 > w 2 holds. The last ossibility is w 1 1, w 2 = w 3 3, and 3w 1 w 2 w 3 > 0. This is art of the (0, 1, 1, 1) case. As the main inequalities found are homogenous we can visualize the result by setting w 3 = n. Then the sace of all weight systems is covered by the five cases as shown in the icture below. Note that the weight systems (0, 0, 1, 1) and (0, 1, 1, 1) are only necessary for very secial cases at the boundary and for small w 3. (0,1,1,1) case (0,n,n,n) (0,2,2,3) case (0,1,2,2) case (0,0,0,1) case (0,0,1,1) case (0,0,0,n) (0,0,n,n) Covering of all weight-systems as shown in the roof

8 ANDREAS-STEPHAN ELSENHANS To get a ractical algorithm one needs a strategy for choosing lattices which is better then an enumeration of all of them. We give a solution for this by a reduction modulo method. The main idea behind this is the following. We have characterized the cases by divisibility conditions of the coefficients. Reduction modulo leads to cubic surfaces with very few monomials. These are known to be singular. A detailed analysis of the singularities is the basis of the algorithm. Proosition 12. Let f = 0 be a model of a cubic surface. Then the following hold. An imrovement with the weight system (0, 0, 0, 1) is ossible if and only if the reduction modulo of f is reducible over Z/Z. If the weight system (0, 0, 1, 1) leads to an imrovement then the reduction modulo of f = 0 has at least a singular line. For > 3 the following holds. Let [t : u] [g 0 (t, u), g 1 (t, u), g 2 (t, u), g 3 (t, u)] be the arameterization of an arbitrary lift of the singular line. An imrovement is ossible if and only if f(g 0, g 1, g 2, g 3 ) 0 holds modulo 2. A necessary condition that the weight system (0, 1, 1, 1) leads to an imrovement is that the reduction modulo is a cone. Further there exists a oint P 0 that reduces to a singular oint and satisfies f(p 0 ) 0 modulo 3. If the cone has a unique aex then P 0 is a lift of this oint. Proof. Without loss of generality the imrovement takes lace in the standard aartment. The weight system (0, 0, 0, 1) works if and only if the coefficients of all monomials without x 3 are divisible by. Equivalently x 3 is a linear factor of the reduction of f. The ossibility of an imrovement with the weight system (0, 0, 1, 1) leads to the following conditions Monomial Coefficient is divisible by x 3 0, x 2 0x 1, x 0 x 2 1, x 3 1 2 x 2 0x 2, x 2 0x 3, x 0 x 1 x 2, x 0 x 1 x 3, x 2 1x 2, x 2 1x 3 On the other hand let h = 0 be a cubic surface such that the reduction modulo has the singular line x 2 = x 3 = 0. This is characterized by h(x 0, x 1, 0, 0) and h. x i x2 =x 3 =0 A sufficient condition for this is that the coefficients of all monomials m with deg x2 (m) + deg x3 (m) 1 are divisible by (for > 3 this is equivalent). These are the monomials listed in the table above.

GOOD MODELS FOR CUBIC SURFACES 9 It remains to look at the extra conditions given by the 2 in the second line of the table. This means 2 f(x 0, x 1, 0, 0). Which is the last statement given above. Note that for singular lines this condition is indeendent of the lift we choose. Now we treat the weight system (0, 1, 1, 1). This leads to f(x 0, x 1, x 2, x 3 ) = f(1, 0, 0, 0)x 3 0 + x 2 0l(x 1, x 2, x 3 ) + 2 x 0 q(x 1, x 2, x 3 ) + O( 3 ) with a linear form l and a quadratic form q. Using 3 f(x 0, x 1, x 2, x 3 ) we get 3 f(1, 0, 0, 0), 2 l, and q. This shows the claim. Proosition 13. Let f = 0 be a model of a cubic surface. Assume that it can be imroved with the weight system (0, 1, 2, 2) or (0, 2, 2, 3) with resect to the standard aartment. Then [1 : 0 : 0 : 0] reduces to a singular oint. Further f(1, 0, 0, 0) 0 holds at least modulo 4. Proof. We can use the conditions 4 f(x 0, x 1, x 2, 2 x 3 ) or 6 f(x 0, 2 x 1, 2 x 2, 3 x 3 ). They imly the claim immediately. Remark 14. A deeer analysis of the singularities of the reduction modulo shows the following. In the (0, 1, 2, 2)-case the quadratic form of the local exansion in the singular oint has rank at most 2. If it has rank 2 then this quadric is the union of two lanes. The common line of the two lanes is contained in the cubic surface. The singularity is of tye A 3 or worse. In the (0, 2, 2, 3)-case the quadratic form of the local exansion in the singular oint has rank at most 1. This means the singularity is of tye D 4 or worse. We will not use these facts for our algorithm. But it seams ossible to imrove the algorithm by using them. Algorithm 15. Given a model of a cubic surface by f = 0 over Q. This algorithm comutes one model of this surface which is otimal at each finite lace. The descrition starts at the to level. The subroutines follow afterwards. i) Check that f = 0 is at least semi-stable. Otherwise terminate the algorithm with an error-message. Outut: Unstable forms can not be treated. ii) Calculate the GCD of the coefficients of f. Divide f by it. iii) Comute all rimes of bad reduction. iv) For each rime of bad reduction reeat the following until no better model is found. If a better model is found start the comutation for the next rime with this model. Try each of the five weight systems to imrove the model. Use the order (0, 0, 0, 1), (0, 0, 1, 1), (0, 1, 1, 1), (0, 1, 2, 2), (0, 2, 2, 3). If a better model is found then restart immediately with the first weight system and the new model.

10 ANDREAS-STEPHAN ELSENHANS v) Return the last model model found as an otimal one. Try the weight system (0, 0, 0, 1) i) Try to find a linear factor l of f in (Z/Z)[x 0, x 1, x 2, x 3 ]. ii) If no factor is found terminate this subroutine. iii) Comute the lattice generated by Z 4 and arbitrary lifts of three indeendent kernel vectors of l. iv) Write a basis of this lattice into the columns of a matrix M. v) Set f := f(mx). Let c be the GCD of the coefficients of f. vi) return 1 c f as a better model. Try the weight system (0, 0, 1, 1) We assume that we are not in the (0, 0, 0, 1)-case. Comute the rimary decomosition of the singular locus of the reduction of f = 0 modulo. For each line l found in the singular locus do the following. i) Choose two oints P 0, P 1 Z 4 that reduce to two different oints on the singular line. Comute the lattice generated by P 0, P 1 and Z 4. ii) Write a basis of this lattice into the columns of a matrix M. iii) Set f := f(mx). Let c be the GCD of the coefficients of f. iv) If 2 c then return 1 c f as a better model. Try the weight systems (0, 1, 1, 1), (0, 1, 2, 2) and (0, 2, 2, 3) We assume that we are not in the (0, 0, 0, 1)-case or the (0, 0, 1, 1)-case. i) Comute the list L 1 of relevant singular oints. (Use the subroutine below.) ii) For each oint S in L 1 do the following: a) Comute the lattice generated by S and Z 4. b) Write a basis of this lattice into the columns of the matrix M. c) Set f := f(mx). Let c be the GCD of the coefficients of f. d) Test 3 c. e) If this divisibility condition is satisfied then return 1 c f as a better model. f) Test 2 c. g) If this divisibility condition is satisfied then try to comlete for then weight systems (0, 1, 2, 2) or (0, 2, 2, 3) starting with the quotient 1 c f. Search for relevant singular oints We assume that we are not in the (0, 0, 0, 1)-case or the (0, 0, 1, 1)-case. i) Comute the rimary decomosition of the singular locus of the reduction of f = 0 modulo. ii) Write all isolated oints (defined over Z/Z) found into a list L 1.

GOOD MODELS FOR CUBIC SURFACES 11 iii) For each line l (defined over Z/Z) found in the singular locus do the following. a) If 3 then write all rational oints of the line into the list L 1. b) If > 3 comute a arameterization of an arbitrary lift of the line [t : u] [g 0 (t, u) : g 1 (t, u) : g 2 (t, u) : g 3 (t, u)]. Solve 1 f(g 0, g 1, g 2, g 3 ) 0 (mod ). (As we are not in the (0, 0, 1, 1)-case this is not the zero equation.) Write [g 0 (t, u) : g 1 (t, u) : g 2 (t, u) : g 3 (t, u)] for all solutions [t : u] into the list L 1. iv) For each Galois orbit of lines found comute the intersection of all these lines. If this leads to a oint defined over Z/Z add it to the list L 1. v) Return arbitrary reresentatives in Z 4 of the oints in L 1 as a list of relevant singular oints. Try to comlete for (0, 1, 2, 2) and (0, 2, 2, 3) Given an intermediate model f = 0 of a cubic surface. i) Factor the reduction of f modulo. ii) If f is irreducible then terminate with the error-message This can not haen. iii) If an irreducible quadratic factor q occurs then do the following. Comute the singular locus of q = 0. If exactly one line is found then treat this line as in the (0, 0, 1, 1)-case. If this does not lead to a GCD of at least 2 then no imrovement is ossible. Terminate the subroutine. Otherwise return the new model. iv) For each linear factors l that occurs do the following: a) Handle l as in the (0, 0, 0, 1) case. Denote the new model by f b) If a GCD of at least 2 occurs then return f this as a new model. c) Factor the reduction of f modulo. If a linear factor occurs then treat this once more as in the (0, 0, 0, 1)-case. Return the resulting model as an imroved one. v) If no linear factor of f occurs multily then terminate the subroutine. vi) Treat the multile factor as in the (0, 0, 0, 1)-case. Call the resulting model f. vii) The reduction of f modulo is irreducible. Search for a singular line in the reduction. viii) If no singular line is found then terminate the subroutine. ix) Treat this singular line as in the (0, 0, 1, 1)-case. x) If the GCD that arises during the comutation is at least 2 then return the result as the new model. xi) If the GCD that arises during the comutation is only then denote the new model by g. xii) Factor the reduction of g modulo.

12 ANDREAS-STEPHAN ELSENHANS xiii) If a linear factor occurs multily then treat this factor as in the (0, 0, 0, 1)- case. If the GCD that arises during the comutation is at least 2 then return the new model as an imroved one. xiv) Return no imrovement is ossible. Remark 16. In [H] Hilbert gave a classification of unstable cubic surfaces. This can be used to erform the first ste. Remark 17. U to now the discussion of the (0, 1, 2, 2)-case and the (0, 2, 2, 3)- case led to a ractical descrition of the first basis element modulo. I.e., it is ossible to change to the lattice generated by e 0, e 1, e 2, e 3. The new coefficients become divisible by 2. In order to exlain the last subroutine of the algorithm let us insect this intermediate model f 1. Note that f 1 can always be imroved with the weight system (0, 0, 0, 1) by scaling the first basis element. This means we return to the original model f. That is why the reduction of f 1 is always reducible. We have to analyze the conditions 2 f 1 (x, y, z, w) and 4 f 1 (x, y, z, 2 w) in this situation. In the first case we find Monomial Coefficient is divisible by x 3, x 2 y, xy 2, y 3 2 x 2 z, x 2 w, xyz, xyw, y 2 z, y 2 w As a return to f is ossible all monomials without x have coefficients divisible by. Consequently the reduction of f 1 has at most the monomials xz 2, xzw, xw 2. This shows that the reduction consists (geometrically) of three lanes. If the three linear factors are defined over Z/Z then we have to choose the right linear factor and treat it as in the (0, 0, 0, 1)-case. Then we do one more ste with the weight system (0, 0, 0, 1). In the case that only one linear factor is defined over Z/Z we have to insect the irreducible quadratic factor q. Then the scheme q = 0 has a singular line. We treat it as in the (0, 0, 1, 1)-case. In the second case we find Monomial Coefficient is divisible by x 3 4 x 2 y, x 2 z 3 x 2 w, xy 2, xyz, xz 2 2 xyw, xzw, y 3, y 2 z, yz 2, z 3 As a return to f is ossible all monomials without x vanish in the reduction. Summarizing the reduction consists of only one monomial xw 2. That means the reduction of f 1 has a multile linear factor. We can treat it as in the (0, 0, 0, 1)-case. If a GCD of at least 2 occurs then we have an imroved

GOOD MODELS FOR CUBIC SURFACES 13 model. If this is not the case then we have to continue with this new intermediate model f 2. It remains to do a ste with the weight system (0, 1, 1, 1). This leads to the following divisibilities: Monomial Coefficient is divisible by x 3 3 x 2 y, x 2 z, x 2 w 2 xy 2, xyz, xyw, xz 2, xzw, xw 2 Note that f 2 = 1 f(x, y, z, 2 w). This ensures the following additional divisibilities: 3 Monomial Coefficient is divisible by w 3 3 w 2 y, w 2 z 2 wz 2, wzy, wy 2, w 2 x Summarizing the reduction of f 2 has only the monomials y 3, y 2 z, yz 2, z 3. If the reduction is reducible over Z/Z then we can treat a linear factor as in the (0, 0, 0, 1)-case. That means we do the same as in the (0, 1, 2, 2)-case and get a better model. This is a very secial situation because we exected the weight system (0, 2, 2, 3) and found that the weight system (0, 1, 2, 2) works. We have to handle irreducible f 2. We exect the weight system (0, 1, 1, 1). But it is simler to do one (0, 1, 1, 0)-ste and one (0, 0, 0, 1)-ste. This avoids a search for relevant singular oints. The set of ossible monomials show that the reduction of f 2 = 0 consists (only geometrically) of three lanes. They meet in a common line which is the singular locus of the reduction. Treating this line as in the (0, 0, 1, 1) case leads to a model f 3. If a GCD of 2 occurs in this ste then we are done. But this is imossible as it would result in the weight system (0, 2, 2, 2). (This can be relaced by (0, 1, 1, 1).) A final ste for f 3 with the weight system (0, 0, 0, 1) has to be done. It must lead to a GCD of at least 2, otherwise no imrovement is ossible. Remark 18. This aroach can be described in the language of Bruhat-Tits buildings. We connect the start and the final lattice with a chain of 1-simlices. Do not be confused with the notion of galleries and geodesics in Bruhat-Tits buildings. These are different objects. 3. The infinite Place Let f = 0 be a cubic surface. Further assume that the model is locally otimal for every finite lace. It remains to choose a matrix M Gl 4 (Z) such that the coefficients of f(mx) are small.

14 ANDREAS-STEPHAN ELSENHANS Naive aroach. For a first try one could do the following: i) Build u a list L of some matrices in Gl 4 (Z). Ensure that these matrices form a generating system. ii) If f(mx) is smaller than f(x) for some M in L change to f(mx). iii) Reeat the last ste until no better equation is found. The main roblem of this aroach is the selection of the list L. Exerimentally we found that all matrices with at most two non-diagonal entries and all entries in {0, ±1} suffice in most cases to find a good equation. Further on this is very slow. Next we describe a faster method with is based on a symbolic reresentation of the surface and the LLL-algorithm. Proosition 19. (Sylvester, Clebsch) Let f = 0 be a general cubic surface, then there exist five linear forms l 1,..., l 5 such that f = l 3 1 + l 3 2 + l 3 3 + l 3 4 + l 3 5. These linear forms are unique u to order and multilication by third roots of unity. This is called the symbolic reresentation of f. Remark 20. For some very secial cubic surfaces this statement does not hold. E.g. the diagonal cubic surface x 3 0 + x 3 1 + x 3 2 + x 3 3 = 0 has infinitely many such reresentations. For these surfaces the aroach will not work. Definition 21. Let f = 0 be a cubic surface. The kernel surface (sometimes called Hessian) is the quartic given by the equation ( ) 2 f det = 0. x i x j Proosition 22. (Clebsch) Let f = l1 3 + l2 3 + l3 3 + l4 3 + l5 3 be a general cubic surface. Choose coefficients a 1,..., a 5 and linear forms k 1,..., k 5 such that k 1 + k 2 + k 3 + k 4 + k 5 = 0 and a i k i = l i. Then the singular oints of the kernel surface of f are the oints given by k i1 = 1, k i2 = 1, k i3 = 0, k i4 = 0, k i5 = 0 for {i 1, i 2, i 3, i 4, i 5 } = {1, 2, 3, 4, 5}. Remark 23. The symbolic reresentation of a cubic surface can be comuted by insecting the singular oints of the kernel surface. The main idea of the reduction algorithm is to do LLL-reduction with the linear forms of the symbolic reresentation. Algorithm 24. Let f = 0 be a general cubic surface this algorithm comutes a reduction of f for the infinite lace. ( i) Set q = det 2 f x i x j )i,j. ii) Comute the singular oints of q = 0. iii) Comute k 1,..., k 5 by solving the linear system of equations for k 1,..., k 5 given by the singular oints of q. i,j

GOOD MODELS FOR CUBIC SURFACES 15 iv) Solve the linear system for the coefficients a i. v) Set l i := 3 a i k i for i = 1,..., 5. vi) Define the hermitian form h(x) = l 1 (x) 2 + l 2 (x) 2 + l 3 (x) 2 + l 4 (x) 2 + l 5 (x) 2 vii) Use the LLL-algorithm to comute a matrix M whose columns are a reduced bases of Z 4 with resect to h. viii) Return f(m 1 x) as reduced olynomial. Examle 25. We start with the irreducible olynomial t 6 + 330t 4 + 1452t 3 + 13705t 2 + 123508t + 835540. Its Galois grou is of order 72. Doing the exlicit Galois descent as described in [EJ] naively leads to a cubic surface S 0 with coefficients having u to 43 digits. On the 27 lines of this surface oerates a Galois grou of order 144. S 0 has bad reduction at = 2, 3, 5, 7, 13, 113, 463, 733, 2141, 9643, 14143, 17278361, 22436341. Choosing better models and running the LLL-based reduction algorithm one gets the new surface S S has bad reduction at 2x 3 + 16x 2 z 12x 2 w 17xy 2 + 61xyz 26xyw 20xz 2 + 95xzw + 18xw 2 + 5y 3 + 33y 2 z + 10y 2 w 25yzw 22yw 2 11z 3 21z 2 w + 50zw 2 52w 3 = 0 = 2, 3, 5, 7, 13, 733, 22436341. Modulo = 3, 5, 7, 13, 22436341 the singularity is one oint of tye A 1. Modulo = 2 the surface has one singular oint of tye A 1 and one of tye A 3. Modulo 733 the surface degenerates to a cone over a smooth curve. References [B] Brown, K. S.: Buildings, Sringer, New York 1998 [EJ] Elsenhans, A.-S. and Jahnel, J.: Cubic surfaces with a Galois invariant double-six, Prerint [H] Hilbert, D.: Über die vollen Invariantensysteme, Math. Ann. 42 (1893), 313 373 [K] Kollar, J.: Polynomials with integral coefficients, equivalent to a given olynomial, Electron. Res. Announc. Amer. Math. Soc. 3 (1997), 17 27 [MFK] Mumford, D.; Fogarty, J.; Kirwan, F.: Geometric invariant theory, Sringer, Berlin 1994