Equimultiple locus of embedded algebroid surfaces and blowing-up in arbitrary characteristic

arXiv:math/0703583v1 [math.AG] 20 Mar 2007 Equimultiple locus of embedded algebroid surfaces and blowing–up in arbitrary characteristic Piedra, R.∗ Depto. de ´ Algebra Universidad de Sevilla Tornero, J.M.† Depto. de ´ Algebra Universidad de Sevilla November, 2005 Abstract This paper extends previous results of the authors, concerning the behaviour of the equimultiple locus of algebroid surfaces under blowing–up, to arbitrary characteristic. Mathematics Subject Classification (2000): 14B05, 32S25. 1 Introduction Let Kbe an algebraically closed field of characteristic pand S= Spec(K[[X, Y, Z]]/(F)) an embedded algebroid surface. With no loss of generality Scan be considered to be defined by a Weierstrass equation F(Z) = Zn+ n−1 X k=0 ak(X, Y )Zk, where nis the multiplicity of S, that is, ord (ak)≥n−kfor all k= 0, ..., n −1. After a change of variables we can assume that either F, the initial form of F, is not a power of a linear form, or F=Zn. ∗Supported by FQM 304 and BFM 2000–1523. †Supported by FQM 218 and MTM2004–07203–C02. 1 From now on, by a Weierstrass equation we will mean an equation like this. In this situation the equimultiple locus of Sis E(S) = nP∈Spec(K[[X, Y, Z]]) |F∈P(n)o, which is never empty, as M= (X, Y, Z) always lies in E(S). Note that all elements of E(S), different from M, can be assumed to have the form P= (Z+G(X, Y ), H(X, Y )). Geometrically speaking, the equimultiple locus represents points at which the multiplicity is the same than in the origin; hence they are the “closest” points to the origin, in (coarse) terms of singularity complexity. We will note by E0(S) the subset of smooth elements of E(S). In our previous paper [5] we prove a theorem relating E0S(1) to E(S), where S(1) is the blowing–up of Scentered in an element of E0(S). Our aim is extending this result to the arbitrary characteristic case. The main difficulty in the positive characteristic case comes from the fact that, if pdivides n, we cannot apply the Tchirnhausen transformation. This deceivingly naive procedure assures us, when it can be carried out, that: (a) If Fis the power of a linear form (that is, if the tangent cone is a plane), then it must be F=Zn. (b) All the elements of E(S) contain Z. (c) If, after blowing–up, the multiplicity remains the same, the above conditions still hold. In our current situation, as we pointed out, one finds a regular parameter verifying (a) with a simple change of coordinates. Moreover, after Mulay’s work ([3]) one can also find a regular parameter satisfying (b). We should note here that the outstanding result of Mulay is not constructive (much in the spirit of Abhyankar’s beautiful theory on good points [1]). Even though one can manage to find a parameter verifying (a) and (b), the preservation of (a) and (b) under certain blowing–ups is false in positive characterisitic, as is well–known (see [4] for how this might be overcome in a Levi–Zariski resolution). The existence of such an amenable parameter would have made the null characteristic proof valid for p > 0. The peculiarities of the 2 case p|nhave forced us to use completely different strategies for most parts of the result, although our proof turned to be characteristic–free, a somehow unusual fact in singularity theory. The main result thus generalizes previous work of Zariski ([7]), Hironaka ([2]), Abhyankar ([1]) and others, who treated specific cases tailored for their purposes on resolution of surface singularities. Note that, in fact, all these results were previous to Mulay’s which is a key stone in our strategy. As the geometric behaviour of the equimultiple locus of three-dimensional varieties remains unknown (see, for instance, [6]), we hope this result can be used as a first step to this much more difficult and intriguing problem. 2 Notation and technical results For the sake of completeness (and for the convenience of the reader), we recall the basic facts and technical results related to quadratic and monoidal transformations that were used in [5] and which will be used afterwards. For all what follows, let Sbe an embedded algebroid surface of multiplicity n, F=Zn+ n−1 X k=0  X i,j aijkXiYj Zk=Zn+ n−1 X k=0 ak(X, Y )Zk a Weierstrass equation of S. We will note N{X,Y,Z}(F) = n(i, j, k)∈N3|aijk 6= 0o, although we will omit the subscript whenever the variables is clear from the context. Definition.– The elements of E(S) different from Mwill be called equimultiple curves. The elements of E0(S) other than Mwill be called permitted curves. Remark.– In particular, any P∈ E0(S) can be assumed to be, for instance (Z, X), after a suitable change of variables. Clearly (Z, X)∈ E0(S) is equivalent to i+k≥nfor all (i, j, k)∈N(F). The monoidal transform of S, centered in (X, Z), in the point corresponding to the direction (α: 0 : γ) (say α6= 0) of the exceptional divisor is the surface S(1) defined by the equation F(1) =Z1+γ αn +X (i,j,k)∈N(F) aijkXi+k−n 1Yj 1Z1+γ αk . 3 Observe that this only makes sense (that is, gives a non–unit) whenever F(α, 0, γ) = 0. The homomorphism πP (α:0:γ):K[[X, Y, Z]] −→ K[[X1, Y1, Z1]] X7−→ X1 Y7−→ Y1 Z7−→ X1Z1+γ α will be called the homomorphism associated to the monoidal transformation in (α: 0 : γ) or, in short, the equations of the monoidal transformation. The overline is because one must privilege a non-zero coordinate, but all the possibilities define associated equations. The quadratic transform (that is, blowing–ups with center M) in the point corresponding to the direction (α:β:γ) (say α6= 0) of the exceptional divisor is the surface S(1) defined by the equation F(1) =Z1+γ αn +X (i,j,k)∈N(F) aijkXi+j+k−n 1Y1+β αjZ1+γ αk . Again this only makes sense whenever F(α, β, γ) = 0. Analogously, the homomorphism πM (α:β:γ):K[[X, Y, Z]] −→ K[[X1, Y1, Z1]] X7−→ X1 Y7−→ X1Y1+β α Z7−→ X1Z1+γ α will be called the homomorphism associated to the quadratic transformation in (α:β:γ) or the equations of the quadratic transformation. Remark.– In the previous situation, consider a change of variables in K[[X, Y, Z]] given by      ϕ(X) = a1X′+a2Y′+a3Z′+ϕ1(X′, Y ′, Z′) ϕ(Y) = b1X′+b2Y′+b3Z′+ϕ2(X′, Y ′, Z′) ϕ(Z) = c1X′+c2Y′+c3Z′+ϕ3(X′, Y ′, Z′) , with ord (ϕi)≥2. 4 Assume also that      α=a1α′+a2β′+a3γ′ β=b1α′+b2β′+b3γ′ γ=c1α′+c2β′+c3γ′ with, say, γ′6= 0. Then there is a unique change of variables ψ: K[[X1, Y1, Z1]] −→ K[[X′ 1, Y ′ 1, Z′ 1]] such that ψπM (α:β:γ)=πM (α′:β′:γ′)ϕ. Definition.– Let Q∈ E(S), with Q= (Z+H(X, Y ), G(X, Y )). Then for u∈P2(K), the ideal M u(Q) =   πM u(Z+H(X, Y )) X1 ,πM u(G(X, Y )) Xord(G) 1  is called the (strict) quadratic transform of Qin the point u. Obviously, this definition makes sense only if the quadratic transform in the direction udoes. There is a natural version of monoidal transform of Qwith center P, for all P∈ E0(S). Notation.– We will note by νthe natural isomorphism ν:K[[X, Y, Z]] −→ K[[X1, Y1, Z1]] sending Xto X1,Yto Y1and Zto Z1. 3 The theorem We will restrict ourselves to the case which is interesting for desingularization issues: that where Sand S(1) have the same multiplicity. This leaves out some situations. Lemma.– If the tangent cone of Sis not a plane, the multiplicity of any monoidal transform is strictly less than n. Proof.– See [5] for a characteristic–free proof. Remark.– In particular, note that if the tangent cone is not a plane, there cannot be more than one permitted curve. In fact, assume we have two curves, Pand Q, and choose Zto be a regular parameter with P= (Z, G(X, Y )), Q= (Z, H(X, Y )). If Fis a Weierstrass equation with the usual form, then Pis permitted if and only if Gn−k|akfor all k= 0, ..., n −1. As the same goes for Q, it is clearly impossible that 5 there exists some akwith ord(ak) = n−k. Hence the tangent cone must be a plane (in fact it must be Z= 0). Theorem.– Let Sbe an algebroid surface and S(1) a quadratic or monoidal transform of Shaving the same multiplicity. (a) Let S(1) be the monoidal transform of Swith center P∈ E0(S) then, either E0S(1)=ν(E0(S)) or E0S(1)=ν(E0(S)\ {P}). (b) Let S(1) be the quadratic transform of Sin the point u. (b.1) If the tangent cone is not a plane then E0S(1)= M u(E0(S)). (b.2) If the tangent cone is a plane then we can find three types of curves in E0S(1): (i) The exceptional divisor of the transform. (ii) Primes M u(Q), with Q∈ E(S)\ E0(S), which are tangent to the exceptional divisor. (iii) Primes M u(Q), with Q∈ E0(S), where both ν(Q) and M u(Q) are transversal to the exceptional divisor. Moreover, if E0S(1)contains a prime of type (ii), then it also contains a prime of type (i). Proof.– Although some partial results are common to the characteristic 0 case, we will repeat them or, at the very least, we will give a detailed outline when appropriate for the convenience of the reader. In what follows let Fbe, as usual, a Weierstrass equation of S. Case (a) This case presents the first (small) differences of argumentation with the characteristic 0, as the reader can check with [5]. We have to prove that, after a monoidal transformation with center P∈ E0(S), Q∈ E0(S) if and only if ν(Q)∈ E0S(1), except maybe for Q=P. Hence assume Zis a parameter verifying that for all I∈ E(S), Z∈I. After a change of variables in K[[X, Y ]], we can assume Pto be (Z, X) and Q(other than P) to be (Z, G(X, Y )) with ord(G) = 1. As we noticed above F=Zn, hence there is only one direction in the exceptional divisor, (1 : 0 : 0). An equation for S(1) is then F(1) =Zn 1+ n−1 X k=0 ak(X1, Y1) Xn−k 1 Zk 1=Zn 1+ n−1 X k=0 a(1) k(X1, Y1)Zk 1, and therefore G(X1, Y1)n−k|a(1) k(X1, Y1) if and only if G(X, Y )n−k|ak(X, Y ). This finishes the case. 6 Case (b.2) Much like in zero characteristic, this case gives the basis for the other one. It is also the point where the main differences between both cases become notorious. Remark.– First of all note that we can restrict ourselves to the case where the tangent cone is Z= 0 (this is as before an easy change of variables) and the direction in the exceptional divisor is (1 : 0 : 0). If this is not the case, for the results at points (1 : α: 0) it suffices considering the (commutative) diagram K[[X, Y, Z]] K[[X′, Y ′, Z′]] K[[X1, Y1, Z1]] K[[X′ 1, Y ′ 1, Z′ 1]] ✲ ✲ ❄ ❄ ϕ ψ πM (1:α:0) πM (1:0:0) with ϕgiven by      ϕ(X) = X′ ϕ(Y) = Y′+αX′ ϕ(Z) = Z′ Of course the results at (0 : 1 : 0) are clearly symmetric. Remark.– Assume that we have a permitted curve in S(1) of the general type, say, Q= (G1, G2), with G1=α1X1+β1Y1+γ1Z1+G′ 1(X1, Y1, Z1),where ord (G′ 1)>1 G2=α2X1+β2Y1+γ2Z1+G′ 2(X1, Y1, Z1),where ord (G′ 2)>1 As the multiplicity remains the same, the monomial Zn 1must appear in F(1). Hence either γ1or γ2must be non zero. Let us suppose it is γ16= 0 and so we can substitute G1by its associated Weierstrass polynomial with respect to Z1, of the form Z1+a(X1, Y1) with ord(a)≥1. Now we change Z1by −a(X1, Y1) in G2to obtain Q= (Z1+a(X1, Y1), αX1+βY1+b(X1, Y1)) , with ord(a),ord(b)>1. We will look first at permitted curves in S(1) which are transversal to the exceptional divisor. 7 Lemma.– Under the hypothesis of case (b.2) assume that there is a permitted curve Q∈ E0S(1)which is transversal to the exceptional divisor. Then there exists some P∈ E0(S) such that Q=M (1:0:0)(P) and ν(P) is transversal to the exceptional divisor. Proof.– In the above notation, Qis transversal whenever β6= 0. In this case we can change αX1+βY1+b(X1, Y1) for its associated Weierstrass polynomial with respect to Y1and then make the corresponding substitution in a(X1, Y1) to obtain Q=Z1+a′(X1), Y1+b′(X1), with ord (a′)>1, ord (b′)>0. Consider then the following diagram K[[X, Y, Z]] K[[X′, Y ′, Z′]] K[[X1, Y1, Z1]] K[[X′ 1, Y ′ 1, Z′ 1]] ✲ ✲ ❄ ❄ ϕ ψ πM (1:0:0) πM (1:0:0) with changes of variables      ϕ(X) = X′ ϕ(Y) = Y′−X′b′(X′) ϕ(Z) = Z′−X′a′(X′) ,     ψ(X1) = X′ 1 ψ(Y1) = Y′ 1−b′(X′ 1) ψ(Z1) = Z′ 1−a′(X′ 1) As ψ(Q) = (Z′ 1, Y ′ 1), we know that (Z′ 1, Y ′ 1) is permitted in S(1). But, looking at the equations of the transformation πM (1:0:0) this clearly implies that (Z′, Y ′) was permitted in S. Therefore P=ϕ−1Z′, Y ′=Z+Xa′(X), Y +Xb′(X) was permitted in S. It is clear that ν(P) is transversal to the exceptional divisor and M (1:0:0)(Q) = P. This proves the lemma. Now we will prove that the existence of permitted curves tangent to the exceptional divisor implies that the exceptional divisor lies in E0S(1). We will do it without using the fact that Fis the power of a linear form, and so it will be still valid for case (b.1). 8 Lemma.– Under the hypothesis of case (b) assume that there is a permitted curve P∈ E0S(1)which is tangent to the exceptional divisor. Then the exceptional divisor lies in E0S(1). Proof.– From the remarks at the beginning of this case we can already assume that Phas the form P= (Z1+a(Y1), X1+b(Y1)) ,with ord(a),ord(b)≥2. The proof is considerably different, depending on whether a(Y1) is zero or not. Let us assume a(Y1) = 0 (this is the easy part) and let us write F(1) as usual: F(1) =Zn 1+ n−1 X k=0 a(1) k(X1, Y1)Zk 1, with the following decomposition a(1) k(X1, Y1) = bk(X1, Y1) (X1+b(Y1))n−k, k = 0, ..., n −1, where bkdoes not divide X1+b(Y1). Let us take any k∈ {0, ..., n−1}, let t= ord(b)≥2 and let us call Xr 1Ys 1the smallest monomial with respect to the lexicographic order appearing in bk(X1, Y1). Then the monomial Xr 1Ys+t(n−k) 1Zk 1must occur in a(1) k(X1, Y1)Zk 1. Now, as this monomial appears after a quadratic transform in the point (1 : 0 : 0) of the exceptional divisor it is clear that it must hold r≥s+t(n−k) + k−n≥s+ 2(n−k)−(n−k)≥n−k, and therefore Xn−k 1|bk(X1, Y1) for all k. This proves that the exceptional divisor is also permitted. Let us move now to the more complicated case, when a(Y1)6= 0. Now we write F(1) in the following form F(1) = (Z1+a(Y1))n+ n−1 X k=0 bk(X1, Y1) (X1+b(Y1))n−k(Z1+a(Y1))k, for some bk(X1, Y1)∈K[[X1, Y1]]. Hence the summand a(Y1)n, which is a power series in K[[Y1]] of order strictly greater than n, appears in the independent term. But as F(1) comes from Fafter a quadratic transformation at the point (1 : 0 : 0) of the exceptional divisor, there can be no monomials 9 For proving that the quadratic transform cannot have new permitted curves note that, in (b.2), we have shown that, if a new permitted curve appears, so does the exceptional divisor (whether Fis the power of a linear form or not). But (Z, Y ) cannot be a permitted curve, since F(1) contains monomials in K[X, Z] other than Zn. It is also clear that the quadratic transform does not erase permitted curves either. If there is a permitted curve we may take it to be (Z, X), after a change of variables which does not affect (0 : 1 : 0). Clearly this curve cannot disappear from the equimultiple locus after a quadratic transform on (0 : 1 : 0). This finishes the proof of the theorem. References [1] S.S. Abhyankar: Good points of a hypersurface. Adv. Math. 68 (1988), 87–256. [2] H. Hironaka: Desingularization of excellent surfaces. Notes by B.M. Bennet (1967). 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