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Abstract In M Artin proved that any formal power series solution of a system of analytic equations may be approximated by convergent power series solutions Motivated by this result and a similar result of P oski he con jectured that this remains true when we replace the ring of convergent power series by a more general ring This paper presents the state of the art on this problem aimed at non experts In particular we put a slant on the Artin Approximation Problem with con straints

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ARTIN APPROXIMATION GUILLAUME ROND Abstract. In 1968, M. Artin proved that any formal power series solution of a system of analytic equations may be approximated by convergent power series solutions. Motivated by this result and a similar result of P?oski, he con- jectured that this remains true when we replace the ring of convergent power series by a more general ring. This paper presents the state of the art on this problem, aimed at non-experts. In particular we put a slant on the Artin Approximation Problem with con- straints. Contents 1. Introduction 1 2. Artin Approximation 13 2.1. The analytic case 13 2.2. Artin Approximation and Weierstrass Division Theorem 22 2.3. Néron's desingularization and Popescu's Theorem 24 3. Strong Artin Approximation 27 3.1. Greenberg's Theorem: the case of a discrete valuation ring 27 3.2. Strong Artin Approximation Theorem: the general case 31 3.3. Ultraproducts and proofs of Strong Approximation type results 33 3.4. Effectivity of the behaviour of Artin functions: some examples 34 4. Examples of Applications 39 5. Approximation with constraints 41 5.1. Examples 42 5.2. Nested Approximation in the algebraic case 44 5.3. Nested Approximation in the analytic case 48 5.4. Other examples of approximation with constraints 53 Appendix A. Weierstrass Preparation Theorem 56 Appendix B. Regular morphisms and excellent rings 56 Appendix C. Étale morphisms and Henselian rings 58 References 61 1.

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Abstract.In 1968, M. Artin proved that any formal power series solution of a system of analytic equations may be approximated by convergent power series solutions. Motivated by this result and a similar result of Płoski, he con-jectured that this remains true when we replace the ring of convergent power series by a more general ring. This paper presents the state of the art on this problem, aimed at non-experts. In particular we put a slant on the Artin Approximation Problem with con-straints.
1. Introduction 2. Artin Approximation 2.1. The analytic case 2.2. Artin Approximation and Weierstrass Division Theorem 2.3. Néron’s desingularization and Popescu’s Theorem 3. Strong Artin Approximation 3.1. Greenberg’s Theorem: the case of a discrete valuation ring 3.2. Strong Artin Approximation Theorem: the general case 3.3. Ultraproducts and proofs of Strong Approximation type results 3.4. Effectivity of the behaviour of Artin functions: some examples 4. Examples of Applications 5. Approximation with constraints 5.1. Examples 5.2. Nested Approximation in the algebraic case 5.3. Nested Approximation in the analytic case 5.4. Other examples of approximation with constraints Appendix A. Weierstrass Preparation Theorem Appendix B. Regular morphisms and excellent rings Appendix C. Étale morphisms and Henselian rings References
1 13 13 22 24 27 27 31 33 34 39 41 42 44 48 53 56 56 58 61
The aim of this paper is to present the Artin Approximation Theorem and some related results. The problem we are interested in is to find analytic solutions of some system of equations when this system admits formal power series solutions and the Artin Approximation Theorem yields a positive answer to this problem. We begin
Date: June 6, 2012.
this paper by giving several examples explaining what this sentence means exactly. Then we will present the state of the art on this problem. There is essentially three parts: the first part is dedicated to present the Artin Approximation Theorem and its generalizations; the second part presents a stronger version of Artin Approxima-tion Theorem; the last part is mainly devoted to explore the Artin Approximation Problem in the case of constraints. An appendix presents the algebraic material used in this paper (Weierstrass Preparation Theorem, excellent rings, étales mor-phisms and Henselian rings). We do not give the proofs of all the results presented in this paper but, at least, we always try to outline the proofs and give the main arguments. Example 1.Let us consider the following curveC:={(t3, t4, t5), tC}inC3. This curve is an algebraic set which means that it is the zero locus of polynomials in three variables. Indeed, we can check thatCis the zero locus of the polynomials f:=y2xz,g:=yzx3andh:=z2x2y. If we consider the zero locus of any two of these polynomials we always get a set larger thanC. The complex dimension of the zero locus of one non-constant polynomial in three variables is 2 (such a set is called a hypersurface ofC3). HereCis the intersection of the zero locus of three hypersurfaces and not of two of them, but its complex dimension is 1. In fact we can see this phenomenon as follows: we call an algebraic relation between f,gandhany element of the kernel of the linear mapϕ:C[x, y, z]3−→C[x, y, z] defined byϕ(a, b, c) :=af+bg+ch. Obviouslyr1:= (g,f,0),r2:= (h,0,f) andr3:= (0, h,g)Ker(ϕ) are called the trivial relations between. Thesef,g andh in our case there is one more relation which is. Butr0:= (z, y,x)andr0 cannot be written asa1r1+a2r2+a3r3witha1,a2anda3C[x, y, z], which means thatr0is not in the sub-C[x, y, z]-module ofC[x, y, z]3generated byr1,r2andr3. On the other hand we can prove that Ker(ϕ)is generated byr0,r1,r2andr3. LetXbe the common zero locus offandg. If(x, y, z)Xandx6= 0, then h=zf+yg= 0thus(x, y, z)C. If(x, y, z)Xandx= 0, theny= 0. Geometri-x cally this means thatXis the union ofCand thez-axis, i.e. the union of two curves.
Now let us denote byCJx, y, zKthe ring of formal power series with coefficients inC. We can also consider formal relations betweenf,gandh, that is elements of the kernel of the mapCJx, y, zK3−→CJx, y, zKinduced byϕ element. Any of the forma0r0+a1r1+a2r2+a3r3is a formal relation as soon asa0,a1,a2, a3CJx, y, zK. In fact any formal relation is of this form, i.e. the algebraic relation generate the formal and analytic relations. We can show this as follows: we can assign the weights 3 tox, 4 toyand 5 toz this case. Inf,g,hare homogeneous polynomials of weights8,9and10andr0,r1,r2andr3are homogeneous relations of weights (5,4,3),(9,8,0),(10,0,8)and(0,10,9). If(a, b, c)CJx, y, zK3is a formal relation then we can writea=Pi=0ai,b=Pi=0biandc=Pi=0ciwhereai,biandciare homogeneous polynomials of degreeiwith respect to the previous weights. Then saying thataf+bg+ch= 0is equivalent to
aif+bi1g+ci2h= 0iN with the assumptionbi=ci= 0fori <0. Thus(a0,0,0),(a1, b0,0)and any (ai, bi1, ci2), for2i, are in Ker(ϕ), thus are homogeneous combinations ofr0, r1,r2andr3. Hence(a, b, c)is a combination ofr0,r1,r2andr3with coefficients
inCJx, y, zK.
Now we can investigate the same problem by replacing the ring of formal power series byC{x, y, z}, the ring of convergent power series with coefficients inC, i.e. C{x, y, z}:=i,j,XkNai,j,kxiyjzk/ρ >0,i,jX,k|ai,j,k|ρi+j+k<
We can also consider analytic relations betweenf,gandh, that is elements of the kernel of the mapC{x, y, z}3−→C{x, y, z}induced byϕ the formal case. From we see that any analytic relationris of the forma0r0+a1r1+a2r2+a3r3with aiCJx, y, zKfor0i4. In fact we can prove thataiC{x, y, z}for0i4. Let us remark that, saying thatr=a0r0+a1r1+a2r2+a3r3is equivalent to say thata0,...,a3satisfy a system of three affine equations with analytic coefficients. This is the first example of the problem we are interested in: if we some equations with analytic coefficients have formal solutions do they have analytic solutions? Artin Approximation Theorem yields an answer to this problem. Here is the first theorem proven by M. Artin in 1968:
Theorem 1.1(Artin Approximation Theorem).[Ar68]Letf(x, y)be a vector of convergent power series overCin two sets of variablesxandy given a. Assume formal power series solutionyb(x),
f(x, yb(x)) = 0.
Then there exists, for anycN, a convergent power series solutiony(x),
f(x, y(x)) = 0
which coincides withyb(x)up to degreec, y(x)yb(x)modulo(x)c.
We can define a topology onCJxKby saying that two power series are close if their difference is in a high power of the maximal ideal(x). Thus we can reformulate Theorem 1.1 as: formal power series solutions of a system of analytic equations may be approximated by convergent power series solutions.
Example 2.A special case of Theorem 1.1 and a generalization of Example 1 occurs whenfis linear iny, sayf(x, y) =Pfi(x)yi, wherefi(x)is a vector of convergent power series withrcoordinates for anyi solution. Ay(x)off(x, y) = 0 is a relation between thefi(x). In this case the formal relations are linear combi-nations of analytic combinations with coefficients inCJxK term of commutative. In algebra, this is expressed as the flatness of the ring of formal power series over the ring of convergent powers series, a result which can be proven via the Artin-Rees Lemma. It means that ifyb(x)is a formal solution off(x, y) = 0, then there exist analytic so-b lutions off(x, y) = 0denoted byyei(x),1is, and formal power seriesb1(x),..., b b bs(x), such thatyb(x) =Pibi(x)yei(x) by replacing in the previous sum the. Thus, b bi(x)by their truncation at orderc, we obtain an analytic solution off(x, y) = 0 coinciding withyb(c)up to degreec. If thefi(x)are vectors of polynomials then the formal relations are also linear’s combinations of algebraic relations since the ring of formal power series is flat over
the ring of polynomials, and Theorem 1.1 remains true iff(x, y)is linear inyand C{x}is replaced byC[x].
Example 3.A slight generalization of the previous example is whenf(x, y)is a vector of polynomials inyof degree one with coefficients inC{x}(resp.C[x]), say m f(x, y) =Xfi(x)yi+b(x) i=1 where thefi(x)’s andb(x) polynomi-are vectors of convergent power series (resp. als). Herexandyare multi-variables Ifyb(x)is a formal power series solution of f(x, y) = 0, then(yb(x),1)a formal power series solution ofis g(x, y, z) = 0where m g(x, y, z) :=Xfi(x)yi+b(x)z i=1 andzis a single variable. using the flatness of ThusCJxKoverC{x}(resp.C[x]) (Example 2), we can approximate(yb(x),1) poly-by a convergent power series (resp. nomial) solution(ye(x), ze(x))which coincides with(yb(x),1)up to degreec. In order to obtain a solution off(x, y) = 0we would like to be able to divideye(x)byze(x) sinceye(x)ze(x)1would be a solution off(x, y) = 0aroppmaxingtiyb(x) can. We remark that, ifc1, thenze(0) = 1thusze(x)is not the ideal(x). ButC{x}is a local ring. We call a local ring any ringA This isthat has only one maximal ideal. equivalent to say thatAis the disjoint union of one ideal (its only maximal ideal) and of the set of units inA. In particularz(x)1is invertible inC{x}, hence we e can approximate formal power series solutions off(x, y) = 0by convergent power series solutions. In the case(ye(x), ze(x))is a polynomial solution ofg(x, y, z) = 0,ze(x)is not invert-ible in general inC[x] Forsince it is not a local ring. instance set f(x, y) := (1x)y1 wherexandy Thenare single variables.y(x) :=Xxn1=1is the only formal x n=0 power series solution off(x, y) = 0, buty(x) Thus we cannotis not a polynomial. approximate the roots offinCJxKby roots offinC[x]. But instead of working inC[x]we can work inC[x](x)which is the ring of rational functions whose denominator does not vanish at 0. This ring is a local ring. Since ze(0)6= 0, thenye(x)ze(x)1is a vector of rational function ofC[x](x) particular. In any system of polynomial equations of degree one with coefficients inC[x]which has solutions inCJxKhas solutions inC[x](x). Example 4.The next example we are looking at is the following: setfAwhere A=C[x]orC[x](x)orC{x}. When do there existg,hAsuch thatf=gh? First of all, we can takeg= 1andh=for, more generally,ga unit inAand h=g1f. These are trivial cases and thus we are looking for non unitsgandh. Of course, if there exist non unitsgandhinAsuch thatf=gh, thenf= (bug)(bu1h)for any unitubCJxK let us assume that. But is the following true: b b there existbg,hCJxKsuch thatf=gbh. Then do there exist non unitsg,hA such thatf=gh? Let us remark that this question is equivalent to the following: if(Af)is an integral domain, is(fC)JCxJKxKstill an integral domain?
The answer to this question is no in general: setA:=C[x, y]and setf:= x2y2(1 +y). Thenfis irreducible as a polynomial sincey2(1 +y)is not a square inC[x, y]. Butf= (x+y1 +y)(xy1 +y)where1 +yis a formal power series such that1 +y2= 1 +y. Thusfis not irreducible inCJx, yKnor in C{x, y}but it is irreducible inC[x, y]orC[x, y](x,y). In fact it is easy to see thatx+y1 +yandxy1 +yare power series which are algebraic overC[x, y] are roots of polynomials with coefficients in they, i.e.C[x, y]. The set of such algebraic power series is a subring ofCJx, yKand it is denoted by Chx, yi. In general ifxmultivariable the ring of algebraic power seriesis a Chxiis the following:
Chxi:={fCJxK/P(z)C[x][z], P(f) = 0}.
It is not difficult to prove that the ring of algebraic power series is a subring of the ring of convergent power series and is a local ring. In 1969, M. Artin proved an analogue of Theorem 2.1 for the rings of algebraic power series [Ar69]. Thus iffChxi(orC{x}) is irreducible then it remains irreducible inCJxK, this is a consequence of Artin Approximation Theorem. From this theorem we can also deduce that iffChIxi(orC{Ix}), for some idealI, is irreducible, then it remains irreducible inICCJJxxKK. Example 5.Let us strengthen the previous question. Let us assume that there b b existbg,hCJxKsuch thatf=gbhwithfAwithA=ChxiorC{x}. Then does there exist a unitbuCJxKsuch thatubbgAandub1bhA? The answer to this question is positive ifA=ChxiorC{x}, this is a non trivial corollary of Artin Approximation Theorem (see Corollary 4.4). But it is negative in general forChIxiorC{xI}ifI The following example is due to S. Izumiis an ideal. [Iz92]: SetA:=(Cy{2xx,,y3z)}. Setϕb(z) :=Pn=0n!zn(this is a divergent power series) and set fb:=x+b(z),bg:= (xb(z))(1b(z)2)1CJx, y, zK. Then we can check thatx2=fbbgmodulo(y2x3) let us assume that. Now there exists a unitbuCJx, y, zKsuch thatbufbC{x, y, z}modulo(y2x3). ThusP:=bufb(y2x3)hbC{x, y, z}for somehCJx, y, zK. We can check easily thatP(0,0,0) = 0andxP(0,0,0) =ub(0,0,0)6= 0 by the Implicit. Thus Function Theorem for analytic functions there existsψ(y, z)C{y, z}, such that P(ψ(y, z), y, z) = 0andψ(0,0) = 0. This yields ψ(y, z) +b(z)(y2ψ(y, z)3)hb(ψ(y, z), y, z)bu1(ψ(y, z), y, z) = 0. uting 0 forywe obta)3b By substit inψ(0, z) +ψ(0, z k(z) = 0for some power series b k(z)CJzK. Sinceψ(0,0) = 0, this gives thatψ(0, z) = 0, thusψ(y, z) =(y, z) withθ(y, z)C{y, z} we obtain. Thus θ(y, z) +ϕb(z)(yy2θ(y, z)3)hb(ψ(y, z), y, z)bu1(ψ(y, z), y, z) = 0
and by substituting 0 fory, we see thatϕb(z) =θ(0, z)C{z}which is a contra-diction.
6 GUILLAUME ROND Thusx2=fbgbmodulo(y2x3)but there is no unitbuCJx, y, zKsuch that bC{x, y, z}modulo(y2x3 bf). u
Example 6. ifA similar question is the following:fAwithA=C[x],C[x](x), ChxiorC{x}and if there exist a non unitbgCJxKand an integermNsuch thatbgm=f, does there exist a non unitgAsuch thatgm=f? A weaker question is the following: if(fA)is reduced, is(fC)JCxKJxK Indeed,still reduced? ifgbm=ffor some non unitbgthen(fC)JCxJKxK if the answer to Thus,is not reduced. the second question is positive, then there exists a non unitgAand a unituA such thatugk=ffor some integerk.
As before, the answer to the first question is positive forA=ChxiandA=C{x} by Artin Approximation Theorem. IfA=C[x]orC[x](x), the answer to this question is negative. Indeed let us con-siderf=xm+xm+1. Thenf=gbmwithgb:=xm1 +xbut there is nogAsuch thatgm=f.
Nevertheless, the answer to the second question is positive in the casesA=C[x] orC[x](x) deep result is due to D. Rees (see [H-S06] for instance).. This Example 7.Using the same notation as in Example 4 we can ask a stronger question: setA=ChxiorC{x}and letfbe inA. If there existgandhC[x], vanishing at 0, such thatf=ghmodulo a large power of the ideal(x), do there existgandhinAsuch thatf=gh example 4 there is no hope, if? Bygandh exist, to expect thatgandhC[x]. We have the following theorem:
Theorem 1.2(Strong Artin Approximation Theorem).[Ar69]Letf(x, y)be a vector of polynomials overCin two sets of variablesxandy. Then there exists a functionβ:N−→N, such that for any integercand any given approximate solutiony(x)at orderβ(c), β(c) f(x, y(x))0modulo(x), there exists an algebraic power series solutiony(x), f(x, y(x)) = 0 which coincides withy(x)up to degreec, y(x)y(x)modulo(x)c. In particular, ifghf0modulo(x)β(1), whereβis the function of the previous theorem for the polynomialy1y2f, and ifg(0) =h(0) = 0, then there exist non unitsgandhChxisuch thatghf= 0. A natural question is: givenfC[x]how to computeβor, at least,β(1)? That is, up to what order do we have to check that the equationy1y2f= 0has an approximate solution in order to be sure that this equation as solutions? For instance, iff:=x1x2xd3thenfis irreducible butx1x2f0modulo(x)dfor anydN, so obviouslyβ(1)really depends onf. In fact, in Theorem 1.2 M. Artin proved thatβcan be chosen independently of the degree of the components of the the vectorf(x, y) it is still an open problem. But to find effective bounds onβ(see Section 3.4).