Finiteness results concerning algebraic power series
aa r X i v : . [ m a t h . A C ] F e b FINITENESS RESULTS CONCERNING ALGEBRAIC POWERSERIES
FUENSANTA AROCA, JULIE DECAUP, AND GUILLAUME ROND
Abstract.
We construct an explicit filtration of the ring of algebraic powerseries by finite dimensional constructible sets, measuring the complexity ofthese series. As an application, we give a bound on the dimension of the setof algebraic power series of bounded complexity lying on an algebraic varietydefined over the field of power series.
The ring of polynomials over a field k is a k -vector space filtered by finite dimen-sional vector spaces, namely the k -vector spaces of polynomials of degree less than d , for every d ∈ N . In many situations, the description of the ring of polynomials asa ring filtered by finite dimensional vector spaces is very useful, both for theoreticaland applied aspects. The main advantage of this, is that the set of polynomials ofdegree bounded by d is fully described by a finite number of data. One common useof this fact is when one tries to approximate objects by polynomials. For example,this is the case in analysis, by the use of Weierstrass approximation theorem, or inalgebra when one deals with approximations by polynomials of solutions of differ-ential or functional equations, or transcendental estimations of such solutions.In several cases, dealing with polynomials is not practical or effective enough, inparticular because the implicit function theorem does not hold in the polynomialsetting. To avoid this problem, one can replace the ring of polynomials by the ringof algebraic power series over k . This is the set of formal power series that arealgebraic over the ring of polynomials. This set is a ring (satisfying the implicitfunction theorem), in particular it is a k -vector space, but there is no explicit oreffective description of a filtration of this ring by finite dimensional spaces.In this note, we give an explicit description of a filtration of the ring of alge-braic power series by constructible sets. A constructible set is a subset of an affinespace k n which is described by polynomial equalities and inequalities. Here, thesets of this filtration are the sets A ( d, h ) of algebraic power series whose minimalpolynomial P ( x, T ) over k [ x ], is such thatdeg x ( P ) ≤ h, deg T ( P ) ≤ d. We prove that these sets can be seen as constructible subsets of k N ( d,h ) for someconstant N ( d, h ) depending only on d and h (see Corollary 2.8), and the dimen-sion of this constructible set is computed. Roughly speaking, the idea is to try toidentify an algebraic power series with its minimal polynomial. But this cannot Mathematics Subject Classification.
Key words and phrases. algebraic power series rings, constructible set.The third author is deeply grateful to the UMI LASOL of CNRS where this project has beenpartially carried out. work directly, since two distinct algebraic power series may have the same minimalpolynomial. Moreover, an irreducible polynomial in x and T has no power seriesroot in general. Therefore there is no correspondence between algebraic power se-ries and irreducible polynomials. To overcome this problem, we first describe thesubset of A ( d, h ) of algebraic power series whose minimal polynomial satisfies theimplicit function theorem as a constructible subset of some k N (see Theorem 2.6).And from this we deduce the same kind of result for A ( d, h ). Let us mention thatthis kind of approach has already been used in [FS98] and [HM17].As an application, we give a bound on the dimension of the set of solutions ofpolynomial equations with coefficients in k (( x )) whose entries are in A ( d, h ), when k is algebraically closed (see Theorem 3.1). This problem is a non-archimedeananalogue of the problem of bounding the number of Q -points of a complex alge-braic variety. Acknowledgment.
We are very grateful to Michel Hickel and Micka¨el Matusinskifor their comments and suggestions on a previous version of our paper. In particularthey pointed out that there was an essential gap in the proof of Theorem 2.8.1.
Preliminaries
In the whole paper k will always denote a field, and x will denote a singleindeterminate. The ring of algebraic power series will be denoted by k h x i . Forevery integer n , A n k will denote the affine space of dimension n over k . Definition 1.1.
Let f ∈ k h x i . The morphism ψ : k [ x, T ] −→ k h x i defined by ψ ( P ( x, T )) = P ( x, f ) is not injective and its kernel is a height one prime ideal of k [ x, T ]. Therefore it is generated by one polynomial. If P ( x, T ) is such a generator,any other generator of this ideal is a multiple of P ( x, T ) by a non-zero element of k . Any such a generator is called a minimal polynomial of f . But, by abuse oflanguage, we will always refer to such an element by the minimal polynomial of f . Definition 1.2. [AB13][Ro15] Let P ( T ) ∈ k [ x, T ]. The maximum of the degreesof the coefficients of P ( T ) seen as a polynomial in T is called the height of P .For f ∈ k h x i , the height of its minimal polynomial is called the height of f , and isdenoted by H( f ). The degree of f is the degree of the field extension k ( x ) −→ k ( x, f )or, equivalently, the degree of its minimal polynomial seen as a polynomial in T . Itis denoted by Deg( f ). Definition 1.3.
We define the following sets: • A ( d, h ) denotes the set of algebraic power series of degree ≤ d and height ≤ h . • A ( d, h ) denotes the set of algebraic series of A ( d, h ) that vanish at 0. • I ( d, h ) denotes the set of algebraic power series of A ( d, h ) whose minimalpolynomial satisfies the implicit function theorem.That is, f ∈ I ( d, h ), if and only if its minimal polynomial P ( x, T ) satisfies P (0 ,
0) = 0 and ∂P∂T (0 , = 0 . In particular we have I ( d, h ) ⊂ A ( d, h ) ⊂ A ( d, h ) for every d , h . Remark . It is straightforward to check that A ( d, h ) = A k × A ( d, h ) . INITENESS RESULTS CONCERNING ALGEBRAIC POWER SERIES 3
Remark . Let f be an algebraic power series with f (0) = 0 and assume thatthere is P ∈ k [ x, T ] such that P (0 ,
0) = 0, ∂P∂T (0 , = 0 and P ( x, f ) = 0. Thenthe minimal polynomial of f satisfies the implicit function theorem. Indeed sucha minimal polynomial is denoted by Q and should divide P : P = QR for somepolynomial R . Then ∂P∂T = ∂Q∂T R + Q ∂R∂T .
Since f (0) = 0 and Q ( x, f ( x )) = 0 then Q (0 ,
0) = 0. Hence (cid:16) ∂Q∂T R (cid:17) (0 , = 0 and ∂Q∂T (0 , = 0.2. Filtration of the ring of algebraic series by constructible sets
Definition 2.1.
Let k be a field and let d , h , e ∈ N . We define A ( d, h, e ) to be thesubset of algebraic power series f ∈ A ( d, h ) such that if P denotes the minimalpolynomial of f , we have ord (cid:18) ∂P∂T ( x, f ) (cid:19) = e. So we have I ( d, h ) = A ( d, h,
0) and A ( d, h ) = ∪ e ∈ N A ( d, h, e ). Indeed in positivecharacteristic the minimal polynomial of an algebraic power series is separable since k h x i is a separable extension of k [ x ]. This comes from the fact that k h x i is thehenselization of the local ring k [ x ] ( x ) . Moreover this union is finite by the followinglemma: Lemma 2.2. [HM17, Lemma 3] If e > dh then A ( d, h, e ) = ∅ . Lemma 2.3.
We have an injective map ϕ d,h,e : A ( d, h, e ) −→ A e k × I ( d, h + e ( d − defined as follows: let f ∈ A ( d, h, e ) and let us write f = f (0) + x e f (1) where f (0) is a polynomial in x of degree ≤ e and vanishing at 0, and f (1) is analgebraic power series vanishing at 0. Then f (1) ∈ I ( d, h + e ( d − and f (0) ∈ A e k by identifying the set of polynomials of degree ≤ e vanishing at 0 with A e k . Thereforewe define ϕ d,h,e ( f ) := ( f (0) , f (1) ) . Proof. let f ∈ A ( d, h, e ) and let P = a d T d + · · · + a T + a be its minimal polynomial. We can write f = f (0) + x e f (1) where f (0) is a polynomial in x of degree ≤ e and f (1) is an algebraic power seriesvanishing at 0.We have that P ( f (0) + Z ) = P ( f (0) ) + ∂P∂T ( f (0) ) Z + Q ( x, Z )where Q ( x, Z ) is a polynomial in x and Z , divisible by Z . Moreover if we write P ( f (0) + Z ) = b d Z d + · · · + b we have that deg( b i ) ≤ h + e ( d − i ) . Let us set R ( T ) := P ( f (0) + x e T ) x e = P ( f (0) ) x e + 1 x e ∂P∂T ( f (0) ) T + Q ( x, x e T ) x e . We have Q ( x,x e T ) x e ∈ k [ x, T ] since Q ( x, T ) is divisible by T . Moreover since f (0) − f ∈ ( x e +1 ) we have ∂P∂T ( f (0) ) − ∂P∂T ( f ) ∈ ( x e +1 ), and because ∂P∂T ( f ) is of order e then ∂P∂T ( f (0) ) has order e . Finally R ( f (1) ) = 0 so P ( f (0) ) x e ∈ k [ x ]. This proves that R ( T ) ∈ k [ x, T ].We have that R ( T ) = c d T d + · · · + c with deg( c i ) = deg( b i ) + ie − e ≤ h + e ( d − ∂R∂T (0 ,
0) = c (0) = 0. So f (1) is the only power series solution of R = 0vanishing at 0 by the implicit function theorem and f (1) ∈ I ( d, h + e ( d − . This construction gives an injective map A ( d, h, e ) −→ A e k × I ( d, h + e ( d − (cid:3) This proves that k h x i = [ e,d,h A e k × I ( d, h )where A e k is identified with polynomials in x of degree ≤ e vanishing at 0 and theinclusion A e k × I ( d, h ) ⊂ k h x i is given by( f, g ) f + x e g. Lemma 2.4.
We have A e k + I ( d, h ) ⊂ A ( d, h + ed ) for all integers d , h , e .Proof. Let e ∈ N and let us consider a polynomial R ( Z ) ∈ k [ x, Z ], of degree d ′ ≤ d ,such that ∂R∂Z (0 , = 0. Assume that deg x R ≤ h . By the implicit function theorem, R = 0 has a unique algebraic power series solution denoted by f (1) . Now let f (0) ∈ k [ x ] be a polynomial of degree ≤ e vanishing at 0. We set P ( T ) = x ed ′ R (cid:18) T − f (0) x e (cid:19) . Then P ( f (0) + x e f (1) ) = 0. Moreover if R ( T ) = c d ′ T d ′ + · · · + c then P ( T ) = c d ′ ( T − f (0) ) d ′ + c d ′ − x e ( T − f (0) ) d ′ − + · · · + c x d ′ e = a d ′ T d ′ + · · · + a wheredeg( a i ) ≤ max j ≥ i n deg( c j ) + e ( d ′ − j ) + deg( f (0) )( j − i ) o ≤ h + e ( d ′ − i ) . So f := f (0) + x e f (1) ∈ A ( d ′ , h + ed ′ ) ⊂ A ( d, h + ed ). (cid:3) INITENESS RESULTS CONCERNING ALGEBRAIC POWER SERIES 5
Lemma 2.5.
Let k be a field. Let f ∈ I ( d, h ) and let P = P i ≤ h,j ≤ d a i,j x i T j bea polynomial satisfying the implicit function theorem and vanishing at f . Let usconsider P d,h := X i ≤ h,j ≤ d A i,j x i T j where the A i,j are new indeterminates and A , is assumed to be 0. Then P d,h hasa unique power series solution f d,h ∈ ( x ) k (cid:28) A i,j A , , x (cid:29) where ( i, j ) runs over { , . . . , d } × { , . . . , h }\{ (0 , , (0 , } . Moreover, the coeffi-cients of the x k in the expansion of f d,h ( x ) are in k h A i,j A , i , and we have f d,h (cid:18) a i,j a , , x (cid:19) = f ( x ) . Proof.
The existence and the unicity of f d,h comes from the implicit function the-orem.First we prove that f d,h ∈ k h A i,j A , i [[ x ]]. Indeed we have that P ( d, h ) (cid:18) A i,j A , , x, (cid:19) ∈ ( x ) and ∂P∂T (cid:18) A i,j A , , x, (cid:19) / ∈ ( x ) . So by Hensel Lemma the root f d,h belongs to the completion of k h A i,j A , , x i withrespect to the ideal ( x ), i.e. f d,h ∈ k h A i,j A , i [[ x ]]. This implies that f d,h (cid:16) a i,j a , , x (cid:17) iswell defined.Finally we conclude that f d,h (cid:16) a i,j a , , x (cid:17) = f ( x ) since f d,h is the unique solution of P d,h = 0 vanishing at x = 0 and f is the unique solution of P = 0 vanishing at x = 0. (cid:3) Since there are ( d + 1)( h + 1) − dh + d + h − A i,j A , theproposition defines a surjective map f d,h : A dh + d + h − k −→ I ( d, h ) . This map is not injective since different polynomials can have the same power seriessolution.In fact let us identify the set of polynomials of k [ x, T ] (up to multiplication witha nonzero constant of k ) of degree in x less than h and of degree in T less than d with P ( d +1)( h +1) − k with homogeneous coordinates A i,j . Then A dh + d + h − k is iden-tified with the set of polynomials P ( x, T ) such that P (0 ,
0) = 0 and ∂P∂T (0 , = 0.Thus here A dh + d + h − k is the intersection of the affine open chart A , = 0 with thehypersurface A , = 0.We can state our first main result concerning the structure of I ( d, h ): Theorem 2.6.
Assume that k is algebraically closed. We have the following prop-erties: FUENSANTA AROCA, JULIE DECAUP, AND GUILLAUME ROND For every d ≥ and every h ≥ , there is an injective map ψ d,h : I ( d, h ) −→ A dh + d + h − k whose image is a constructible subset C d,h that contains a non empty opensubset of A dh + d + h − k .Here we identify A dh + d + h − k with the set of polynomials P ( x, t ) such that P (0 ,
0) = 0 , ∂P∂t (0 ,
0) = 1 , deg x ( P ) ≤ h, deg t ( P ) ≤ d, and the map ψ d,h is defined by identifying I ( d, h ) with the subset of irre-ducible polynomials in A dh + d + h − k .Moreover, we have f d,h ◦ ψ d,h = id I ( d,h ) . For every d ′ ≥ d , h ′ ≥ h , we denote by π (1) d ′ ,d,h ′ ,h : I ( d ′ , h ′ ) −→ I ( d, h ) andby π (2) d ′ ,d,h ′ ,h : A d ′ h ′ + d ′ + h ′ − k −→ A dh + d + h − k the canonical projection maps.Then we have that ψ d,h ◦ π (1) d ′ ,d,h ′ ,h = π (2) d ′ ,d,h ′ ,h ◦ ψ d ′ ,h ′ . Remark . We can see in the proof that the map ψ d,h is the map sending aseries of I ( d, h ) onto the vector of coefficients of its minimal polynomial, which isnormalized in the sense that the coefficient of x T is 1. Proof.
The map f d,h : A dh + d + h − k −→ I ( d, h ) is not injective since different poly-nomials may have the same root. To get an injective map, we need to restrictthe map to the set of irreducible polynomials. Indeed, by Remark 1.5, the minimalpolynomial of f ∈ I ( d, h ) satisfies the implicit function theorem. Therefore we haveto remove from A dh + d + h − k the points corresponding to the reducible polynomials.We can do it as follows:For every integers d , d , h , h with d + d ≤ d and h + h ≤ h set Q d ,h = X i ≤ h ,j ≤ d q i,j x i T j , R d ,h = X i ≤ h ,j ≤ d r i,j x i T j , for some variables q i,j and p i,j . Then the product Q d ,h R d ,h is a polynomial P = P i ≤ h,j ≤ d a i,j x i T j where the a i,j are polynomials in the q i,j and r i,j . Theproduct defines a rational mapΦ d ,d ,h ,h : A ( d +1)( h +1) k × A ( d +1)( h +1) k −→ A ( d +1)( h +1) k whose image can be identified with the polynomials P , with deg T ( P ) ≤ h + h and deg x ( P ) ≤ d + d , that are the product of 2 polynomials of degrees less than( h , d ) and ( h , d ). In fact it is straightforward to check that this map is definedby bi-homogeneous polynomials so it induces a rational map P k Φ d ,d ,h ,h : P ( d +1)( h +1) − k × P ( d +1)( h +1) − k −→ P ( d +1)( h +1) − k Here we want to consider only polynomials P whose constant term is zero. If sucha polynomial is the product of two polynomials Q and R then the constant term of Q or of R has to be zero. Thus here we set C d ,d ,h ,h = Im( P k Φ d ,d ,h ,h | P ( d h − k × P ( d h − k ) ∪∪ Im( P k Φ d ,d ,h ,h | P ( d h − k × P ( d h − k ) INITENESS RESULTS CONCERNING ALGEBRAIC POWER SERIES 7 and the intersection of this set with A dh + d + h − k corresponds exactly to the set ofpolynomials P with P (0 ,
0) = 0 and ∂P∂T (0 , = 0 that are the product of twopolynomials whose degrees are coordinatewise less than or equal to ( d , h ) and( d , h ).Let us remark that the dimension of P ( d +1)( h +1) − k × P ( d +1)( h +1) k is( d + 1)( h + 1) + ( d + 1)( h + 1) − . But we have that( d + d )( h + h ) + d + d + h + h − − (( d + 1)( h + 1) + ( d + 1)( h + 1) − d h + d h ≥ d + d < d or h + h < h then dim( C d ,d ,h ,h ) < ( d + 1)( h + 1).If d + d = d and h + h = h we have equality in (2.1) if and only if( d , h ) or ( d , h ) = 0 , ( d , d ) = 0 , or ( h , h ) = 0 . In the first case Q or R is a nonzero constant of k . In the second case we have that P = QR is of degree 0 in T and so it does not correspond to a point in A dh + d + h − k .In the last case P = QR is of degree 0 in x so its roots are in the algebraic closureof k , and the only algebraic power series vanishing at 0 which is in k is 0, and thiscase cannot occur if h ≥ C d ,d ,h ,h ) < dh + d + h − . So, when h ≥
1, we can identify I ( d, h ) with C d,h := A dh + d + h − k \ [ d ,d ,h ,h d d ≤ d,h h ≤ hd h d h > C d ,d ,h ,h which is a constructible set by Chevalley’s Theorem. Finally this former set con-tains an open subset of A dh + d + h − k by (2.2). This proves 1).Let d ′ ≥ d and h ′ ≥ h . From the construction of f d ′ ,h ′ and ψ d ′ ,h ′ , it is straightfor-ward to see that f d ′ ,h ′| A dh + d + h − k = f d,h and ψ d ′ ,h ′| A dh + d + h − k = ψ d,h . This proves 2). (cid:3)
Our second result describes the structure of A ( d, h ) as a constructible set. Theorem 2.8.
Assume that k is algebraically closed. Let d , h , and e ∈ N . Thenwe have the following properties:i) The image C d,h,e of the injective map (cid:0) Id A e k × ψ d,h + e ( d − (cid:1) ◦ ϕ d,h,e : A ( d, h, e ) −→ A ek × C d,h + e ( d − is constructible in A ek × C d,h + e ( d − . FUENSANTA AROCA, JULIE DECAUP, AND GUILLAUME ROND ii) The set A ( d, h ) can be identified with the constructible set [ e ≤ dh C d,h,e ⊂ A N ( d,h ) k for some N ( d, h ) ∈ N .iii) The dimension of this constructible set is dh + d + h − .Remark . In particular, the map (cid:0)(cid:0) Id A e k × ψ d,h + e ( d − (cid:1) ◦ ϕ d,h,e (cid:1) − : C d,h,e −→ A ( d, h, e )is a regular map in the sense that all the coefficients of the series f ∈ A ( d, h, e ) arepolynomial functions into the coordinates on A N ( d,h ) k (by Lemma 2.5) and are welldefined on C d,h,e . Remark . In the proof we show that we can choose N ( d, h ) = 2 dh ( d − d + 1) + 3 dh + d + h − . Proof of Theorem 2.8.
We do the following: let f (0) ∈ k e , f (1) ∈ I ( d, h + e ( d − f := f (0) + x e f (1) . We want to find necessary and sufficient conditions for f to be in A ( d, h, e ), that is we want to describe the image of the map ϕ d,h,e definedin Lemma 2.3.Let R ( Z ) ∈ k [ x, Z ] be a polynomial such thatdeg Z ( R ) ≤ d, deg x ( R ) ≤ h + e ( d − ,R ( f (1) ) = 0 , and ∂R∂Z (0 , = 0 . We assume that R ( Z ) is irreducible. For such a f ∈ A ( d, h, e ), we set P ( T ) = x ed R (cid:18) T − f (0) x e (cid:19) . Then P ( f (0) + x e f (1) ) = 0, i.e P vanishes at f . Moreover deg x ( P ) ≤ h + ed . Themap ( f (0) , R ( T )) P ( T ) := x ed R (cid:18) T − f (0) x e (cid:19) is a polynomial map from A e k × P ( d +1)( h + e ( d − − k to P ( h + ed +1)( d +1) − k .Let S ( T ) be an irreducible polynomial vanishing at f . Since the following fieldextensions degrees are equal:[ k (( x ))( f ) : k (( x ))] = [ k (( x ))( f (1) ) : k (( x ))] , we have deg T ( S ) = deg T ( R ). Since deg T ( P ) = deg T ( R ), there is a polynomial a ( x ) ∈ k [ x ] such that(2.3) P ( x, T ) = a ( x ) S ( x, T ) , (2.4) ∂P∂T ( x,
0) = a ( x ) ∂S∂T ( x, . Then, f ∈ A ( d, h, e ) if and only if S and R are irreducible, deg x ( S ) ≤ h andord (cid:0) ∂S∂T ( x, f (0) ) (cid:1) = e . INITENESS RESULTS CONCERNING ALGEBRAIC POWER SERIES 9
We denote by P d k (resp. P ( d +1)( d +1) − k ) the projective space of nonzero poly-nomials of k [ x ] (resp. k [ x, T ]) of degree ≤ d (resp. ≤ d in x and ≤ d in T )modulo multiplication by elements of k ∗ . For every d , d , d , e , h , we denote by V d ,d ,d,e,h the algebraic set of elements( a ( x ) , S ( x, T ) , f (0) , R ( T )) ∈ P d k × P ( d +1)( d +1) − k × A e k × P ( d +1)( h + e ( d − − k such that(2.5) a ( x ) S ( x, T ) − x ed R (cid:18) T − f (0) x e (cid:19) = 0 . Recall that f ∈ A ( d, h, e ) is identified with ( f (0) , R ( x, T )) ∈ A e k × P ( d +1)( h + e ( d − − k .Therefore f ∈ A ( d, h, e ) if and only if there is ( a ( x ) , S ( x, T )) ∈ P d k × P ( d +1)( d +1) − k ,for some d , d with d + d ≤ h + ed and d ≤ h , such that R ( x, T ) and S ( x, T ) are irreducible, R (0 ,
0) = 0 , ∂R∂T (0 , = 0 , ( a ( x ) , S ( x, T ) , f (0) ( x ) , R ( x, T )) ∈ V d ,d ,d,e,h and ord (cid:0) ∂S∂T ( x, f (0) ) (cid:1) = e. The set of irreducible polynomials S ( x, T ), with deg x ( S ) ≤ d , deg T ( S ) ≤ d , isa constructible subset C d ,d ⊂ P ( d +1)( d +1) − k as shown in the proof of Lemma2.6. Moreover, by Lemma 2.6, the set of irreducible polynomials R ( x, T ) such that R (0 ,
0) = 0, ∂R∂T (0 , = 0, deg x ( R ) ≤ h , deg T ( R ) ≤ d , is a constructible subset C d,h of P ( d +1)( h + e ( d − − k .The condition ord (cid:0) ∂S∂T ( x, f (0) ) (cid:1) = e defines a constructible set C ′ d ,d,e ⊂ P ( d +1)( d +1) − k × A e k . Now, we consider the following projections: π , : P d k × P ( d +1)( d +1) − k × A e k × P ( d +1)( h + e ( d − − k −→ A e k × P ( d +1)( h + e ( d − − k π , : P d k × P ( d +1)( d +1) − k × A e k × P ( d +1)( h + e ( d − − k −→ P ( d +1)( d +1) − k × A e k and we set E d ,d ,d,e,h := (cid:16) P d k × C ′ d ,d,e × P ( d +1)( h + e ( d − − k (cid:17) ∩ V d ,d ,d,e,h ∩ (cid:16) P d k × C d ,d × A e k × C d,h (cid:17) . Then we have that A ( d, h, e ) is equal to [ d d ≤ h + edd ≤ h π , ( E d ,d ,d,e,h )that is a constructible subset C d,h,e of A ek × A ( d +1)( h + e ( d − − k .We claim that π , | Ed ,d ,d,e,h is injective. Indeed, for ( f (0) , R ( x, T )) ∈ A ( d, h, e ),there is a unique couple ( a ( x ) , S ( x, T )) such that ( a ( x ) , S ( x, T ) , f (0) , R ( x, T )) ∈ E d ,d ,d,e,h , by (2.5) and because S ( x, T ) has to be irreducible. On the other hand, the fiber of π , over ( S ( x, T ) , f (0) ) ∈ C ′ d ,d,e ∩ (cid:16) C d ,d × A e k (cid:17) isfinite. Indeed, if there is ( a ( x ) , R ( x, T )) ∈ P d k × P ( d +1)( h + e ( d − − k such that a ( x ) S ( x, T ) − x ed R (cid:18) T − f (0) x e (cid:19) = 0 , then, f (0) + x e f (1) is a root of S ( x, T ) where f (1) is the unique solution of R ( x, T ) =0 vanishing at x = 0. Therefore π − , (( S, f (0) )) is finite because S ( x, T ) has a finitenumber of roots.Therefore the dimension of E d ,d ,d,e,h is equal to the dimension of its image under π , . First let us assume that ( S ( x, T ) , f (0) ) ∈ π , ( E d ,d ,d,e,h ). Indeed, by (2.5),we have that S ( x, f (0) + x e f (1) ) = 0 where f (1) is the unique solution of somepolynomial equation R ( x, T ) = 0 with R (0 ,
0) = 0 and ∂R∂T (0 , = 0. Therefore S (0 ,
0) = 0. We denote by C d ,d the set of polynomials S such that S (0 ,
0) = 0.Then π , ( E d ,d ,d,e,h ) is included in C ′ d ,d,e ∩ ( C d ,d × A e k ).We are going to bound the dimension of C ′ d ,d,e ∩ ( C d ,d × A e k ) as follows:We denote by F k the coefficient of x k in the expansion of f (0) , for 1 ≤ k ≤ e .We denote by S i,j the coefficient of x i T j in S ( x, T ), for 0 ≤ i ≤ d and 0 ≤ j ≤ d . The coefficient of x l in S ( x, f (0) ) is a polynomial G l with integer coefficientsdepending on the indeterminates F k for k ≤ l , and S i,j for i ≤ l −
1. Therefore C ′ d ,d,e ∩ ( C d ,d × A e k ) is defined by the equations : S , = 0 G l ( F k , S i,j ) k ≤ l,i ≤ l = 0 , l < eG e ( F k , S i,j ) k ≤ e,i ≤ e − = 0and the ideal defining the Zariski closure of C ′ d ,d,e ∩ ( C d ,d × A e k ) is the radical idealof the ideal generated by S , and the G l for l < e . For any p < e , we set I p := h S , , G , . . . , G p i and I − = h S , i . These are ideals depending only on the following indeterminates (if p ≥ F k , for k ≤ p, and S i,j for i ≤ p. Moreover G p has the form G p = S p, + e G p ( F k , S i,j ) k ≤ p,i ≤ p − . Therefore we have, for all p ≤ e :ht( I p − ) ≤ ht( I p ∩ Q [ F k , S i,j ] k ≤ p − ,i ≤ p − ) < ht( I p ) . In particular we have that ht( I e ) ≥ e + 2. Hencedim( C ′ d ,d,e ∩ ( C d ,d × A e k )) ≤ [ e + ( d + 1)( d + 1) − − e − ≤ ( d + 1)( h + 1) − . Therefore dim( A ( d, h, e )) < ( d + 1)( h + 1) − . Moreover, when e = 0, we know that A ( d, h,
0) = I ( d, h ) and by Theorem 2.6, thisset is a constructible set of dimension ( d + 1)( h + 1) −
2. Thus, A ( d, h ) = [ e ≤ dh A ( d, h, e ) INITENESS RESULTS CONCERNING ALGEBRAIC POWER SERIES 11 is a constructible set of dimension equal to ( d + 1)( h + 1) − (cid:3) Points of bounded complexity in varieties over k (( x ))Let E be a subset of k (( x )) n . For all non negative integers d and h , we set E d,h := E ∩ A ( d, h ) n and we denote by n d,h ( E ) the dimension of the Zariski closure of E d,h in the affinespace A nN ( d,h ) k where N ( d, h ) is given in Remark 2.10. Theorem 3.1.
Let k be an algebraically closed field and let X be an algebraicsubvariety of A n k (( x )) of dimension m . Then n d,h ( X ) ≤ m ( dh + d + h ) . Remark . This result is an analogue of Lemma [CCL15, 5.1.1] (that gives thesame kind of bound on the dimension of the polynomial solutions of degree ≤ d ).The proof is based on the use of linear projections. But, on the contrary of thepolynomials of degree ≤ d , the difficulty comes from the fact that the sets A ( d, h )are not stable by linear change of coordinates with coefficients in k (since these setsare not k -vector spaces). Proof of Theorem 3.1.
We will prove the statement by induction on m . If m = 0,then X is finite and n ( d, h ) = 0.Now assume that the statement is true for every integer less than m and let X be of dimension m . For every E ⊂ { , . . . , n } with Card( E ) = m , we denote by π : A n −→ A m the projection defined by π E ( x , . . . , x n ) := ( x i ) i ∈ E . For such aset E , we define B E to be the subset of points of π E ( X ) where π E | X is not finite.Then π E ( X ) \ B E contains an open set of the Zariski closure of π E ( X ) by the uppersemi-continuity of the dimension of the fibers of π E .Therefore the dimension of the Zariski closure of the set of A ( d, h )-points of X \ π − E ( B E ) has dimension less or equal to dim( A ( d, h ) m ), which is equal to m ( dh + d + h ) by Remark 1.4 and Theorem 2.8. Thus we can replace X by X ∩ π − E ( B E ).We repeat this operation for every E as above and we assume that X ⊂ \ E ⊂{ ,...,n } , Card( E )= m π − E ( B E ) . By considering the different irreducible components of the B E separately, we mayassume that all the B E are irreducible. Therefore the ideal of π − E ( B E ) is a primeideal J E generated by polynomials depending only on the x i for i ∈ E .We fix such a set E that we denote by E . We also fix an index i ∈ E suchthat J E is generated by polynomials of K [ x i , i ∈ E ] but not by polynomials of K [ x i , i ∈ E \ { i } ].Then we pick E ⊂ { , . . . , n } \ { i } such that Card( E ) = m , and we fix an index i ∈ E such that J E is generated by polynomials of K [ x i , i ∈ E ] but not bypolynomials of K [ x i , i ∈ E \ { i } ].We repeat this process and construct a sequence E , . . . , E n − m +1 ⊂ { , . . . , n } of subsets of cardinal m , and a sequence i ∈ E \ ( ∪ n − m +1 i =2 E i ) , i ∈ E \ ( ∪ n − m +1 i =3 E i ) , . . . , i n − m +1 ∈ E n − m +12 FUENSANTA AROCA, JULIE DECAUP, AND GUILLAUME ROND such that J E k is generated by polynomials of K [ x i , i ∈ E k ] but not by polynomialsof K [ x i , i ∈ E k \ { i k } ]. Hence, by Krull’s principal ideal theorem we haveht (cid:0) J E + J E + . . . + J E n − m +1 (cid:1) ≥ n − m + 1 . Equivalently we have dim( ∩ n − m +1 i =1 π − E i ( B E i )) ≤ m − . Therefore, the result follows by induction. (cid:3)
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Ann. Sci. ´Ec. Norm. Sup´er. , (2013), 963-1004.[CCL15] R. Cluckers, G. Comte, F Loeser, Non-Archimedean Yomdin-Gromov parametrizationsand points of bounded height, Forum Math. Pi , , (2015), e5, 60 pp.[FS98] P. Flajolet and M. Soria, Coefficients of algebraic series, Algorithms seminar 1997-1998,Tech. Report, INRIA, 27-30.[HM17] M. Hickel, M. Matusinski, On the algebraicity of Puiseux series, Rev. Mat. Complut. , ,(2017), no. 3, 589-620.[Ro15] G. Rond, Local zero estimates and effective division in rings of algebraic power series, J.Reine Angew. Math. , , (2018), 111-160. E-mail address : [email protected] Instituto de Matem´aticas, Universidad Nacional Aut´onoma de M´exico (UNAM), Mex-ico
E-mail address : [email protected] Instituto de Matem´aticas, Universidad Nacional Aut´onoma de M´exico (UNAM), Mex-ico
E-mail address : [email protected]@univ-amu.fr