A geometric framework for the subfield problem of generic polynomials via Tschirnhausen transformation
aa r X i v : . [ m a t h . N T ] O c t A GEOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEMOF GENERIC POLYNOMIALS VIA TSCHIRNHAUSENTRANSFORMATION
AKINARI HOSHI AND KATSUYA MIYAKE
Abstract.
Let k be an arbitrary field. We study a general method to solvethe subfield problem of generic polynomials for the symmetric groups over k viaTschirnhausen transformation. Based on the general result in the former part,we give an explicit solution to the field isomorphism problem and the subfieldproblem of cubic generic polynomials for S and C over k . As an applicationof the cubic case, we also give several sextic generic polynomials over k . Introduction
Let k be a fixed base field of arbitrary characteristic and G a finite group. Let k ( t ) be the rational function field over k with m variables t = ( t , . . . , t m ). Apolynomial F ( t , . . . , t m ; X ) ∈ k ( t )[ X ] is called k -generic for G if the Galois groupof F ( t ; X ) over k ( t ) is isomorphic to G and every G -Galois extension L/M with M ⊃ k and M = ∞ can be obtained as L = Spl M F ( a ; X ), the splitting fieldof F ( a ; X ) over M , for some a = ( a , . . . , a m ) ∈ M m . By Kemper’s Theorem[Kem01], furthermore, every H -Galois extension for a subgroup H of G over aninfinite field M is also given by a specialization of F ( t ; X ). Examples of genericpolynomials for various G are found, for example, in [JLY02]. We also give severalsextic k -generic polynomials in Section 6. The fact that a k -generic polynomial for G covers all H -Galois extensions ( H ⊂ G ) over M ⊃ k by specializing parametersnaturally raises a problem; namely, Subfield problem of a generic polynomial.
Let F ( t ; X ) be a k -generic poly-nomial for G . For a field M ⊃ k and a , b ∈ M m , determine whether Spl M F ( b ; X ) is a subfield of Spl M F ( a ; X ) or not. When we restrict ourselves to G -Galois extensions over M for a fixed group G ,we are to consider a special case of the subfield problem: Field isomorphism problem of a generic polynomial.
Determine whether
Spl M F ( a ; X ) and Spl M F ( b ; X ) are isomorphic over M or not for a , b ∈ M m . Mathematics Subject Classification.
Primary 11R16, 12E25, 12F10, 12F12, 12F20.This work was partially supported by Grant-in-Aid for Scientific Research (C) 19540057 ofJapan Society for the Promotion of Science.
In this paper, we develop a method to solve the field isomorphism problem of k -generic polynomial for the symmetric group S n of degree n via Tschirnhausentransformation.In Section 2, we investigate a geometric interpretation of Tschirnhausen trans-formations. Our main idea, which we describe here briefly, is as follows: Let f ( X )and g ( X ) be monic separable polynomials of degree n in k [ X ] for a field k withroots α , . . . , α n and β , . . . , β n , respectively, in a fixed algebraic closure of k . Apolynomial g ( X ) ∈ k [ X ] is called a Tschirnhausen transformation of f ( X ) over M ( ⊃ k ) if g ( X ) is of the form g ( X ) = n Y i =1 (cid:0) X − ( c + c α i + · · · + c n − α n − i ) (cid:1) , c i ∈ M. (1)Two polynomials f ( X ) and g ( X ) in k [ X ] are Tschirnhausen equivalent over M ifthey are Tschirnhausen transformations over M of each other. For two irreducibleseparable polynomials f ( X ), g ( X ) ∈ k [ X ], the following two conditions are equiv-alent: (i) f ( X ) and g ( X ) are Tschirnhausen equivalent over M ; (ii) the quotientfields M [ X ] / ( f ( X )) and M [ X ] / ( g ( X )) are isomorphic over M .Now we replace the roots α = ( α , . . . , α n ) and β = ( β , . . . , β n ) by independentvariables x = ( x , . . . , x n ) and y = ( y , . . . , y n ) respectively. If we take u i := u i ( x , y ) ∈ M [ x , y , ∆ − s ] as u i ( x , y ) = ∆ − s · det x · · · x i − y x i +11 · · · x n − x · · · x i − y x i +12 · · · x n − ... ... ... ... ... ...1 x n · · · x i − n y n x i +1 n · · · x n − n where ∆ s = Q ≤ i 1, the following basic properties:(i) ( L s L t ) g − Hg = K ( u g , . . . , u gn − ) = K ( u gi ) for g ∈ G s , t , (ii) L s ∩ K ( u gi ) = L t ∩ K ( u gi ) = K for g ∈ G s , t , (iii) L s L t = L s ( u gi ) = L t ( u gi ) for g ∈ G s , t , (iv) L s L t = K ( u gi | g ∈ H \ G s , t ) . In Section 3, we study a specialization of parameters ( s , t ) ( a , b ) ∈ M n × M n of polynomials f n ( s ; X ) = Q ni =1 ( X − x i ) and f n ( t ; X ) = Q ni =1 ( X − y i ). We always EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 3 assume that such a specialization ( s , t ) ( a , b ) ∈ M n × M n satisfy the condition∆ a · ∆ b = 0, i.e. both of f n ( a ; X ) ∈ M [ X ] and f n ( b ; X ) ∈ M [ X ] are separable.For i , 0 ≤ i ≤ n − 1, we take a polynomial F i ( a , b ; X ) := Y g ∈ H \ S n × S n ( X − u gi ( α , β )) = Y g ∈ H \ S n × S n ( X − c gi ) ∈ M [ X ] , of degree n !. Therefore we obtain the following: Theorem (Theorem 3.8) . For a fixed j, ≤ j ≤ n − , and a , b ∈ M n with ∆ a · ∆ b = 0 , assume that the polynomial F j ( a , b ; X ) has no multiple root. Then F j ( a , b ; X ) has a root in M if and only if M [ X ] / ( f n ( a ; X )) and M [ X ] / ( f n ( b ; X )) are M -isomorphic. Corollary (Corollary 3.9) . Let j and a , b ∈ M n be as above. Assume that both of Gal( f n ( a ; X ) /M ) and Gal( f n ( b ; X ) /M ) are isomorphic to a transitive subgroup G of S n and that all subgroups of G with index n are conjugate in G . Then F j ( a , b ; X ) has a root in M if and only if Spl M f n ( a ; X ) and Spl M f n ( b ; X ) coincide. In Sections 4 and 5, based on the general result in Sections 2 and 3, we givean explicit solution to the field isomorphism problem and the subfield problem of k -generic polynomials for S and for C . Here we display some results to k -genericpolynomials g S ( s ; X ) := X + sX + s for S and g C ( s ; X ) := X − sX − ( s +3) X − C .For g S ( s ; X ), we take coefficients c , c , c of a Tschirnhausen transformationfrom g S ( a ; X ) to g S ( b ; X ) and we put u := 3 c /c . Then we get H ( a, b ; X ) :=( a − b ) · Q g ∈ H \ S n × S n ( X − u g ) where H ( a, b ; X ) = a ( X + 9 X − a ) − b ( X − aX − aX − a − a ) . Theorem (Theorem 4.10) . Assume that char k = 3 . For a, b ∈ M with a = b ,the decomposition type of irreducible factors h µ ( X ) of H ( a, b ; X ) over M gives ananswer to the subfield problem of X + sX + s as on Table in Section 4. Inparticular, two splitting fields of X + aX + a and of X + bX + b over M coincideif and only if there exists u ∈ M such that b = a ( u + 9 u − a ) ( u − au − au − a − a ) . For g C ( s ; X ), we obtain the following theorem which is an analogue to the resultsof Morton [Mor94] and Chapman [Cha96]. Theorem (Theorem 5.4) . Assume that char k = 2 . For m, n ∈ M , two splittingfields of X − mX − ( m + 3) X − and of X − nX − ( n + 3) X − over M coincide if and only if there exists z ∈ M such that either n = m ( z − z − − z ( z + 1) mz ( z + 1) + z + 3 z − or n = − m ( z + 3 z − 1) + 3( z − z − mz ( z + 1) + z + 3 z − . AKINARI HOSHI AND KATSUYA MIYAKE By applying Hilbert’s irreducibility theorem (cf. for example [JLY02, Chapter3]) and Siegel’s theorem for curves of genus 0 (cf. [Lan78, Theorem 6.1], [Lan83,Chapter 8, Section 5], [HS00, Theorem D.8.4]) to the preceding theorems respec-tively, we get the following corollaries: Corollary (Corollary 4.11 and Corollary 5.5) . Let g G ( a ; X ) = X + aX + a ( resp. X − aX − ( a + 3) X − be as above with given a ∈ M , and suppose that M ⊃ k is Hilbertian ( e.g. a number field ) . Then there exist infinitely many b ∈ M suchthat Spl M g G ( a ; X ) = Spl M g G ( b ; X ) . Corollary (Corollary 4.12 and Corollary 5.6) . Let M be a number field and O M thering of integers in M . For g G ( a ; X ) = X + aX + a ( resp. X − aX − ( a +3) X − asabove with a given integer a ∈ O M , there exist only finitely many integers b ∈ O M such that Spl M g G ( a ; X ) = Spl M g G ( b ; X ) . In Section 6, as an application of the cubic case, we also give several sextic k -generic polynomials.The calculations in this paper were carried out with Mathematica [Wol03].2. Tschirnhausen transformation (geometric interpretation) Let n ≥ x = ( x , . . . , x n ) and y = ( y , . . . , y n ) be2 n independent variables over k . Let f n ( s ; X ) = f n ( s , . . . , s n ; X ) ∈ k ( s )[ X ] (resp. f n ( t ; X ) = f n ( t , . . . , t n ; X ) ∈ k ( t )[ X ]) be a monic polynomial of degree n whoseroots are x , . . . , x n (resp. y , . . . , y n ). Then we have f n ( s ; X ) = n Y i =1 ( X − x i ) = X n − s X n − + s X n − + · · · + ( − n s n ,f n ( t ; X ) = n Y i =1 ( X − y i ) = X n − t X n − + t X n − + · · · + ( − n t n , where s i (resp. t i ) is the i -th elementary symmetric function in n variables x , . . . , x n (resp. y , . . . , y n ). We put K := k ( s , t );it is naturally regarded as the rational function field over k with 2 n variables. Weput L s := Spl K f n ( s ; X ) = K ( x , . . . , x n ) ,L t := Spl K f n ( t ; X ) = K ( y , . . . , y n ) . Then we have L s ∩ L t = K and L s L t = k ( x , y ). The field extension k ( x , y ) /K isa Galois extension whose Galois group is isomorphic to S n × S n , where S n is the EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 5 symmetric group of degree n . Put G s := Gal( L s L t /L t ) , G t := Gal( L s L t /L s )and G s , t := G s × G t . Then we have G s , t ∼ = Gal( L s L t /K ) which acts on k ( x , y ) from the right. For g = ( σ, τ ) ∈ G s , t , we take anti-isomorphisms ϕ : G s → S n , σ ϕ ( σ ) ,ψ : G t → S n , τ ψ ( τ ) , and we regard the group G s , t as S n × S n by the rule x σi = x ϕ ( σ )( i ) , y σi = y i , x τi = x i , y τi = y ψ ( τ )( i ) , ( i = 1 , . . . , n ) . In a fixed algebraic closure of K , there exist n ! Tschirnhausen transformationsfrom f n ( s ; X ) to f n ( t ; X ). We first study the field of definition of Tschirnhausentransformations from f n ( s ; X ) to f n ( t ; X ). Let D := x x · · · x n − x x · · · x n − ... ... ... . . . ...1 x n x n · · · x n − n be a so-called Vandermonde matrix of size n . The matrix D is invertible becausedet D = ∆ s , where ∆ s := Y ≤ i The Galois group G s , t acts on the orbit { u ( σ,τ ) i | ( σ, τ ) ∈ G s , t } via regular represen-tation from the right. However this action is not faithful. We put H := { ( σ, τ ) ∈ G s , t | ϕ ( σ ) = ψ ( τ ) } ∼ = S n . Let g = Hg be a right coset of H in G s , t . If ( σ, τ ) ∈ H then we have u ( σ,τ ) i = u i for i = 0 , . . . , n − G s , t acts on the set { u gi | g ∈ H \ G s , t } transitively from the right through the action on the set H \ G s , t of right cosets.We see that the set { (1 , τ ) | (1 , τ ) ∈ G s , t } (resp. { ( σ, | ( σ, ∈ G s , t } ) form acomplete residue system of H \ G s , t . Indeed, for g = ( σ, τ ) ∈ G s , t , g = H ( σ, τ ′ ) − g = H (1 , ( τ ′ ) − τ ) if ϕ ( σ ) = ψ ( τ ′ ). Lemma 2.1. For ( σ, τ ) ∈ G s , t and i , ≤ i ≤ n − , we have u ( σ,τ ) i = u i if andonly if ϕ ( σ ) = ψ ( τ ) ; that is, H = Stab G s , t ( u i ) , the stabilizer of u i in G s , t .Proof. Let A i,j be the ( i, j )-cofactor of the matrix D . Thus we have the cofactorexpansion, u i ( x , y ) = ∆ − s n X j =1 A j,i +1 y j , ( i = 0 , . . . , n − . Hence for each i, ≤ i ≤ n − 1, we see u i ( x , y ) ( σ, = ε ( σ ) ∆ − s n X j =1 A ( σ, j,i +1 y j = ε ( σ ) ∆ − s n X j =1 ε ( σ ) A ϕ ( σ )( j ) ,i +1 y j where ε ( σ ) is the signature of ϕ ( σ ) ∈ S n , and also u i ( x , y ) (1 ,τ − ) = ∆ − s n X j =1 A j,i +1 y τ − j = ∆ − s n X j =1 A j,i +1 y ψ ( τ − )( j ) = ∆ − s n X j =1 A ψ ( τ )( j ) ,i +1 y j . Therefore we have u i ( x , y ) ( σ,τ ) = u i ( x , y ) if and only if A ϕ ( σ )( j ) ,i +1 = A ψ ( τ )( j ) ,i +1 for j = 1 , . . . , n because the variables y , . . . , y n are linearly independent over L s .Since A j,i +1 ∈ k [ x , . . . , x j − , x j +1 , . . . , x n ], we finally have u i ( x , y ) ( σ,τ ) = u i ( x , y ) ifand only if σ = τ . (cid:3) Hence, in particular, we see that H \ G s , t = n ! and that the subgroups G s and G t of G s , t act on the set { u gi | g ∈ H \ G s , t } transitively.For g = (1 , τ ), we obtain the following equality from the definition (2): y ψ ( τ )( i ) = u g + u g x i + · · · + u gn − x n − i for i = 1 , . . . , n. This means that the set { ( u g , . . . , u gn − ) | g ∈ H \ G s , t } gives coefficients of n !different Tschirnhausen transformations from f n ( s ; X ) to f n ( t ; X ) each of which isrespectively defined over K ( u g , . . . , u gn − ). EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 7 Definition. For each g ∈ G s , t , we call K ( u g , . . . , u gn − ) a field of Tschirnhausencoefficients from f n ( s ; X ) to f n ( t ; X ).We put v i ( x , y ) := u i ( y , x ) , for i = 0 , . . . , n − . We write v i = v i ( x , y ) for simplicity. Then K ( v g , . . . , v gn − ) gives a field of Tschirn-hausen coefficients from f n ( t ; X ) to f n ( s ; X ). We obtain Proposition 2.2. We have ( L s L t ) g − Hg = K ( u g , . . . , u gn − ) = K ( v g , . . . , v gn − ) = K ( u gi ) = K ( v gi ) and [ K ( u gi ) : K ] = n ! for each i, ≤ i ≤ n − , and for each g ∈ G s , t .Proof. We have L s L t ⊃ ( L s L t ) g − Hg ⊃ K ( u g , . . . , u gn − ) ⊃ K ( u gi ) ⊃ K, || || || L s L t ⊃ ( L s L t ) g − Hg ⊃ K ( v g , . . . , v gn − ) ⊃ K ( v gi ) ⊃ K. Hence the assertion follows from Stab G s , t ( u gi ) = Stab G s , t ( v gi ) = g − Hg . (cid:3) Corollary 2.3. We have Spl K ( u gi ) f n ( s ; X ) = Spl K ( u gi ) f n ( t ; X ) for every g ∈ G s , t .Proof. The polynomials f n ( s ; X ) and f n ( t ; X ) are Tschirnhausen equivalent over K ( u g , . . . , u gn − ) = K ( u gi ) = K ( v gi ). Hence the quotient fields K ( u gi )[ X ] / ( f n ( s ; X ))and K ( u gi )[ X ] / ( f n ( s ; X )) are isomorphic over K ( u gi ). (cid:3) Proposition 2.4. We have (i) L s ∩ K ( u gi ) = L t ∩ K ( u gi ) = K for g ∈ G s , t ;(ii) L s L t = L s ( u gi ) = L t ( u gi ) for g ∈ G s , t . Proof. (i) We should show that ( g − Hg ) G t = ( g − Hg ) G s = G s , t . We may suppose g = (1 , τ ) without loss of generality. Then, for any ( σ ′ , τ ′ ) ∈ H , there existsan element ( σ ′ , τ − τ ′ τ )( σ ′− , 1) = (1 , τ − τ ′ τ ) in ( g − Hg ) G s . Hence the equality( g − Hg ) G s = G s , t follows from { }× G t ⊂ ( g − Hg ) G s . The assertion ( g − Hg ) G t = G s , t is obtained by a similar way because we may replace g by an element of theform ( σ, g − Hg ∩ G s = g − Hg ∩ G t = { } . An element c in g − Hg canbe described as c = ( σ ′ , τ − τ ′ τ ) where ϕ ( σ ′ ) = ψ ( τ ′ ). If c ∈ G s , then we obtain c = ( σ ′ , τ − τ ′ τ ) = ( σ ′ , ∈ G s × { } . Then ϕ ( σ ′ ) = ψ ( τ ′ ) = 1, and hence c = (1 , G s , t . If c ∈ G t , then we have σ ′ = 1. Hence follows c = (1 , ∈ G s , t . (cid:3) Moreover we obtain Proposition 2.5. L s L t = K ( u gi | g ∈ H \ G s , t ) for every i, ≤ i ≤ n − . AKINARI HOSHI AND KATSUYA MIYAKE Proof. We should show that T g ∈ H \ G s , t g − Hg = { } because Stab G s , t ( u gi ) = g − Hg .Suppose ( σ ′ , τ ′ ) ∈ T g ∈ H \ G s , t g − Hg . From H \ G s , t = { (1 , τ ) | (1 , τ ) ∈ G s , t } , wehave ϕ ( σ ′ ) = ψ ( τ − τ ′ τ ) for every τ ∈ G t . Since ϕ ( σ ′ ) = ψ ( τ ) ψ ( τ ′ ) ψ ( τ ) − = ψ ( τ ) ϕ ( σ ′ ) ψ ( τ ) − , we see that ϕ ( σ ′ ) is in the center of the symmetric group S n .Thus we have ( σ ′ , τ ′ ) = (1 , 1) because the center of S n is trivial. (cid:3) We define a polynomial of degree n ! by F i ( s , t ; X ) := Y g ∈ H \ G s , t ( X − u gi ) ∈ K [ X ] , ( i = 0 , . . . , n − . We see from Proposition 2.2 that F i ( s , t ; X ) , ( i = 0 , . . . , n − k ( s , t ). From Proposition 2.5, furthermore, we have the following theorem: Theorem 2.6. The polynomial F i ( s , t ; X ) ∈ k ( s , t )[ X ] is k -generic for S n × S n .Proof. The assertion follows from Spl K F i ( s , t ; X ) = K ( u gi | g ∈ H \ G s , t ) = L s L t and the S n -genericness of f n ( s ; X ) and f n ( t ; X ). (cid:3) In case of char k = 2, we have k ( s )(∆ s ) = k ( s ) because ∆ s ∈ k ( s ). Hence, weuse the results of Berlekamp [Ber76] and take the Berlekamp discriminant β s := X i If char k = 2 ( resp. char k = 2) , the polynomial F i ( s , t ; X ) splitsinto two irreducible factors F + i ( X ) and F − i ( X ) of degree n ! / over the quadraticextension K (∆ s / ∆ t ) ( resp. K ( β s + β t )) of K .Proof. Let G + s (resp. G + t ) be the subgroup of G s (resp. G t ) with index two whichis isomorphic to the alternating group A n of degree n . If char k = 2, we havea sequence of subgroups G + s × G + t ⊂ Stab G s × G t (∆ s / ∆ t ) ⊂ G s , t each of whoseindices is two. It follows from H ⊂ Stab G s , t (∆ s / ∆ t ) that F + i ( X ) | F i ( X ) and F + i ( X ) ∈ K (∆ s / ∆ t ). The case of char k = 2 may easily be obtained by a similarmanner. (cid:3) EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 9 Specialization of parameters into an infinite field Let M ( ⊃ k ) be an infinite field. We assume that we always take a specializationof parameters ( s , t ) ( a , b ) ∈ M n × M n so that both of the polynomials f n ( a ; X )and f n ( b ; X ) are separable over M , (i.e. ∆ a = 0 and ∆ b = 0). Put L a =Spl M f n ( a ; X ) and L b = Spl M f n ( b ; X ); these splitting fields are taken in a fixedalgebraic closure of M . We denote the Galois groups of f n ( a ; X ) and f n ( b ; X ) over M by G a and G b respectively, that is, G a = Gal( L a /M ) and G b = Gal( L b /M ).We put G a , b := Gal( L a L b /M ). Let α := ( α , . . . , α n ) (resp. β := ( β , . . . , β n ))be the roots of f n ( a ; X ) (resp. f n ( b ; X )) in the algebraic closure of M . Oncewe fix the orders of the roots as α and β , then each element of G a , b inducessubstitutions on the two sets of indices. Then we may identify G a , b as a subgroupof G s , t . More precisely, we express each element h ∈ G a , b as h = ( σ, τ ) ∈ G s , t through the conditions, α hi = α ϕ ( σ )( i ) and β hi = β ψ ( τ )( i ) for i = 1 , . . . , n . We put for g = ( σ, τ ) ∈ G s , t ( c g , . . . , c gn − ) := ( u g ( α , β ) , . . . , u gn − ( α , β )) , (4) ( d g , . . . , d gn − ) := ( u g ( β , α ) , . . . , u gn − ( β , α )) . Then we have β ψ ( τ )( i ) = c g + c g α ϕ ( σ )( i ) + · · · + c gn − α n − ϕ ( σ )( i ) , (5) α ϕ ( σ )( i ) = d g + d g β ψ ( τ )( i ) + · · · + d gn − β n − ψ ( τ )( i ) (6)for each i = 1 , . . . , n . For each g ∈ G s , t , there exists a Tschirnhausen trans-formation from f n ( a ; X ) to f n ( b ; X ) over its field of Tschirnhausen coefficients M ( c g , . . . , c gn − ), and the n -tuple ( d g , . . . , d gn − ) gives the coefficients of a transfor-mation of the inverse direction.From the assumption ∆ a · ∆ b = 0, we first see Lemma 3.1. Let M ′ /M be a field extension. If f n ( b ; X ) is a Tschirnhausen trans-formation of f n ( a ; X ) over M ′ , then f n ( a ; X ) is a Tschirnhausen transformationof f n ( b ; X ) over M ′ . In particular, we have M ( c g , . . . , c gn − ) = M ( d g , . . . , d gn − ) forevery g = ( σ, τ ) ∈ G s , t .Proof. Suppose M ′ contains M ( c g , . . . , c gn − ). We will take a ring homomorphism ρ : M ′ [ X ] / ( f n ( b ; X )) −→ M ′ [ X ] / ( f n ( a ; X )) , X c g + c g X + · · · + c gn − X n − and show that the map ρ is an isomorphism over M ′ .We first assume that f n ( a ; X ) (resp. f n ( b ; X )) splits into irreducible factors g µ ( X ) (resp. h ν ( X )) for 1 ≤ µ ≤ m (resp. 1 ≤ ν ≤ m ′ ) over M ′ . Since f n ( a ; X )and f n ( b ; X ) have no multiple root, the quotient algebras M ′ [ X ] / ( f n ( a ; X )) and M ′ [ X ] / ( f n ( b ; X )) are semi-simple and direct sums of the fields M ′ [ X ] / ( g µ ( X )) and M ′ [ X ] / ( h ν ( X )), respectively. We put L ′ = Spl M ′ f n ( a ; X ) Spl M ′ f n ( b ; X ). Fix i, ≤ i ≤ n − 1, and suppose that α ϕ ( σ )( i ) and β ψ ( τ )( i ) are roots of the irreducible factors g µ ( X ) and h ν ( X ), respectively. Then we have embeddings Φ i : M ′ [ X ] / ( g µ ( X )) → L ′ with Φ i ( X ) = α ϕ ( σ )( i ) and Ψ i : M ′ [ X ] / ( h ν ( X )) → L ′ with Ψ i ( X ) = β ψ ( τ )( i ) . Bythe equality (5) we seeΨ i ( X mod( h ν ( X ))) = Φ i ( c g + c g X + · · · + c gn − X n − mod( g µ ( X ))) . Thus we get an injective homomorphism from M ′ [ X ] / ( h ν ( X )) to M ′ [ X ] / ( g µ ( X ))by assigning c g + c g X + · · · + c gn − X n − mod( g µ ( X )) to X mod( h ν ( X )). Since thequotient algebras M ′ [ X ] / ( f n ( a ; X )) and M ′ [ X ] / ( f n ( b ; X )) are direct sums of thefields M ′ [ X ] / ( g µ ( X )) and M ′ [ X ] / ( h ν ( X )), respectively, the homomorphism ρ asabove is well defined and injective. Then this has to be an isomorphism over M ′ because the dimensions of the two algebras over M ′ coincide. Therefore, in partic-ular, the corresponding fields M ′ [ X ] / ( g µ ( X )) and M ′ [ X ] / ( h ν ( X )) are isomorphicover M ′ . In particular, each irreducible factor g µ ( X ) of f n ( a ; X ) corresponds to anirreducible factor h ν ( X ) of f n ( b ; X ) in a one-to-one manner. The degrees of thetwo factors are equal. Hence we also see that the number of irreducible factors of f n ( a ; X ) is same as that of f n ( b ; X ), that is, m = m ′ .We take the inverse isomorphism from M ′ [ X ] / ( f n ( a ; X )) to M ′ [ X ] / ( f n ( b ; X ))over M ′ . Then the image of X mod( f n ( a ; X )) is expressed as e + e X + · · · + e n − X n − mod( f n ( b ; X )) with e , . . . , e n − ∈ M ′ . The homomorphisms Φ i and Ψ i now give us α ϕ ( σ )( i ) = e + e β ψ ( τ )( i ) + · · · + e n − β n − ψ ( τ )( i ) for i = 1 , . . . , n . Since ∆ b = 0, therefore, these equalities together with (6) show d gj = e j ∈ M ′ for j = 0 , . . . , n − M ′ = M ( c g , . . . , c gn − ) then we see M ( c g , . . . , c gn − ) ⊃ M ( d g , . . . , d gn − ). Conversely if we take M ′ = M ( d g , . . . , d gn − ) then we have M ( d g , . . . , d gn − ) ⊃ M ( c g , . . . , c gn − ). (cid:3) Our main idea of this paper is to study the behavior of the field M ( c g , . . . , c gn − )of Tschirnhausen coefficients from f n ( a ; X ) to f n ( b ; X ). Proposition 3.2. Under the assumption, ∆ a · ∆ b = 0 , we have the following twoassertions :(i) Spl M ( c g ,...,c gn − ) f n ( a ; X ) = Spl M ( c g ,...,c gn − ) f n ( b ; X ) for each g ∈ G s , t ;(ii) L a L b = L a M ( c g , . . . , c gn − ) = L b M ( c g , . . . , c gn − ) for each g ∈ G s , t . Proof. Put M ′ = M ( c g , . . . , c gn − ). Then, by Lemma 3.1, M ′ [ X ] / ( f n ( a ; X )) and M ′ [ X ] / ( f n ( b ; X )) are isomorphic over M ′ . Hence follows the assertion (i). By (i)we see that L a M ( c g , . . . , c gn − ) = L b M ( c g , . . . , c gn − ). Thus the assertion (ii) alsoholds. (cid:3) EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 11 It follows from Proposition 2.2 and Proposition 2.5 that, for a fixed j, (0 ≤ j ≤ n − K ( u g , . . . , u gn − ) = K ( u gj ) for g ∈ G s , t , (7) L s L t = K ( u gj | g ∈ H \ G s , t ) . and [ K ( u gj ) : K ] = n !. After the specialization as in (4) we may only have inclusionrelations M ( c g , . . . , c gn − ) ⊃ M ( c gj ) for g ∈ G s , t ,L a L b ⊃ M ( c gj | g ∈ H \ G s , t ) . Whether the equality M ( c g , . . . , c gn − ) = M ( c gj ) holds or not depends on the spe-cialization ( s , t ) ( a , b ) ∈ M n × M n . By (7) there exists P i,j ( s , t ; X ) ∈ K [ X ]such that u i = P i,j ( s , t ; u j ) and deg X ( P i,j ( s , t ; X )) < n ! . Then we take polynomials P i,j ( s , t ; X ) ∈ k [ s , t ][ X ] and D i,j ( s , t ) ∈ k [ s , t ] whichsatisfy u i = 1 D i,j ( s , t ) P i,j ( s , t ; u j ) , and deg X ( P i,j ( s , t ; X )) < n !(8)by extracting the minimal multiple of the denominators of the coefficients of thepolynomial P i,j ( s , t ; X ) in X . Then the following lemma is clear. Lemma 3.3. For a fixed j, ≤ j ≤ n − , and a , b ∈ M n , the condition D i,j ( a , b ) =0 for i = 0 , . . . , n − implies M ( c g , . . . , c gn − ) = M ( c gj ) for each g ∈ G s , t . After the specialization as in (4), we also utilize the polynomial F i ( a , b ; X ) = Y g ∈ H \ G s , t ( X − c gi ) ∈ M [ X ] , ( i = 0 , . . . , n − n ! which is not necessary irreducible. Lemma 3.4. If F j ( a , b ; X ) has no multiple root for a fixed j, ≤ j ≤ n − and a , b ∈ M n with ∆ a · ∆ b = 0 then D i,j ( a , b ) = 0 for i = 0 , . . . , n − .Proof. By (8), if D i,j ( a , b ) = 0 for an i , then we should have P i,j ( a , b ; c gj ) = 0 for g ∈ G s , t . From the assumption c gj = c hj ( g = h ), n ! different c gj satisfy the equality P i,j ( a , b ; c gj ) = 0 of degree less than n !. This is a contradiction. (cid:3) Proposition 3.5. For a fixed j, ≤ j ≤ n − , and a , b ∈ M n with ∆ a · ∆ b = 0 ,suppose D i,j ( a , b ) = 0 for i = 0 , . . . , n − . Then F j ( a , b ; X ) has no multiple root. Proof. We first note that { c gj | g ∈ H \ G s , t } = { c (1 ,τ ) j | τ ∈ G t } . It follows from thecondition, ∆ a · ∆ b = 0, the roots β , . . . , β n of f n ( b ; X ) are distinct. For τ, τ ′ ∈ G t ,therefore, ( β ψ ( τ )(1) , . . . , β ψ ( τ )( n ) ) = ( β ψ ( τ ′ )(1) , . . . , β ψ ( τ ′ )( n ) ) if and only if τ = τ ′ . Since y τ = ( y ψ ( τ )(1) , . . . , y ψ ( τ )( n ) ), we have, by the definition, c (1 ,τ )0 c (1 ,τ )1 ... c (1 ,τ ) n − = α α · · · α n − α α · · · α n − ... ... ... . . . ...1 α n α n · · · α n − n − β ψ ( τ )(1) β ψ ( τ )(2) ... β ψ ( τ )( n ) . We also have the equation for τ ′ if we replace τ by τ ′ . On the other hand, we have c (1 ,τ ) i = 1 D i,j ( a , b ) P i,j ( a , b ; c (1 ,τ ) j ) ,c (1 ,τ ′ ) i = 1 D i,j ( a , b ) P i,j ( a , b ; c (1 ,τ ′ ) j ) , from the condition D i,j ( a , b ) = 0 for i = 0 , . . . , n − 1. If c (1 ,τ ) j = c (1 ,τ ′ ) j , therefore,we have ( c (1 ,τ )0 , . . . , c (1 ,τ ) n − ) = ( c (1 ,τ ′ )0 , . . . , c (1 ,τ ′ ) n − ). Hence from the above equalities, wehave ( β ψ ( τ )(1) , . . . , β ψ ( τ )( n ) ) = ( β ψ ( τ ′ )(1) , . . . , β ψ ( τ ′ )( n ) ). Thus we see τ = τ ′ and theassertion of the proposition. (cid:3) Before we go on farther analysis, we explain the actions of G s , t and G a , b on theset of values { c gi | g ∈ H \ G s , t } , ≤ i ≤ n − 1, where g ∈ G s , t runs over a set ofrepresentatives of the cosets H \ G s , t . We defined c gi by c gi = u gi ( α , β ). Since therational function u i ( x , y ) of 2 n variables x = ( x , . . . , x n ) , y = ( y , . . . , y n ) belongsto L s L t = k ( x , y ), the Galois group G s , t = Gal( L s L t /k ( s , t )) naturally acts on u i ( x , y ); since g induces substitutions on the sets { x , . . . , x n } and { y , . . . , y n } , weexpress g = ( σ, τ ) and u gi ( x , y ) = u i ( x g , y g ) with x g = ( x ϕ ( σ )(1) , . . . , x ϕ ( σ )( n ) ) and y g = ( y ψ ( τ )(1) , . . . , x ψ ( τ )( n ) ). However, G s , t does not directly act on the set of values { c gi | g ∈ H \ G s , t } of u gi . We consider the collection of the values c gi , g ∈ H \ G s , t , asa function c i ( g ) := c gi defined on the cosets H \ G s , t because u gi and hence c gi dependonly on the coset Hg . Then for h ∈ G s , t , c hi ( g ) := c ghi is the composition of c i withthe translation on H \ G s , t by h . Therefore each h ∈ G s , t induces a substitution ofthe set of values { c gi | g ∈ H \ G s , t } . This is the way by which G s , t acts on the set.Hence, in particular, its action is transitive. Thus if { c gi | g ∈ H \ G s , t } = n ! thenwe have Stab G s , t ( c gi ) = Stab G s , t ( u gi ( x , y )) = g − Hg because H \ G s , t ) = n !.As for G a , b , the situation is different. It acts on the set of values { c gi | g ∈ H \ G s , t } as the Galois group Gal( L a L b /M ) because every c gi is contained in the field L a L b .Under the assumption, ∆ a · ∆ b = 0, furthermore, we regarded G a , b as a subgroupof G s , t . EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 13 Lemma 3.6. The action of G a , b on the set { c gi | g ∈ H \ G s , t } as the Galois group Gal( L a L b /M ) coincides with the one as a subgroup of G s , t through the action onthe set of cosets H \ G s , t .Proof. For g = ( σ, τ ) ∈ G s , t , we have by the definition c g c g ... c gn − = α ϕ ( σ )(1) α ϕ ( σ )(1) · · · α n − ϕ ( σ )(1) α ϕ ( σ )(2) α ϕ ( σ )(2) · · · α n − ϕ ( σ )(2) ... ... ... . . . ...1 α ϕ ( σ )( n ) α ϕ ( σ )( n ) · · · α n − ϕ ( σ )( n ) − β ψ ( τ )(1) β ψ ( τ )(2) ... β ψ ( τ )( n ) . An element h ∈ G a , b is identified as an element h = ( σ ′ , τ ′ ) ∈ G s , t through ( α i ) h = α ϕ ( σ ′ )( i ) and ( β i ) h = β ψ ( τ ′ )( i ) for i = 1 , . . . , n . Since ϕ and ψ are anti-isomorphisms,we have ϕ ( σ ′ ) ϕ ( σ ) = ϕ ( σσ ′ ) and ψ ( τ ′ ) ψ ( τ ) = ψ ( τ τ ′ ). Hence the action of h on c gi , i = 0 , . . . , n − 1, as an element of Gal( L a L b /M ) is given by ( c g ) h ( c g ) h ...( c gn − ) h = α ϕ ( σσ ′ )(1) α ϕ ( σσ ′ )(1) · · · α n − ϕ ( σσ ′ )(1) α ϕ ( σσ ′ )(2) α ϕ ( σσ ′ )(2) · · · α n − ϕ ( σσ ′ )(2) ... ... ... . . . ...1 α ϕ ( σσ ′ )( n ) α ϕ ( σσ ′ )( n ) · · · α n − ϕ ( σσ ′ )( n ) − β ψ ( ττ ′ )(1) β ψ ( ττ ′ )(2) ... β ψ ( ττ ′ )( n ) . Therefore we see ( c gi ) h = c ghi , i = 0 , . . . , n − (cid:3) Proposition 3.7. Assume as above that ∆ a · ∆ b = 0 for a , b ∈ M n . Suppose thatthe polynomial F j ( a , b ; X ) has no multiple root for some j, ≤ j ≤ n − . Thenthe following two assertions hold :(i) M ( c g , . . . , c gn − ) = M ( c gj ) for each g ∈ G s , t ;(ii) L a L b = M ( c gj | g ∈ H \ G s , t ) .Proof. The assertion (i) follows from Lemma 3.3 and Lemma 3.4. Here we give analternative proof of (i) and a proof of (ii) from the viewpoint of above discussion.(i) If F j ( a , b ; X ) has no multiple root, then { c gj | g ∈ H \ G s , t } = n !. Therefore,we have Stab G a , b ( c gj ) = g − Hg ∩ G a , b for each g ∈ G s , t . It follows from Lemma 2.1that g − Hg ∩ G a , b is contained in Stab G a , b ( c gi ) for every i, ≤ i ≤ n − 1. Thereforewe see Stab G a , b ( c gj ) ⊂ Stab G a , b ( c gi ) for every i, ≤ i ≤ n − 1, and M ( c gj ) ⊃ M ( c gi )for every i, ≤ i ≤ n − 1. This shows (i).(ii) By the definition of c gj , we have L a L b ⊃ M ( c gj | g ∈ H \ G s , t ) ⊃ M = L a L b G a , b . Let h be an element of G a , b and suppose that h is trivial on M ( c gj | g ∈ H \ G s , t ). As we saw above, we have Stab G a , b ( c gj ) = g − Hg ∩ G a , b for each g ∈ G s , t .Therefore h belongs to T g ∈ H \ G s , t g − Hg which is equal to { } as we showed it inthe proof of Proposition 2.5. Hence we have h = 1 and then Gal( L a L b /M ( c gj | g ∈ H \ G s , t )) = { } . This shows (ii). (cid:3) Theorem 3.8. For a fixed j, ≤ j ≤ n − , and a , b ∈ M n with ∆ a · ∆ b = 0 , assumethat the polynomial F j ( a , b ; X ) has no multiple root. Then F j ( a , b ; X ) has a rootin M if and only if M [ X ] / ( f n ( a ; X )) and M [ X ] / ( f n ( b ; X )) are M -isomorphic.Proof. First suppose that one of the roots of F j ( a , b ; X ), c gj for some g ∈ G s , t , isin M . By the preceding proposition, we see c g , . . . , c gn − ∈ M . Express g = ( σ, τ )as usual. Then for the root α ϕ ( σ )( i ) of f n ( a ; X ), β ψ ( τ )( i ) := c g + c g α ϕ ( σ )( i ) + · · · + c gn − α n − ϕ ( σ )( i ) is a root of f n ( b ; X ) in M ( α ) for i = 1 , . . . , n . It is now easy toadopt the argument in the proof of Lemma 3.1 and to obtain an isomorphismfrom M [ X ] / ( f n ( b ; X )) to M [ X ] / ( f n ( a ; X )) over M by assigning c g + c g X + · · · + c gn − X n − mod( f n ( a ; X )) to X mod( f n ( b ; X )).Now suppose conversely that M [ X ] / ( f n ( a ; X )) and M [ X ] / ( f n ( b ; X )) are M -isomorphic. Then as in the proof of Lemma 3.1, the irreducible factors of f n ( a ; X )and those of f n ( b ; X ) perfectly correspond. Moreover, each pair of correspondingsimple components are isomorphic over M . Denote the image of X mod( f n ( b ; X ))by the isomorphism from M [ X ] / ( f n ( b ; X )) to M [ X ] / ( f n ( a ; X )) by η := e + e X + · · · + e n − X n − mod( f n ( a ; X )) with e , . . . , e n − ∈ M . Then we find τ ∈ G t sothat we have β ψ ( τ )( i ) = e + e α i + · · · + e n − α n − i for i = 1 , . . . , n . Hence for g = (1 , τ ) ∈ G s , t , we must have e ν = c gν for ν = 0 , . . . , n − 1. In particular, we seethat c gj is a root of F j ( a , b ; X ) in M . (cid:3) In the case where G a and G b are isomorphic to a transitive subgroup G of S n and every subgroups of G with index n are conjugate in G , the condition that M [ X ] / ( f n ( a ; X )) and M [ X ] / ( f n ( b ; X )) are isomorphic over M is equivalent to thecondition that Spl M f n ( a ; X ) and Spl M f n ( b ; X ) coincide. Hence we get an answerto the field isomorphism problem via F j ( a , b ; X ). Corollary 3.9. Let j and a , b ∈ M n be as in Theorem 3.8. Assume that both of G a and G b are isomorphic to a transitive subgroup G of S n and that all subgroupsof G with index n are conjugate in G . Then F j ( a , b ; X ) has a root in M if andonly if Spl M f n ( a ; X ) and Spl M f n ( b ; X ) coincide. We note that if G is one of the symmetric group S n , ( n = 6), the alternatinggroup A n of degree n , ( n = 6), and solvable transitive subgroups of S p of primedegree p , then all subgroups of G with index n or p , respectively, are conjugate in G (cf. [Hup67], [BJY86]).Let H and H be subgroups of S n . As an analogue to Theorem 2.6, we obtaina k -generic polynomial for H × H , the direct product of groups H and H . Theorem 3.10. Let M = k ( q , . . . , q l , r , . . . , r m ) , (1 ≤ l, m ≤ n − be therational function field over k with ( l + m ) variables. For a ∈ k ( q , . . . , q l ) n , b ∈ k ( r , . . . , r m ) n , we assume that f n ( a ; X ) ∈ M [ X ] and f n ( b ; X ) ∈ M [ X ] be k -generic polynomials for H and H respectively. For a fixed j, ≤ j ≤ n − , EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 15 assume that F j ( a , b ; X ) ∈ M [ X ] has no multiple root. Then F j ( a , b ; X ) is a k -generic polynomial for H × H which is not necessary irreducible.Proof. It follows from Proposition 3.7 that M ( c gj | g ∈ H \ G s , t ) = L a L b . Hencethe assertion follows from the H -genericness of f n ( a ; X ) and the H -genericnessof f n ( b ; X ). (cid:3) In each Tschirnhausen equivalence class, we can always choose a specialization s a ∈ M n of the polynomial f n ( s ; X ) which satisfy a = 0 and a n − = a n (see[JLY02, § g n ( q , . . . , q n − ; X ):= ( − n · f n (0 , q , . . . , q n − , q n − , q n − ; − X )= X n + q X n − + · · · + q n − X + q n − X + q n − is k -generic for S n with ( n − 2) parameters q , . . . , q n − over an arbitrary field k .Indeed if the characteristic of the field k is prime to n , we obtain q , . . . , q n − interms of s , . . . , s n as follows: we put X := ( X , . . . , X n ), X := x − s /n, X := x − s /n, . . . , X n := x n − s /n :then we have k ( X ) := k ( X , . . . , X n − ) ⊂ k ( x ) and X + X + · · · + X n = 0 . The S n -action on k ( x ) induces an action on k ( X ) which is linear and faithful.We also have k ( X ) S n = k ( S ) := k ( S , S , . . . , S n ) where S i is the i -th elementarysymmetric functions in X , . . . , X n . Note S = 0. The polynomials f n ( S ; X ) and f n ( s ; X ) are Tschirnhausen equivalent over k ( s ), and f n ( S ; X ) generates the fieldextension k ( X ) /k ( X ) S n . By Kemper-Mattig’s theorem [KM00], f n ( S ; X ) is k -generic for S n with parameters S , . . . , S n . Define q := S n /S n − , q i := S i /q i , ( i = 2 , . . . , n − . Then we obtain k ( S ) = k ( q , . . . , q n − ) and g n ( q , . . . , q n − ; X ) = ( − /q ) n f n ( S ; − q X ) . The polynomials g n ( q , . . . , q n − ; X ) and f n ( S ; X ) are Tschirnhausen equivalentover k ( S ). Since deg( q ) = 1 , deg( q i ) = 0 , ( i = 2 , . . . , n − g n ( q , . . . , q n − ; X ) is a generating polynomial of the degree-zero field k ( X ) := k ( X /X , . . . , X n − /X n ) ⊂ k ( X ) over k ( X ) S n = k ( q , . . . , q n − ), (cf. [Kem96],[KM00, Theorem7]). Corollary 3.11. Let M = k ( q , . . . , q n − , r , . . . , r n − ) be the rational functionfield with n − variables. Let a = (0 , q , . . . , q n − , q n − ) ∈ M n and b =(0 , r , . . . , r n − , r n − ) ∈ M n . For a fixed j, ≤ j ≤ n − , assume that F j ( a , b ; X ) ∈ M [ X ] has no multiple root. Then F j ( a , b ; X ) is a k -generic polynomial for S n × S n with n − parameters q , . . . , q n − , r , . . . , r n − . In order to obtain an answer to the subfield problem of generic polynomials, westudy the degrees of the fields of Tschirnhausen coefficients M ( c g , . . . , c gn − ) over M for g ∈ G s , t . The factorization pattern of the polynomial F i ( a , b ; X ) over M givesus information about the degeneration of Galois groups under the specialization( s , t ) ( a , b ) and about the intersection of root fields of f n ( a ; X ) and f n ( b ; X )over M through the degrees of M ( c g , . . . , c gn − ) over M as in Proposition 3.2. Proposition 3.12. Assume that the characteristic of M is not equal to resp. isequal to . If ∆ a / ∆ b ∈ M ( resp. β a + β b ∈ M ) , then the polynomial F i ( a , b ; X ) splits into two factors of degree n ! / over M which are not necessary irreducible.Proof. The assertion follows from Proposition 2.7. (cid:3) Corollary 3.13. If G a , G b ⊂ A n , then F i ( a , b ; X ) splits into two factors of degree n ! / over M which are not necessary irreducible. Cubic case In this section, we treat the cubic case of the subfield problem of generic poly-nomial via general Tschirnhausen transformations. We take f ( s ; X ) := X − s X + s X − s ∈ k ( s )[ X ]where s = x + x + x ,s = x x + x x + x x ,s = x x x . As in the previous section, we take the discriminant ∆ s in general and especiallythe Berlekamp discriminant β s if char k = 2:∆ s := ( x − x )( x − x )( x − x ) ,β s := x x + x + x x + x + x x + x (9) = x x + x x + x x + x x x x x + x x + x x + x x + x x + x x . Then we have ∆ s = s s − s − s s + 18 s s s − s ,β s ( β s + 1) = s + s s + s s s + s s s + s . EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 17 The field k ( s )(∆ s ) (resp. k ( s )( β s )) is a quadratic extension of k ( s ) if char k = 2(resp. char k = 2). For g = (1 , τ ) ∈ H \ G s , t , we put u g u g u g := x x x x x x − y τ y τ y τ . By the definition we evaluate ( u , u , u ) as u = ∆ − s · det y x x y x x y x x = x x y ( x − x )( x − x ) − x x y ( x − x )( x − x ) + x x y ( x − x )( x − x ) ,u = ∆ − s · det y x y x y x = − ( x + x ) y ( x − x )( x − x ) + ( x + x ) y ( x − x )( x − x ) − ( x + x ) y ( x − x )( x − x ) ,u = ∆ − s · det x y x y x y = y ( x − x )( x − x ) − y ( x − x )( x − x ) + y ( x − x )( x − x ) . A general form of Tschirnhausen transformations of f ( s ; X ) is given by g ( s , u , u , u ; X ):= Resultant Y (cid:0) f ( s ; Y ) , X − ( u + u Y + u Y ) (cid:1) = X + ( − u − s u − s u + 2 s u ) X + (3 u + 2 s u u + s u + 2 s u u − s u u + s s u u − s u u + s u − s s u ) X − u − s u u − s u u − s u − s u u + 2 s u u − s s u u u + 3 s u u u − s s u u − s u u + 2 s s u u − s s u u − s u . By the definition, the elements u , u , u satisfy f ( t ; X ) = g ( s , u g , u g , u g ; X ) for g ∈ G s , t . (10) We put A s := s − s ,B s := 2 s − s s + 27 s , (11) C s := s − s s + s + 6 s s ,D s := Disc X f ( s ; X ) = s s − s − s s + 18 s s s − s (= ∆ s ) . We can check the equality 4 A s − B s = 27 D s (12)by direct calculation.With the aid of computer algebra, we get the sextic polynomials F i ( s , t ; X ) = Y g ∈ H \ G s , t ( X − u gi ) ∈ K [ X ] , ( i = 1 , F ( s , t ; X ) := X − A t C s D s X − ( s s − s ) B t D s X + A t C s D s X + ( s s − s ) A t B t C s D s X + ( s s − s ) A t D s − C s D t D s ,F ( s , t ; X ) := X − A s A t D s X + B t D s X (13) + A s A t D s X − A s A t B t D s X + A t D s − A s D t D s , where A s , B s , C s and D s are defined in (11). Here we omit the explicit description ofthe polynomial F ( s , t ; X ) because whose roots u g are available from F ( s , t ; X ) and F ( s , t ; X ) by (15) and (19) below. The discriminant of the polynomial F ( s , t ; X )with respect to X is given by D s , t := B s D t ( A s B t − A t D s ) D s . (14)We note that A s B t − A t D s is invariant under the action ( x , x , x ) ↔ ( y , y , y ).Indeed, by using (12), we can obtain the following symmetric representation: A s B t − A t D s = 4 A s A t − A t D s + A s D t ) . In the case of char k = 2, we also see that A s B t − A t D s = B s B t − D s D t . By Proposition 2.7, we have the decomposition F ( s , t ; X ) = F +2 ( X ) F − ( X ) , EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 19 where F +2 ( X ) and F − ( X ) are elements of K (∆ s / ∆ t )[ X ] and of K ( β s + β t )[ X ] inthe case of char k = 2. In the case of char k = 2, we have, by the definition (3), F +2 ( X ) = Y (1 ,τ ) ∈ H \ G s , t ψ ( τ ) ∈ A ( X − u (1 ,τ )2 ) = X − A s A t D s X + B t − B s (∆ t / ∆ s )2 D s ,F − ( X ) = Y (1 ,τ ) ∈ H \ G s , t ψ ( τ ) A ( X − u (1 ,τ )2 ) = X − A s A t D s X + B t + B s (∆ t / ∆ s )2 D s . If char k = 2, we have F +2 ( X ) = X + A s A t D s X + s A s B t + t A t B s + B s B t ( β s + β t ) B s D s ,F − ( X ) = X + A s A t D s X + s A s B t + t A t B s + B s B t ( β s + β t + 1) B s D s . When we specialize parameters ( s , t ) ( a , b ) ∈ M × M for a fixed field M ( ⊃ k ) with infinite elements, we assume that f ( a ; X ) and f ( b ; X ) are separableover M (i.e. D a · D b = 0). We define L a := Spl M f ( a ; X ) , L b := Spl M f ( b ; X ) ,G a := Gal( L a /M ) , G b := Gal( L b /M )and suppose G a ≥ G b . We also suppose that f ( a ; X ) is irreducible over M .Then G a is isomorphic to either S or A = C , and G b is isomorphic to one of S , C , C or { } . We shall give an answer to the subfield problem of f ( s ; X ) via F j ( s , t ; X ). More precisely, we give a necessary and sufficient condition to have L a ⊇ L b for a , b ∈ M .4.1. The case of char k = 3 . First we study the case of char k = 3. The case ofchar k = 3 will be studied in Subsection 4.4.By comparing the coefficients of (10) with respect to X , we obtain u = t − s u − s u + 2 s u , u = Q , ( s , t ; u ) D , ( s , t ; u )(15)where Q , ( s , t ; u ) := 3 A s B t − A t (6 A s − B s + 2 A s B s s ) u + 6 D s ( A s + B s s ) u ,D , ( s , t ; u ) := 3 B s ( A s A t − D s u ) . We also have u = ( t − s u + 2 s u ) D , ( s , t ; u ) − s Q , ( s , t ; u )3 D , ( s , t ; u ) . (16)From (15) and (16), we can directly check K ( u g , u g , u g ) = K ( u g ) for every g ∈ G s , t as in Proposition 2.2 where K = k ( s , t ). As in (8), we obtain D , ( s , t ) ∈ k [ s , t ] and D , ( s , t ) ∈ k [ s , t ] as follows: first we see that there exist those D , ( s , t ) and h i ( s , t ) ∈ k [ s , t ] which satisfy1 D , ( s , t ; u ) = 13 B s ( A s A t − D s u ) = 1 D , ( s , t ) X i =0 h i ( s , t ) u i . Actually, we get, with the aid of computer algebra, D , ( s , t ) := 3 B s ( A s B t − A t D s ) and { h ( s , t ) , . . . , h ( s , t ) } = n A s A t ( A s B t + 27 A t D s − B t D s ) , B t D s (4 A s A t + 9 A t D s − A s D t ) , − A s A t D s (5 A s B t + 135 A t D s − B t D s ) , − A s A t B t D s , D s ( A s B t + 27 A t D s ) , A s A t B t D s o . We define D , ( s , t ) := 3 · D , ( s , t ) ∈ k [ s , t ] on account of (16). Hence we obtainthose P i, ( s , t ; X ) ∈ k [ s , t ][ X ] , ( i = 0 , u i = 1 D i, ( s , t ) P i, ( s , t ; u ) , and deg X ( P i, ( s , t ; X )) = 5 . After specializing parameters ( s , t ) ( a , b ) ∈ M × M with D a D b = 0, we seethe following lemma: Lemma 4.1. (i) If f ( a ; X ) is irreducible over M then B a = 0 ;(ii) If A a = 0 , then f ( a ; X ) and X − B a are Tschirnhausen equivalent. Hencethe Galois group of f ( a ; X ) over M ( √− is cyclic of order If A a = 0 , then f ( a ; X ) and Y − Y − ( B a + 1 /B a ) are Tschirnhausenequivalent over M .Proof. The statements of (i) and (ii) follow from the equality3 · f ( s ; X ) = (3 X − s ) − A s (3 X − s ) − B s . In the case of X − B a = 0, we put Y = X + 1 /X = X (1 + X/B a ) and obtain Y − Y − ( B a + 1 /B a ) = 0. Thus we have (iii). (cid:3) From Lemma 4.1 (i), the assumption that f ( a ; X ) is irreducible over M implies B a = 0. From Lemma 4.1 (iii), we may assume that A a = 0 and A b = 0 withoutloss of generality.By (14) and the assumptions B a = 0 and D b = 0, we see that the polynomial F ( a , b ; X ) has multiple roots if and only if A a B b − A b D a = 0. EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 21 Lemma 4.2. Assume that f n ( a ; X ) is irreducible and A a A b = 0 . (i) If A a B b − A b D a = 0 then F ( a , b ; X ) splits into the following form over M : F ( a , b ; X ) = (cid:16) X − A b A a B b (cid:17) (cid:16) X + 6 A b A a B b (cid:17)(cid:16) X − A b XA a B b − A b (2 A b − B b ) A a B b (cid:17) with a simple root − A b / ( A a B b ) ;(ii) If A a B b − A b D a = 0 then multiple roots c = 3 A b / ( A a B b ) of F ( a , b ; X ) satisfy A a A b − D a c = 0 . Conversely if a root c of F ( a , b ; X ) satisfies A a A b − D a c = 0 , then A a B b − A b D a = 0 .Proof. (i) We have B b = 0 since 27 A b D a = 0. Using D a = A a B b / (27 A b ) and D b = (4 A b − B b ) / 27, we eliminate D a and D b from F ( a , b ; X ) via (13). Then weobtain explicit factors of F ( a , b ; X ) as F ( a , b ; X )= X − A b X A a B b + 27 A b X A a B b + 729 A b X A a B b − A b XA a B b − A b (2 A b − B b ) A a B b = (cid:16) X − A b A a B b (cid:17) (cid:16) X + 6 A b A a B b (cid:17)(cid:16) X − A b XA a B b − A b (2 A b − B b ) A a B b (cid:17) . We also see that − A b / ( A a B b ) is a simple root of F ( a , b ; X ) since 3 A b / ( A a B b ) = − A b / ( A a B b ) andResultant X (cid:16) X + 6 A b A a B b , X − A b XA a B b − A b (2 A b − B b ) A a B b (cid:17) = − A b + B b = − D b = 0 . (ii) For c = 3 A b / ( A a B b ), the condition A a B b − A b D a = 0 implies A a A b − D a c = A a A b − (cid:16) A a B b A b (cid:17)(cid:16) A b A a B b (cid:17) = A a A b − A a A b = 0 . Conversely if ( A a A b − D a c ) = 0 then D , ( a , b ) = 3 B a ( A a B b − A b D a ) = 0. (cid:3) We obtain a solution to the field isomorphism problem of f ( s ; X ) as a specialcase of Theorem 3.8 and Corollary 3.9 in Section 3. Theorem 4.3. Assume that f n ( a ; X ) is irreducible and A a A b = 0 . Then we have (1) If A a B b − A b D a = 0 then Spl M f ( a ; X ) = Spl M f ( b ; X ) ;(2) If A a B b − A b D a = 0 then the following two conditions are equivalent :(i) Spl M f ( a ; X ) = Spl M f ( b ; X ) ;(ii) The sextic polynomial F ( a , b ; X ) has a root in M .Proof. (1) Suppose A a B b − A b D a = 0; then it follows from Lemma 4.2 (i) that F ( a , b ; X ) has a simple root c g = − (6 A b ) / ( A a B b ) ∈ M for some g ∈ G s , t . Then D , ( a , b ; c g ) = 3 B a ( A a A b − D a ( c g ) ) = − B a A a A b = 0. Thus, by (15) and (16),we have M ( c g , c g , c g ) = M ( c g ) = M , and hence the assertion follows. (2) If A a B b − A b D a = 0 then D , ( a , b ) · D , ( a , b ) = 0. Thus the assertionfollows from Theorem 3.8 (or Corollary 3.9). (cid:3) Example 4.4. We give two examples which satisfy A a B b − A b D a = 0.(1) We take M = Q and a = (0 , , − b = (3 , − , ∈ M . Then f ( a ; X ) = X + 3 X + 2 , f ( b ; X ) = X − X − X − . We have ( A a , B a , C a , D a ) = ( − , − , , − A b , B b , D b ) = (18 , , − f ( a ; X ) and f ( b ; X ) over Q are isomorphic to S because f ( a ; X ) and f ( b ; X ) are irreducible over Q , and neither of D a = − · and D b = − · is a square in M . We also have A a B b − A b D a = 0. By Lemma4.2 (i) we have F ( a , b ; X ) = (cid:16) X + 12 (cid:17) (cid:16) X − (cid:17)(cid:16) X − X − (cid:17) . Hence we take c g = 1 and get ( c g , c g , c g ) = (3 , − , 1) by (15) and (16). Thus we seethat Q [ X ] /f ( a ; X ) ∼ = Q Q [ X ] /f ( b ; X ). An explicit Tschirnhausen transformationfrom f ( a ; X ) to f ( b ; X ) over Q is given by f ( b ; Y ) = Resultant X ( f ( a ; X ) , Y − (3 − X + X )) . We also obtain F ( a , b ; X ) = (cid:16) X − X + 74 (cid:17)(cid:16) X + 1 (cid:17)(cid:16) X + 3 X (cid:17) ,F ( a , b ; X ) = X ( X − X − X − . (2) Take M = Q and a = ( − , − , − b = ( − , − , ∈ M . Then f ( a ; X ) = X + 3 X − X + 1 , f ( b ; X ) = X + X − X − . We have ( A a , B a , C a , D a ) = (21 , − , , 49) and ( A b , B b , D b ) = (7 , , f ( a ; X ) and f ( b ; X ) over Q are isomorphic to C because f ( a ; X )and f ( b ; X ) are irreducible over Q and D a = D b = 7 . Since A a B b − A b D a = 0,we have, by Lemma 4.2 (i), F ( a , b ; X ) = (cid:16) X − (cid:17) (cid:16) X + 2 (cid:17)(cid:16) X − X − (cid:17) . Thus we have c g = − 2. We obtain ( c g , c g , c g ) = (4 , − , − 2) by (15) and (16), and Q [ X ] /f ( a ; X ) ∼ = Q Q [ X ] /f ( b ; X ). An explicit Tschirnhausen transformation from f ( a ; X ) to f ( b ; X ) over Q is given by f ( b ; Y ) = Resultant X ( f ( a ; X ) , Y − (4 − X − X )) . We also obtain F ( a , b ; X ) = (cid:16) X − (cid:17)(cid:16) X − (cid:17)(cid:16) X + 7 (cid:17)(cid:16) X − X − (cid:17) ,F ( a , b ; X ) = (cid:16) X + 3 (cid:17)(cid:16) X + 2 (cid:17)(cid:16) X − (cid:17)(cid:16) X + X − X + 717 (cid:17) . EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 23 Hence we have the other two Tschirnhausen transformations from f ( a ; X ) to f ( b ; X ) over Q by (15) as f ( b ; Y ) = Resultant X ( f ( a ; X ) , Y − ( − X + X )) ,f ( b ; Y ) = Resultant X ( f ( a ; X ) , Y − ( − X + X )) . Theorem 4.5. For a , b ∈ M , we assume that f n ( a ; X ) is irreducible, A a A b = 0 and A a B b − A b D a = 0 . The decomposition type of irreducible factors h µ ( X ) of F ( a , b ; X ) over M gives an answer to the subfield problem of f ( s ; X ) ason Table . Moreover a root field M µ of each h µ ( X ) satisfies Spl M µ f ( a , X ) =Spl M µ f ( b , X ) . Table 1 G a G b ( d µ ) , d µ = deg( h µ ( X )) L a = L b , L a ∩ L b = M (6) S L a = L b , [ L a ∩ L b : M ] = 2 (3)(3) L a = L b (1)(2)(3) S C L a ∩ L b = M (6) C L a L b (6) L a ⊃ L b (3)(3) { } L a ⊃ L b (6) C L a = L b (3)(3) C L a = L b (1)(1)(1)(3) C L a ∩ L b = M (6) { } L a ⊃ L b (3)(3) Proof. Form Proposition 3.7 and the assumption A a B b − A b D a = 0, we have L a L b = L a M ( c g ) = L b M ( c g ) for g ∈ G s , t . (17)Hence two polynomials f ( a ; X ) and f ( b ; X ) are Tschirnhausen equivalent over aroot field M µ of each irreducible factor h µ ( X ) of F ( a , b ; X ).(i) The case of G a ∼ = S .(i-1) If L a ∩ L b = M then it follows from (17) that [ M ( c g ) : M ] = 6. Hence F ( a , b ; X ) is irreducible over M .(i-2) If [ L a ∩ L b : M ] = 2 then, by Proposition 3.12, F ( a , b ; X ) splits into twocubic factors F +2 ( X ) and F − ( X ) over M , and each cubic factor is irreducible over M because we have [ M ( c g ) : M ] ≥ G b ∼ = S and L a = L b then F ( a , b ; X ) also splits into two cubic factors over M , and one of them must has a linear factor from Theorem 4.3. By Proposition3.7 (ii), we see that the factorization pattern of F ( a , b ; X ) is equal to (1)(2)(3). (ii) The case of G a ∼ = C .(ii-1) If G b ∼ = C and L a = L b then, by Proposition 3.12, F ( a , b ; X ) splits intotwo cubic factors over M and each cubic factor is irreducible over M because wehave [ M ( c g ) : M ] ≥ G b ∼ = C and L a = L b then F ( a , b ; X ) also splits into two cubic factorsover M , and one of them must has a linear factor from Theorem 4.3. By Proposition3.7 (ii), we see that the factorization pattern of F ( a , b ; X ) is equal to (1)(1)(1)(3).(ii-3) If G b ∼ = C then it follows from (17) that [ M ( c g ) : M ] = 6. Hence F ( a , b ; X )is irreducible over M .(ii-4) If G b ∼ = { } then, by Proposition 3.12, F ( a , b ; X ) splits into two cubicfactors F +2 ( X ) and F − ( X ) over M , and each cubic factor is irreducible over M because we have [ M ( c g ) : M ] = 3. (cid:3) Special case X + S X − S . Assume that char k = 3. We treat a k -genericpolynomial of the form f (0 , S , S ; X ) = X + S X − S . Define X := ( X , X , X ), X := x − s / , X := x − s / , X := x − s / . Then k ( X ) := k ( X , X , X ) ⊂ k ( x , x , x ) and X + X + X = 0. The actionof S on k ( x , x , x ) induces an action on k ( X ) which is linear and faithful. Wesee k ( X ) S = k ( S ) where S = ( S , S , S ) and S i is the i -th elementary symmetricfunction in X , X , X . We have S = 0 , S = − A s − ( s − s )3 , S = B s 27 = 2 s − s s + 27 s . The polynomials f (0 , S , S ; X ) and f ( s ; X ) are Tschirnhausen equivalent over k ( s ). Moreover, f ( S ; X ) generates the field extension k ( X ) /k ( X ) S . After spe-cializing parameters s a = ( a , a , a ) ∈ M , the polynomials f ( a ; X ) = X − a X + a X − a and f (0 , A , A ; X ) = X + A X − A are Tschirnhausenequivalent over M where A := − A a / , A := B a / 27. Put T := (0 , T , T ). Thenwe have D S = Disc X f (0 , S , S ; X ) = − S − S ,A S B T − A T D S = − S T + 27 S T + 27 S T ) . EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 25 We also obtain F ( S , T ; X ) = X − S T D S X + 8 S T D S X + 16 S T D S X − S T T D S X + 64 S ( S T − S T ) D S ,F ( S , T ; X ) = X + 6 S T D S X + 27 S T D S X + 9 S T D S X + 81 S S T T D S X (18) + 4 S T + 108 S S T + 729 S T + 27 S T D S ,F ( S , T ; X ) = X − S T D S X + 27 T D S X + 81 S T D S X − S T T D S X + 729( S T − S T ) D S . Put F ( S , S , T , T ; X ) := F ( S , T ; X ). Then we have Theorem 4.6. For ( A , A ) , ( B , B ) ∈ M with A B + 27 A B + 27 A B = 0 ,the decomposition type of irreducible factors h µ ( X ) of F ( A , A , B , B ; X ) over M gives an answer to the subfield problem of X + S X − S as on Table . Special case X + sX + s . Assume that char k = 3. Define A := − A a / , A := B a / 27 as in the previous subsection. For a = ( a , a , a ) ∈ M with A a = 0 and B a = 0, the polynomials f ( a ; X ) and f (0 , a, − a ; X ) = X + aX + a are Tschirnhausen equivalent over M , where a := A A = − A a B a = − a − a ) (2 a − a a + 27 a ) . This follows from the equality X + A X − A = − A A (cid:16)(cid:16) − A XA (cid:17) + a (cid:16) − A XA (cid:17) + a (cid:17) . We take a = (0 , a, − a ) ∈ M and b = (0 , b, − b ) ∈ M . Then D a = Disc X f (0 , a, − a ; X ) = − a (4 a + 27) ,A a B b − A b D a = − a b (4 ab + 27 a + 27 b ) . Remark 4.7. It may be notable that the Tschirnhausen equivalence in the previoustwo subsections is affinely obtained; that is, it is given by linear forms.By Theorem 4.3, we see that if 4 ab + 27 a + 27 b = 0 for a, b ∈ M then X + aX + a and that X + bX + b are Tschirnhausen equivalent over M . Hence X + aX + a and X − a a + 27 X − a a + 27have the same splitting field over M . Example 4.8. For M = Q , we haveSpl Q ( X − X − Q ( X − X − , Spl Q ( X − X − 27) = Spl Q ( X − X − , Spl Q ( X − X − 6) = Spl Q ( X + 54 X + 54) . In our special case, we have F (0 , s, − s, , t, − t ; X ) = X − s tD s X − s tD s X + 16 s t D s X + 32 s t D s X − s t ( s − t ) D s ,F (0 , s, − s, , t, − t ; X ) = X + 6 s tD s X + 27 stD s X + 9 s t D s X + 81 s t D s X + s t (27 s + 729 t + 108 st + 4 s t ) D s ,F (0 , s, − s, , t, − t ; X ) = X − stD s X − tD s X + 81 s t D s X + 243 st D s X − s t ( s − t ) D s where D s = − s (4 s + 27). We put G ( s, t ; X ) := F (0 , s, − s, , t, − t ; X ), and havethe following theorem. Theorem 4.9. For a, b ∈ M with ab + 27 a + 27 b = 0 , the decomposition typeof irreducible factors h µ ( X ) of G ( a, b ; X ) over M gives an answer to the subfieldproblem of X + sX + s as on Table . In the paper [HM07], we gave an answer to the field isomorphism problem of X + sX + s in the case of char k = 3. Here we present a slightly modified versionof the result in [HM07] (cf. Theorem 1 and Theorem 7 in [HM07]). First we notethat if G ( a, b ; X ) has a root zero, i.e. G ( a, b ; 0) = 0, then ab ( a − b ) = 0. Assumethat a = b . Then we have c g = 0 for g ∈ H \ G s , t . Put u := 3 c /c . Then we obtain( c , c ) = (2 ac / , uc / 3) and c = 3( u + 9 u − a ) u − au − au − a − a . Under the condition a (4 a + 27) = 0, we see that u − au − au − a − a = 0.Hence we have M ( c , c , c ) = M ( u ). From the direct computation, we obtain( a − b ) · Q g ∈ H \ G s , t ( X − u g ) =: H ( a, b ; X ) where H ( a, b ; X ) = a ( X + 9 X − a ) − b ( X − aX − aX − a − a ) . Note that the polynomial H ( s, t ; X ) ∈ k ( s, t )[ X ] is k -generic for S × S . We alsosee Disc X H ( a, b ; X ) = a b (4 a + 27) (4 b + 27) . EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 27 Theorem 4.10. For a, b ∈ M with a = b , the decomposition type of irreduciblefactors h µ ( X ) of H ( a, b ; X ) over M gives an answer to the subfield problem of X + sX + s as on Table . In particular, two splitting fields of X + aX + a andof X + bX + b over M coincide if and only if there exists u ∈ M such that b = a ( u + 9 u − a ) ( u − au − au − a − a ) . By applying Hilbert’s irreducibility theorem and Siegel’s theorem for curves ofgenus 0 to Theorem 4.10 respectively, we obtain the following corollaries: Corollary 4.11. If M is a Hilbertian field then for a fixed a ∈ M there existinfinitely many b ∈ M such that Spl M ( X + aX + a ) = Spl M ( X + bX + b ) . Corollary 4.12. Let M be a number field and O M the ring of integers in M . Fora given integer a ∈ O M , there exist only finitely many integers b ∈ O M such that Spl M ( X + aX + a ) = Spl M ( X + bX + b ) .Proof. We may apply Siegel’s theorem (cf. [Lan78, Theorem 6.1], [Lan83, Chapter8, Section 5], [HS00, Theorem D.8.4]) to Theorem 4.10 because the discriminant of u − au − au − a − a with respect to u equals − a (4 a + 27) . (cid:3) Remark 4.13. T. Komatsu [Kom] treated a cubic generic polynomial g ( t, Y ) = Y − t ( Y +1) ∈ k ( t )[ Y ] in the case of char k = 2 , 3. He obtained a sextic polynomial P ( t , t ; Z ) satisfying Spl k ( t ,t ) P ( t , t ; Z ) = Spl k ( t ,t ) g ( t , Y ) · Spl k ( t ,t ) g ( t , Y ) viahis descent Kummer theory (see also [Kom04]). His paper [Kom] treats the subfieldproblem of g ( t, Y ) by using the sextic polynomial P ( t , t ; Z ).4.4. The case of char k = 3 . In this subsection we study the case of char k = 3.We have, in this case, A s = s , B s = − s , D s = s s − s − s s . By comparing the coefficients of (10) with respect to X , we obtain u = s t − s t − s t ( s − s ) u − D s u s t , (19) u = t − ( s + s ) u s . We also get F ( s , t ; X ) = X + s t D s X − t D s X + s t D s X + s t D s X + t D s − s D t D s . The discriminant of F ( s , t ; X ) with respect to X is given by D s , t = B s D t ( A s B t − A t D s ) D s = s D t ( s t ) D s = s t D t D s . Hence we have Theorem 4.14. For a = ( a , a , a ) , b = ( b , b , b ) ∈ M with a b = 0 , thedecomposition type of irreducible factors h µ ( X ) of F ( a , b ; X ) over M gives ananswer to the subfield problem of X − s X + s X − s as on Table . Next we treat the case of a = 0. Without loss of generality we may assume that a = 0 , b = 0 because the polynomial f (0 , s, − s ) = X + sX + s is k -generic for S . Actually for f ( s ; X ) = X − s X + s X − s = 0, we take Y = s − s − s X and have Y + − s s s − s − s s Y + − s s s − s − s s = 0 . Hence for a = ( a , a , a ) ∈ M with a · D a = 0, the polynomials f ( a ; X ) and X + − a a a − a − a a X + − a a a − a − a a are Tschirnhausen equivalent over M .After specializing parameters ( s , t ) = (0 , 0) and then by comparing the coeffi-cients of (10) with respect to X , we obtain u = s ( t − t u − u ) s t , u = 0 . Since the equalities (18) are also valid for char k = 3, we have F (0 , s , s , , t , t ; X ) = X − t X + t X + t X + t t X + s t − s t s ,F (0 , s , s , , t , t ; X ) = (cid:16) X − t s (cid:17) ,F (0 , s , s , , t , t ; X ) = X . We take f (0 , s, − s ; X ) = X + sX + s . Then Disc X f (0 , s, − s ; X ) = − s . Define G ( s, t ; X ) := F (0 , s, − s, , t, − t )= X − tX − tX + t X − t X + t ( s − t ) s . We see that the discriminant of G ( s, t ; X ) with respect to X is equal to t /s . Proposition 4.15. The polynomial G ( s, t ; X ) is k -generic for S × S . Theorem 4.16. For a, b ∈ M with ab = 0 , the decomposition type of irreduciblefactors h µ ( X ) of G ( a, b ; X ) over M gives an answer to the subfield problem of X + sX + s as on Table . EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 29 Cyclic cubic case Let the cyclic substitution σ = (123) ∈ S act on k ( x , x , x ) as σ : x x , x x , x x . Put K = k ( z , z , z ) ⊂ k ( x , x , x ) where z := x − x x − x , z := x − x x − x , z := x − x x − x . Then we have z = − 11 + z , z = − (1 + z ) z . Hence the transcendental degree of K over k is equal to one, and C = h σ i actson K = k ( z ) faithfully as σ : z 11 + z (1 + z ) z z . We consider the C -extension K /K C . Put g C ( e m ; X ) = Y x ∈ Orb h σ i ( z ) ( X − x ) = (cid:16) X − z (cid:17)(cid:16) X + 11 + z (cid:17)(cid:16) X + 1 + z z (cid:17) = X − e mX − ( e m + 3) X − , where e m = z − z − z ( z + 1) = − ( x + x + x − x x − x x − x x + 6 x x x )( x − x )( x − x )( x − x ) . We have K C = k ( e m ) and the splitting field of g C ( e m ; X ) over k ( e m ) equals K .The polynomial g C ( e m ; X ) is k -generic for C and is well-known as Shanks’ simplestcubic [Sha74]. Lemma 5.1. The polynomials f ( s ; X ) = X − s X + s X − s and g C ( e m ; X ) = X − e mX − ( e m + 3) X − are Tschirnhausen equivalent over k ( x , x , x ) C .Proof. We see that the splitting field of g C ( e m ; X ) over k ( x , x , x ) C is k ( x , x , x )as follows:(i) The case of char k = 2. We see k ( x , x , x ) C = k ( s , s , s , ∆ s ) where∆ s = ( x − x )( x − x )( x − x ). We have, furthermore, x i = − ∆ s + s s − s − s z i s − s ) , z i = − ∆ s − s s + 9 s s − ( s − s ) x i ∆ s (20)for i = 1 , , k = 2. We now see k ( x , x , x ) C = k ( s , s , s , β s ) where β s is the Berlekamp discriminant as in (9). We have, for i = 1 , , x i = ( s s + s )( β s + z i ) s + s , z i = β s + ( s + s ) x i s s + s . (cid:3) We express the element e m in terms of s , s , s and ∆ s (resp. β s ) in the case ofchar k = 2 (resp. char k = 2). For an arbitrary field k , we have k ( x , x , x ) C = k ( s , s , s , x x + x x + x x )and e m = − s + 6 s s − s − x x + x x + x x ) − s s + 3 s + 2( x x + x x + x x ) . Then in the case of char k = 2, it follows from x x + x x + x x = (∆ s + s s − s ) / e m = − s + 2 s − s s + 27 s s (cid:16) = − s + B s s (cid:17) . In the case of char k = 2, we have x x + x x + x x = s s + β s s s + β s s , and then e m = s + s s + β s s s + β s s s s + s (cid:16) = s A s + β s B s B s (cid:17) . By applying Theorem 4.5 to M = k ( x , x , x ) C , there exist three Tschirnhausentransformations from f ( s ; X ) to g C ( e m ; X ) which are defined over k ( x , x , x ) C .By specializing parameters ( t , t , t ) ( e m, − ( e m + 3) , ∈ k ( x , x , x ) C of thepolynomial F ( s , t ; X ), we explicitly obtain coefficients ( c g , c g , c g ) of Tschirnhausentransformations from f ( s ; X ) to g C ( e m ; X ) over k ( x , x , x ) C : g C ( e m ; X ) = Resultant Y (cid:0) f ( s ; Y ) , X − ( c g + c g Y + c g Y ) (cid:1) . In the case of char k = 2, we obtain explicit factors of F ( s , e m, − ( e m + 3) , X ) by(13): F ( s , s , s , e m, − ( e m + 3) , X )= X (cid:16) X − A s ∆ s (cid:17)(cid:16) X + A s ∆ s (cid:17)(cid:16) X − A s ∆ s X − A s (2 A s s − s s + 27 s )∆ s (cid:17) . EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 31 Hence we have (cid:8) c g | g = (1 , τ ) ∈ H \ G s , t , ψ ( τ ) ∈ A (cid:9) = { , A s / ∆ s , − A s / ∆ s } .From (20) we see c = 0. By (15) and (16), we obtain c g and c g from c g :( c , c , c ) = (cid:16) E s − ∆ s s , − A s ∆ s , (cid:17) , ( c g , c g , c g ) = (cid:16) A s ( A s s − s + 3 s s ) − ( A s s − E s )∆ s − ∆ s s , A s ( − A s s + E s + ∆ s )2∆ s , A s ∆ s (cid:17) , ( c g , c g , c g ) = (cid:16) − A s ( A s s − s + 3 s s ) − ( A s s − E s )∆ s − ∆ s s , A s (2 A s s − E s + ∆ s )2∆ s , − A s ∆ s (cid:17) where E s = s s − s , g i = (1 , τ i ) ∈ H \ G s , t and ψ ( τ i ) ∈ A \{ } , ( i = 1 , z = x − x x − x = c g + c g x + c g x , z = x − x x − x = c g + c g x + c g x . (21)Hence we get ψ ( τ ) = (123) ∈ A and ψ ( τ ) = (132) ∈ A .In the case of char k = 2, we see A s B t − A t D s = 0 where t = ( t , t , t ) =( e m, − ( e m + 3) , F ( s , s , s , e m, − ( e m + 3) , X )= X (cid:16) X + ( s + s ) ( s s + s ) (cid:17) (cid:16) X + ( s + s ) ( s s + s ) X + ( s + s ) ( s s + s ) (cid:17) . By (15) and (16) we obtain( c , c , c ) = (cid:16) β s , s + s s s + s , (cid:17) . We also obtain the coefficients of two other Tschirnhausen transformations whichsatisfy (21):( c g , c g , c g ) = (cid:16) β s ( s + s ) s s + s , ( s + s )( s + s + β s s s + β s s )( s s + s ) , ( s + s ) ( s s + s ) (cid:17) , ( c g , c g , c g ) = (cid:16) s + s s + β s s + β s s s s + s , ( s + s )( s + s s + β s s s + β s s )( s s + s ) , ( s + s ) ( s s + s ) (cid:17) . Now we assume that the Galois group of f ( a ; X ) = X − a X + a X − a over M is isomorphic to C . By specializing parameters s = ( s , s , s ) a =( a , a , a ) ∈ M with A a = 0 and B a = 0 in Lemma 5.1, we see that f ( a ; X )and g C ( m ; X ) = X − mX − ( m + 3) X − M where m = − a + 2 a − a a + 27 a a , if char k = 2 ,a + a a + β a a a + β a a a a + a , if char k = 2 . (22)From now on we take a = ( m, − ( m + 3) , , b = ( n, − ( n + 3) , f ( a ; X ) = X − mX − ( m + 3) X − , f ( b ; X ) = X − nX − ( n + 3) X − 1, and D a = Disc X f ( a ; X ) = ( m + 3 m + 9) ,A a B b − A b D a = D a D b (2 mn + 3 m + 3 n + 18)(2 mn + 3 m + 3 n − . We have ∆ a = m + 3 m + 9 and ∆ b = n + 3 n + 9 and also F ( a , b ; X ) = F +2 ( m, n ; X ) F − ( m, n ; X )where F +2 ( m, n ; X ) = X − ∆ b ∆ a X − ( m − n )∆ b ∆ a ,F − ( m, n ; X ) = X − ∆ b ∆ a X + ( m + n + 3)∆ b ∆ a and Disc X ( F +2 ( m, n ; X )) = ∆ b (2 mn + 3 m + 3 n + 18) ∆ a , Disc X ( F − ( m, n ; X )) = ∆ b (2 mn + 3 m + 3 n − ∆ a . Note that F − ( m, n ; X ) = F +2 ( m, − n − X ). We see that if m + n + 3 = 0 then X − mX − ( m + 3) X − X − nX − ( n + 3) X − M . Hence X − mX − ( m + 3) X − X + ( m + 3) X + mX − M . By Lemma 4.2 (i) and Theorem 4.3 (1) if(2 mn + 3 m + 3 n + 18)(2 mn + 3 m + 3 n − 9) = 0 then X − mX − ( m + 3) X − X − nX − ( n + 3) X − M because F ( a , b ; X )has multiple roots but also has a simple root. Theorem 5.2. For m, n ∈ M , two splitting fields of X − mX − ( m + 3) X − andof X − nX − ( n + 3) X − over M coincide if and only if F +2 ( m, n ; X ) F − ( m, n ; X ) has a root in M . Example 5.3. We take M = Q . If ( m, n ) ∈ { ( − , , ( − , , (0 , , (5 , } then F +2 ( m, n ; X ) splits completely over Q . If ( m, n ) ∈ { ( − , , (0 , , (1 , , EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 33 (2 , , (3 , , (5 , , (12 , } then F − ( m, n ; X ) splits completely over Q .Hence we see L − = L = L = L , L = L = L , L = L , L = L , where L m = Spl Q ( X − mX − ( m + 3) X − m and n in the range − ≤ m < n ≤ F +2 ( m, n ; X ) F − ( m, n ; X ) has a linearfactor over Q only for the values of ( m, n ) noted above.In the case of char k = 2 , 3, we obtain by using (22) that F +2 ( m, n ; X ) and g + ( m, n ; X ) := X + 3( mn + 6 m − n + 9)2 mn + 3 m + 3 n + 18 X − mn − m + 6 n + 9)2 mn + 3 m + 3 n + 18 X − M . We also see that F − ( m, n ; X ) and g − ( m, n ; X ) := X + 3( mn − m − n − mn + 3 m + 3 n − X − mn + 6 m + 6 n + 9)2 mn + 3 m + 3 n − X − M .Putting Z = ( X − / ( X + 2) we have X = − (2 Z + 1) / ( Z − 1) and h + ( m, n ; Z ) := 13 ( m − n ) g + (cid:16) m, n ; − (2 Z + 1) Z − (cid:17) = Z − mn + 3 n + 9 m − n Z − mn + 3 m + 9 m − n Z − ,h − ( m, n ; Z ) := − ( m + n + 3) g − (cid:16) m, n ; − (2 Z + 1) Z − (cid:17) = Z + mn + 3 m + 3 nm + n + 3 Z + mn − m + n + 3 Z − . We also obtainDisc Z ( h + ( m, n ; Z )) = ∆ a ∆ b ( m − n ) , Disc Z ( h − ( m, n ; Z )) = ∆ a ∆ b ( m + n + 3) . In the case of char k = 3, using the result in Subsection 4.4, we directly seethat F +2 ( m, n ; X ) (resp. F − ( m, n ; X )) and h + ( m, n ; Z ) (resp. h − ( m, n ; Z )) areTschirnhausen equivalent over M . Therefore we get the following theorem whichis an analogue to the results of Morton [Mor94] and Chapman [Cha96]. Theorem 5.4. Assume that char k = 2 . For m, n ∈ M , two splitting fields of X − mX − ( m + 3) X − and of X − nX − ( n + 3) X − over M coincide ifand only if there exists z ∈ M such that either n = m ( z − z − − z ( z + 1) mz ( z + 1) + z + 3 z − or n = − m ( z + 3 z − 1) + 3( z − z − mz ( z + 1) + z + 3 z − . Proof. We should check only the case of ( m − n )( m + n + 3)(2 mn + 3 m + 3 n +18)(2 mn + 3 m + 3 n − 9) = 0. If we take z = 0 then we have n = m or n = − m − z = 1 then we get n = − m + 6) / (2 m + 3) or n = − m − / (2 m + 3)which corresponds to 2 mn + 3 m + 3 n + 18 = 0 or 2 mn + 3 m + 3 n − (cid:3) By applying Hilbert’s irreducibility theorem and Siegel’s theorem for curves ofgenus 0 (cf. [Lan78, Theorem 6.1], [Lan83, Chapter 8, Section 5], [HS00, TheoremD.8.4]) to Theorem 5.4 respectively, we get the following corollaries: Corollary 5.5. If M is a Hilbertian field then for a fixed m ∈ M there existinfinitely many n ∈ M such that two splitting fields of X − mX − ( m + 3) X − and of X − nX − ( n + 3) X − over M coincide. Corollary 5.6. Let M be a number field and O M the ring of integers in M . Fora given integer m ∈ O M , there exist only finitely many integers n ∈ O M such thattwo splitting fields of X − mX − ( m + 3) X − and of X − nX − ( n + 3) X − over M coincide.Proof. We may apply Siegel’s theorem to Theorem 5.4 because the discriminantof the denominator mz ( z + 1) + z + 3 z − z + ( m + 3) z + mz − 1) of theequalities in Theorem 5.4 is given as ( m + 3 m + 9) . (cid:3) Some sextic generic polynomials Assume that char k = 3. Let H and H be subgroups of S . Let k ( s, t ) bethe rational function field over k with variables s, t . We take a k -generic poly-nomial f ( a ; X ) ∈ k ( s )[ X ] for H with one parameter s and a k -generic polyno-mial f ( b ; X ) ∈ k ( t )[ X ] for H also with one parameter t where a ∈ k ( s ) and b ∈ k ( t ) . Assume that g ( H ,H ) ( s, t ; X ) := F ( a , b ; X ) has no multiple root. Then,by Theorem 3.10, g ( H ,H ) ( s, t ; X ) is a k -generic polynomial for H × H with twoparameters s, t . Note that there exists no Q -generic polynomial with one param-eter except for the groups { } , C , C , S (cf. [BR97], [Led07], [CHKZ]). If wetake ( H , H ) ∈ { ( S , S ), ( S , C ), ( S , C ), ( S , { } ), ( C , C ) } , then it followsfrom Theorem 4.5 that g ( H ,H ) ( s, t ; X ) is irreducible over k ( s, t ) because we have L a ∩ L b = k ( s, t ). Hence H × H can be regarded as a transitive subgroup of S naturally.(1) The case of ( H , H ) = ( S , S ). We take a = (0 , s, − s ) , b = (0 , t, − t ). Thenwe have f ( a ; X ) = X + sX + s, f ( b ; X ) = X + tX + t , A a B b − A b D a = − s t (4 st + 27 s + 27 t ), and g ( S , S ) ( s, t ; X ) := 13 F (0 , s, − s, , t, − t ; 3 X )= X + 2 ts (4 s + 27) X + ts (4 s + 27) X + t s (4 s + 27) X + t s (4 s + 27) X + ( s − t ) t s (4 s + 27) ; EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 35 this is a k -generic polynomial for S × S ⊂ S .(2) The case of ( H , H ) = ( S , C ). We take a = (0 , s, − s ) , b = ( t, − t − , f ( a ; X ) = X + sX + s, f ( b ; X ) = X − tX − ( t + 3) X − A a B b − A b D a = 729 s ( t + 3 t + 9) ( t + 3 t + 9 + s ), and g ( S ,C ) ( s, t ; X ) := F (0 , s, − s, t, − t − , X )= X − t + 3 t + 9) s (4 s + 27) X − (2 t + 3)( t + 3 t + 9) s (4 s + 27) X + 9( t + 3 t + 9) s (4 s + 27) X + 3(2 t + 3)( t + 3 t + 9) s (4 s + 27) X + ( t + 3 t + 9) (4 st + 27 t + 12 st + 9 s + 81 t + 243) s (4 s + 27) ;this is a k -generic polynomial for S × C ∼ = C ≀ C ∼ = ( C × C ) ⋊ C ⊂ S .After specializing ( s, t ) ( a, b ) ∈ M , we see that if b + 3 b + 9 + a = 0 then X + aX + a and X − bX − ( b + 3) X − M .More precisely, X − ( b + 3 b + 9) X − ( b + 3 b + 9) and X − bX − ( b + 3) X − M . We also see that if 4 ab + 27 b + 12 ab + 9 a +81 b + 243 = 0 then X + aX + a and X − bX − ( b + 3) X − M . Hence we see X − b + 3 b + 9)(2 b + 3) X − b + 3 b + 9)(2 b + 3) and X − bX − ( b + 3) X − M . Here the equivalence is given by an affineform.(3) The case of ( H , H ) = ( S , C ). We take a = (0 , s, − s ) , b = (0 , − t, f ( a ; X ) = X + sX + s, f ( b ; X ) = X ( X − t ), A a B b − A b D a =729 s t (4 s + 27), and g ( S ,C ) ( s, t ; X ) := 13 F (0 , s, − s, , − t, 0; 3 X )= X − ts (4 s + 27) X + t s (4 s + 27) X + t s (4 s + 27) ;this is a k -generic polynomial for S × C ∼ = D ⊂ S where D is the dihedralgroup of order 12.(4) The case of ( H , H ) = ( S , { } ). Assume that char k = 2. We take a =(0 , s, − s ) , b = (0 , − , f ( a ; X ) = X + sX + s, f ( b ; X ) = X ( X + 1)( X − and A a B b − A b D a = 729 s (4 s + 27). We obtain g ( S , { } ) ( s, t ; X ) := 13 F (0 , s, − s, , − , 0; 3 X )= X − s (4 s + 27) X + 1 s (4 s + 27) X + 1 s (4 s + 27) ;this is a k -generic polynomial for S ⊂ S . We also obtain the following k -genericpolynomial for S with one parameter s : h S ( s ; X ) := ( s (4 s + 27)) g ( S , { } ) (cid:16) s, t ; Xs (4 s + 27) (cid:17) = X − s (4 s + 27) X + s (4 s + 27) X + s (4 s + 27) . Two polynomials f ( a ; X ) = X + sX + s and h S ( s ; X ) have the same splittingfield over k ( s ).(5) The case of ( H , H ) = ( C , C ). We take a = ( s, − s − , , b = (0 , − t, f ( a ; X ) = X − sX − ( s + 3) X − , f ( b ; X ) = X ( X − t ), A a B b − A b D a = − t ( s + 3 s + 9) , and g ( C ,C ) ( s, t ; X ) := F ( s, − s − , , , − t, X )= X − ts + 3 s + 9 X + 9 t ( s + 3 s + 9) X − (2 s + 3) t ( s + 3 s + 9) ;this is a k -generic polynomial for C × C ∼ = C ⊂ S .In the case of char k = 3, we should use F ( a , b ; X ) instead of F ( a , b ; X ) becauseof the result in Subsection 4.4. Here we give only g ( H ,H ) ( s, t ; X ) := F ( a , b ; X )for each ( H , H ) in the case of char k = 3: g ( S , S ) (1 /s, t ; X ) = X − tX − tX + t X − t X − t ( st − ,g ( S ,C ) (1 /s, t ; X ) = X + tX + t ( t + 1) X + ( st − t + 1) X − t ( st − t + 1) X − t ( st + 1) X + s t + st − st + 1 ,g ( S ,C ) (1 /s, t ; X ) = X − tX + t X + st ,g ( S , { } ) (1 /s, t ; X ) = X − X + X + s,g ( C ,C ) (1 /s, t ; X ) = X + t s + s + 1 X + s t s + s + 1 . Acknowledgment. We would like to thank the referee for suggesting many im-provements and corrections to the manuscript. We also thank him/her for providingan improved statement and an alternative proof of Lemma 3.4. EOMETRIC FRAMEWORK FOR THE SUBFIELD PROBLEM 37 References [Ber76] E. R. Berlekamp, An analog to the discriminant over fields of characteristic two , J.Algebra (1976), 315–317.[BJY86] A. A. Bruen, C. U. Jensen, N. Yui, Polynomials with Frobenius groups of prime degreeas Galois Groups II , J. Number Theory (1986), 305–359.[BR97] J. 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