On Tate's Proof of a Theorem of Dedekind - Scientific Research

Open Journal of Discrete Mathematics, 2018, 8, 73-78 http://www.scirp.org/journal/ojdm ISSN Online: 2161-7643 ISSN Print: 2161-7635

On Tate’s Proof of a Theorem of Dedekind Shiv Gupta Department of Mathematics, West Chester University, West Chester, PA, USA

How to cite this paper: Gupta, S. (2018) On Tate’s Proof of a Theorem of Dedekind. Open Journal of Discrete Mathematics, 8, 73-78. https://doi.org/10.4236/ojdm.2018.83007

Abstract

Received: March 13, 2018 Accepted: June 29, 2018 Published: July 2, 2018

Galois Group of a Polynomial

Copyright © 2018 by author and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY 4.0). http://creativecommons.org/licenses/by/4.0/ Open Access

In this note we give a complete proof of a theorem of Dedekind.

Keywords

1. Introduction In this note we give a complete proof of the following theorem of Dedekind. Our proof is a somewhat detailed version of the one given in Basic Algebra by Jacobson, Volume I, [1] and we shall keep the notations used in that proof.

Theorem 1 Let f ( x ) ∈  [ x ] be square-free monic polynomial of degree n

and p be a prime such that p does not divide the discriminant of f ( x ) . Let G ⊂ Sn be the Galois group of f ( x ) over the field  of rational numbers. Suppose that f p= f= f ( mod p ) ∈  p [ x ] factors as: f p= f=

r

∏ fi i =1

where fi are distinct monic irreducible polynomials in  p [ x ] , degree fi = di , 1 ≤ i ≤ r , and d1 + d 2 +  + d r = n. Then there exists an automorphism σ ∈ G which when considered as a permutation on the zeros of f ( x ) is a product of disjoint cycles of lengths d1 , d 2 , , d r .

( )

2. Preliminary Results We shall assume that the reader is familiar with the following well-known results.

1) Let  be a field and f ( x ) ∈  [ x ] be a polynomial of degree n ≥ 2 . Then any two splitting fields of f ( x ) are isomorphic. 2) A finitely generated Abelian group is direct sum of (finitely many) cyclic

DOI: 10.4236/ojdm.2018.83007 Jul. 2, 2018

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groups. (This is the fundamental theorem of finitely generated Abelian groups). 3) A system of n homogeneous equation in m > n variables has a non-trivial solutions. 4) Let  /  be an algebraic extension. Then any subring of  containing

 is a subfield of  . Proof: Let K be a ring such that  ⊂ K ⊂  . Let

α ∈ K −  . As α is algebraic over  ,  (α ) =  [α ] . So α −1 ∈  (α ) ⊂ K . 5) (Dedekind’s Independence Theorem). Distinct characters of a monoid (a set with associative binary operation with an identity element) into a field are linearly independent. That is if χ1 , χ 2 , , χ n are distinct characters of a monoid into a field  , then the only elements ai ∈  , 1 ≤ i ≤ n , such that a1 χ1 ( h ) + a2 χ 2 ( h ) +  + an χ n ( h ) = 0

for all h ∈ H are ai = 0 , 1 ≤ i ≤ n . 6) Let p be a prime and GF ( p m ) be a finite field with p m elements. Then the group Aut GF p m = σ is cyclic of order m and the generating

( ( ))

automorphism σ maps α ∈ GF ( p m ) to α p . 7) If R is a commutative ring with identity and M is a maximal ideal of R then

R/M is a field. 8) Let σ ,η ∈ Sn . Then σ and η −1ση have same cyclic structure. Let f ( x ) ∈  [ x ] be a polynomial of degree n ≥ 1 , and p a prime number. Then f p ( x ) ∈  p [ x ] will denote the polynomial obtained by reducing the coefficients of f ( x ) modulo p. Theorem 2 Let f ( x ) ∈  [ x ] be a monic polynomial of degree n ≥ 1 and p be a prime number which does not divide the discriminant of f ( x ) . Let  be a splitting field of f ( x ) over  . Let  p be a splitting field of f p ( x ) over  p =  / ( p ) . Let f ( x) =

( x − r1 )( x − r2 ) ( x − rn ) ,

ri ∈  ⊂ ,1 ≤ i ≤ n

R = {r1 , r2 , , rn } , = Rp

{r , r ,  , r } ⊂  1

2

n

p

where ri , 1 ≤ i ≤ n are the roots of f p ( x ) ∈  p [ x ] and

(

 = = ( r1 , r2 , , rn ) ,  p  p r1 , r2 , , rn

)

Let D =  [ r1 , r2 , , rn ] be the subring generated by the roots of r1 , r2 , , rn of f ( x ) in  . Then • 1) There exists a homomorphism ψ of D onto  p . • 2) Any such homomorphism ψ gives a bijection of the set R of the roots of f ( x ) in  onto the set R p of the roots of the f p ( x ) in  p . • 3) If ψ and ψ ′ are two such homomorphisms then there exist σ ∈ Aut (  /  ) = Gal ( f ( x ) ) , such that ψ =′ ψ ⋅ σ . (Note that the restriction of σ to D is an automorphism of D). Proof 1) One has that:

=  = ( r1 , r2 , , rn )  [ r1 , r2 , , rn ] DOI: 10.4236/ojdm.2018.83007

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We claim that D =  [ r1 , r2 , , rn ] is a finitely generated (additive) Abelian

group. Since each ri is a root of the monic polynomial f ( x ) ∈  [ x ] of degree n any positive power of ri ,1 ≤ i ≤ n can be expressed as an integral linear combination of 1, ri , ri 2 , ri n−1 . It follows that D=

∑

r1e1 r2e2  rnen .

0≤ei ≤n −1

Therefore D is a finitely generated (additive) Abelian group generated by at most nn elements. By the Fundamental Theorem for Finitely Generated Abelian

Groups D is a direct sum of finitely many cyclic groups. Since D ⊂  , none of these cyclic groups is finite. So D is a direct sum of finitely many infinite cyclic groups. Let {u1 , u2 , , u N } be a set consisting of an independent generating system of D. We have D = u1 ⊕ u2 ⊕  ⊕ u N , We claim that

{u1 , u2 , , u N }

N ≤ nn .

is a basis of  /  . Obviously

{u1 , u2 , , u N }

is linearly independent over  . Let D = ∑1≤i≤ Nui . Then D is a ring and  ⊂ D ⊂  therefore D is a field. Since ri ∈ D for 1 ≤ i ≤ n , by (4) D =  and {u1 , u2 , , u N } is a basis of E /  . As

D = u1 ⊕ u2 ⊕  ⊕ u N ,

= pD  ( pu1 ) ⊕  ( pu2 ) ⊕  ⊕  ( pu N )

is an ideal of D and

D pD=

{a u + a u 1 1

2 2

}

+  + aN u N : 0 ≤ ai ≤ p − 1 .

Therefore the D pD is finite of order p N . Let M be a maximal ideal of D containing pD. That is pD ⊂ M ⊂ D and D M is a finite field of characteristic p and so it has a subfield isomorphic to  p =  / p which we will identify as  p in what follows. As

D M≈

D pD M pD

the order of D M is p m , 1 ≤ m ≤ N . Consider the canonical epimorphism

ν :D→D M whose kernel is M and p ⊂ M . Therefore ν (  ) =  p . We note that as D =  [ r1 , r2 , , rn ] we have for 1 ≤ i ≤ n

ν ( ri ) =ri + M =ri , ν ( D ) = p  r1 , r2 , , rn  As ν is an epimorphism we have

ν= M  p  r1 , r2 , , rn  ( D ) D= is a splitting field of f p ( x ) over  p . As both D M and  p are splitting fields of f p ( x ) over  p they are isomorphic. Let

φ : D M → p be such an isomorphism. Then ψ = φ ⋅ν is a homomorphism of D onto  p . DOI: 10.4236/ojdm.2018.83007

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2) Let ψ : D →  p be a homomorphism. So ψ (1) = 1 . As  ⊂ D , and  p has characteristic p, ψ ( p ) = 0 , so ψ (  = )  p ⊂  p . ψ can be extended to a homomorphism of the polynomial rings D [ x ] →  p [ x ] . Under this mapping f ( x ) → f p ( x ) . As

f ( x) = ( x − r1 )( x − r2 ) ( x − rn )

ψ ( f ( x )) = f p ( x) = ( x −ψ ( r1 ) ) ( x −ψ ( r2 ) ) ( x −ψ ( rn ) ) ,

ψ ( ri ) ,1 ≤ i ≤ n are the roots of the f p ( x ) in  p and therefore the restriction of ψ to R

{

ψ | R : {r1 , r2 , , rn } → r1 , r2 , , rn

}

is a bijection of the set R of roots of f ( x ) in  to the set R p of the roots of f p ( x ) in  p . 3) We have seen that given a homomorphism ψ : D →  p , and σ ∈ Gal ( f ) = Aut (  /  ) , ψ =′ ψ ⋅ σ is also a homomorphism from D to  p . We note that the restriction of σ ∈ Aut (  /  ) to D =  [ r1 , r2 , , rn ] is also an automorphism of the ring D. Since [  : ] = N , the group Aut (  /  ) has order N. Let

Aut (  /  ) = {σ 1 , σ 2 , , σ N } So given a non-trivial homomorphism ψ : D →  p , we get N distinct homomorphisms ψ j= ψ ⋅ σ j , 1 ≤ j ≤ N , from D to  p . We claim that these are all the homomorphisms from D to  p . Suppose that there is a homomorphism from D to  p which is different from ψ j , 1 ≤ j ≤ N . Let us denote it by ψ N +1 . By Dedekind Independence Theorem the set {ψ 1 ,ψ 2 , ,ψ N ,ψ N +1} of N + 1 homomorphisms from D to  p is linearly independent over the field  p . Consider the following system of N homogeneous equations in N + 1 variables { x1 , x2 , , xN , xN +1} , N +1

∑ xiψ i ( u = j)

0, 1 ≤ j ≤ N .

i =1

Since there are more variables than the equations this system of equations has ai ∈  p , 1 ≤ i ≤ N + 1 . a non-trivial solution. Let this non-trivial solution be x= i So we have i= N +1

∑ aiψ i ( u = j)

0, 1 ≤ j ≤ N .

i =1

Let y ∈ D = u1 ⊕ u2 ⊕  ⊕ u N . So y = n1u1 + n2u2 +  + nN u N , nk ∈  , 1 ≤ k ≤ N . Then for 1 ≤ i ≤ N + 1 we have

ψ i= uN ) ( y ) n1ψ i ( u1 ) + n2ψ i ( u2 ) +  + nNψ i (=

j=N

∑ n jψ i ( u j ) j =1

n j + ( p ) . We shall show that where n= j i= N +1

∑ aiψ i ( y ) = 0, i =1

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which will contradict the linear independence of {ψ 1 ,ψ 2 , ,ψ N ,ψ N +1} over

p .

=i N +1

∑ aiψ i ( y ) i =1

=

=i N +1

j=N

∑ ai ∑ n jψ i ( u j )

=i 1 =j 1

=

=i N +1

∑ ai ( n1ψ i ( u1 ) + n2ψ i ( u2 ) +  + nNψ i ( uN ) ) i =1

= n1

i= N +1

i= N +1

i= N +1

∑ aiψ i ( u1 ) + n2 ∑ aiψ i ( u2 ) +  + nN ∑ aiψ i ( uN )

=i 1 =i 1

=i 1

= 0.

3. Proof of the Main Theorem Since the field  p has order p m , the group Aut (  p ) has order m and

π : p → p

where

π (a) = a p

for

a∈p ,

all

is

the

generating

automorphism of Aut (  p ) . So if ψ : D →  p is any homomorphism then so

is π ⋅ψ . Since ψ and π ⋅ψ are two homomorphisms from D to  p there

exist σ ∈ Aut (  /  ) such that π ⋅ψ = ψ ⋅ σ or ψ −1 ⋅ π ⋅ψ = σ . This proves

{r1 , r2 , , rn }

that the action on σ on

is similar to the action of π on

{r , r ,, r } . Note: In the following diagram the mapping 1

2

n

σ →D D 

is the restriction of σ ∈ Aut (  /  ) to D and we are only concerned with the

{r1 , r2 , , rn }

effect of the mappings σ , ψ and π on

and

Clearly

{r , r ,  , r } . 1

2

n

σ → {r1 , r2 , , rn } {r1 , r2 , , rn } 

{r , r ,, r } →{r , r ,, r } {r , r , , r } → {r , r , , r } 1

2

n

1

2

n

π

ψ

1

2

n

1

2

n

As ψ −1 ⋅ π ⋅ψ = σ and ψ ⋅ σ ⋅ψ −1 = π the effect of σ on similar to the effect of π on following:

{r1 , r2 , , rn }

is

{r , r ,, r } . This is further illustrated by the 1

2

n

()

rj ⇒ π ri = rj σ ( ri ) = −1

ψ σ ψ ri  → ri  → rj  → rj

()

π ri = rj ⇒ σ ( ri ) = rj −1

ψ π ψ ri  → ri  → rj  → rj

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References [1]

DOI: 10.4236/ojdm.2018.83007

Jacobson, N. (2014) Basic Algebra. 2nd Edition, Dover Publications, Inc., Mineola, New York.

78
