OJDMOpen Journal of Discrete Mathematics2161-7635Scientific Research Publishing10.4236/ojdm.2018.83007OJDM-85772ArticlesPhysics&Mathematics On Tate’s Proof of a Theorem of Dedekind ShivGupta1*Department of Mathematics, West Chester University, West Chester, PA, USA* E-mail:vnitica@wcupa.edu090520180803737813, March 201829, June 2018 2, July 2018© Copyright 2014 by authors and Scientific Research Publishing Inc. 2014This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

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

Galois Group of a Polynomial
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,  and we shall keep the notations used in that proof.

Theorem 1 Let f ( x ) ∈ ℤ [ 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 ⊂ S n be the Galois group of f ( x ) over the field ℚ of rational numbers. Suppose that f p = f ¯ = f ( mod p ) ∈ ℤ p [ x ] factors as:

f p = f ¯ = ∏ i = 1 r   f i ¯

where f i ¯ are distinct monic irreducible polynomials in ℤ p [ x ] , degree ( f i ¯ ) = d i , 1 ≤ i ≤ r , and d 1 + 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 d 1 , d 2 , ⋯ , d r .

2. Preliminary Results

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

1) Let F be a field and f ( x ) ∈ F [ 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 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 E / F be an algebraic extension. Then any subring of E containing F is a subfield of E . Proof: Let K be a ring such that F ⊂ K ⊂ E . Let α ∈ K − F . As α is algebraic over F , F ( α ) = F [ α ] . So α − 1 ∈ F ( 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 χ 1 , χ 2 , ⋯ , χ n are distinct characters of a monoid into a field F , then the only elements a i ∈ F , 1 ≤ i ≤ n , such that

a 1 χ 1 ( h ) + a 2 χ 2 ( h ) + ⋯ + a n χ n ( h ) = 0

for all h ∈ H are a i = 0 , 1 ≤ i ≤ n .

6) Let p be a prime and G F ( p m ) be a finite field with p m elements. Then the group Aut ( G F ( p m ) ) = 〈 σ 〉 is cyclic of order m and the generating automorphism σ maps α ∈ G F ( 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 ∈ S n . Then σ and η − 1 σ η have same cyclic structure.

Let f ( x ) ∈ ℤ [ x ] be a polynomial of degree n ≥ 1 , and p a prime number. Then f p ( x ) ∈ ℤ p [ x ] will denote the polynomial obtained by reducing the coefficients of f ( x ) modulo p.

Theorem 2 Let f ( x ) ∈ ℤ [ 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 E be a splitting field of f ( x ) over ℚ . Let E p be a splitting field of f p ( x ) over ℤ p = ℤ / ( p ) . Let

f ( x ) = ( x − r 1 ) ( x − r 2 ) ⋯ ( x − r n ) ,   r i ∈ E ⊂ ℂ ,1 ≤ i ≤ n

R = { r 1 , r 2 , ⋯ , r n } , R p = { r 1 ¯ , r 2 ¯ , ⋯ , r n ¯ } ⊂ E p

where r i ¯ , 1 ≤ i ≤ n are the roots of f p ( x ) ∈ ℤ p [ x ] and

E = ℚ ( r 1 , r 2 , ⋯ , r n ) ,       E p = ℤ p ( r 1 ¯ , r 2 ¯ , ⋯ , r n ¯ )

Let D = ℤ [ r 1 , r 2 , ⋯ , r n ] be the subring generated by the roots of r 1 , r 2 , ⋯ , r n of f ( x ) in ℂ . Then

• 1) There exists a homomorphism ψ of D onto E p .

• 2) Any such homomorphism ψ gives a bijection of the set R of the roots of f ( x ) in E onto the set R p of the roots of the f p ( x ) in E p .

• 3) If ψ and ψ ′ are two such homomorphisms then there exist σ ∈ A u t ( E / ℚ ) = G a l ( 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:

E = ℚ ( r 1 , r 2 , ⋯ , r n ) = ℚ [ r 1 , r 2 , ⋯ , r n ]

We claim that D = ℤ [ r 1 , r 2 , ⋯ , r n ] is a finitely generated (additive) Abelian group. Since each r i is a root of the monic polynomial f ( x ) ∈ ℤ [ x ] of degree n any positive power of r i , 1 ≤ i ≤ n can be expressed as an integral linear combination of 1, r i , r i 2 , ⋯ r i n − 1 . It follows that

D = ∑ 0 ≤ e i ≤ n − 1 ℤ r 1 e 1 r 2 e 2 ⋯ r n e n .

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 ⊂ ℂ , none of these cyclic groups is finite. So D is a direct sum of finitely many infinite cyclic groups. Let { u 1 , u 2 , ⋯ , u N } be a set consisting of an independent generating system of D. We have

D = ℤ u 1 ⊕ ℤ u 2 ⊕ ⋯ ⊕ ℤ u N ,   N ≤ n n .

We claim that { u 1 , u 2 , ⋯ , u N } is a basis of E / ℚ . Obviously { u 1 , u 2 , ⋯ , u N } is linearly independent over ℚ . Let ℚ D = ∑ 1 ≤ i ≤ N ℚ u i . Then ℚ D is a ring and ℚ ⊂ ℚ D ⊂ E therefore ℚ D is a field. Since r i ∈ D for 1 ≤ i ≤ n , by (4) ℚ D = E and { u 1 , u 2 , ⋯ , u N } is a basis of E / ℚ . As D = ℤ u 1 ⊕ ℤ u 2 ⊕ ⋯ ⊕ ℤ u N ,

p D = ℤ ( p u 1 ) ⊕ ℤ ( p u 2 ) ⊕ ⋯ ⊕ ℤ ( p u N )

is an ideal of D and

D / p D = { a 1 u 1 + a 2 u 2 + ⋯ + a N u N ¯ : 0 ≤ a i ≤ p − 1 } .

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

D / M ≈ D / p D M / p D

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 ℤ ⊂ M . Therefore ν ( ℤ ) = ℤ p . We note that as D = ℤ [ r 1 , r 2 , ⋯ , r n ] we have for 1 ≤ i ≤ n

ν ( r i ) = r i + M = r i ¯ ,   ν ( D ) = ℤ p [ r 1 ¯ , r 2 ¯ , ⋯ , r n ¯ ]

As ν is an epimorphism we have

ν ( D ) = D / M = ℤ p [ r 1 ¯ , r 2 ¯ , ⋯ , r n ¯ ]

is a splitting field of f p ( x ) over ℤ p . As both D / M and E p are splitting fields of f p ( x ) over ℤ p they are isomorphic. Let

ϕ : D / M → ℤ p

be such an isomorphism. Then ψ = ϕ ⋅ ν is a homomorphism of D onto E p .

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

f ( x ) = ( x − r 1 ) ( x − r 2 ) ⋯ ( x − r n )

ψ ( f ( x ) ) = f p ( x ) = ( x − ψ ( r 1 ) ) ( x − ψ ( r 2 ) ) ⋯ ( x − ψ ( r n ) ) ,

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

ψ | R : { r 1 , r 2 , ⋯ , r n } → { r 1 ¯ , r 2 ¯ , ⋯ , r n ¯ }

is a bijection of the set R of roots of f ( x ) in E to the set R p of the roots of f p ( x ) in E p .

3) We have seen that given a homomorphism ψ : D → E p , and σ ∈ G a l ( f ) = A u t ( E / ℚ ) , ψ ′ = ψ ⋅ σ is also a homomorphism from D to E p . We note that the restriction of σ ∈ A u t ( E / ℚ ) to D = ℤ [ r 1 , r 2 , ⋯ , r n ] is also an automorphism of the ring D. Since [ E : ℚ ] = N , the group A u t ( E / ℚ ) has order N. Let

A u t ( E / ℚ ) = { σ 1 , σ 2 , ⋯ , σ N }

So given a non-trivial homomorphism ψ : D → E p , we get N distinct homomorphisms ψ j = ψ ⋅ σ j , 1 ≤ j ≤ N , from D to E p . We claim that these are all the homomorphisms from D to E p . Suppose that there is a homomorphism from D to E 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 E p is linearly independent over the field E p .

Consider the following system of N homogeneous equations in N + 1 variables { x 1 , x 2 , ⋯ , x N , x N + 1 } ,

∑ i = 1 N + 1   x i ψ i ( u j ) = 0 ,   1 ≤ j ≤ N .

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

∑ i = 1 i = N + 1 a i ψ i ( u j ) = 0 ,   1 ≤ j ≤ N .

Let y ∈ D = ℤ u 1 ⊕ ℤ u 2 ⊕ ⋯ ⊕ ℤ u N . So y = n 1 u 1 + n 2 u 2 + ⋯ + n N u N , n k ∈ ℤ , 1 ≤ k ≤ N . Then for 1 ≤ i ≤ N + 1 we have

ψ i ( y ) = n 1 ¯ ψ i ( u 1 ) + n 2 ¯ ψ i ( u 2 ) + ⋯ + n N ¯ ψ i ( u N ) = ∑ j = 1 j = N n j ¯ ψ i (uj)

where n j ¯ = n j + ( p ) . We shall show that

∑ i = 1 i = N + 1 a i ψ i ( y ) = 0 ,

which will contradict the linear independence of { ψ 1 , ψ 2 , ⋯ , ψ N , ψ N + 1 } over E p .

∑ i = 1 i = N + 1 a i ψ i ( y ) = ∑ i = 1 i = N + 1 a i ∑ j = 1 j = N n j ¯ ψ i ( u j ) = ∑ i = 1 i = N + 1 a i ( n 1 ¯ ψ i ( u 1 ) + n 2 ¯ ψ i ( u 2 ) + ⋯ + n N ¯ ψ i ( u N ) ) = n 1 ¯ ∑ i = 1 i = N + 1 a i ψ i ( u 1 ) + n 2 ¯ ∑ i = 1 i = N + 1 a i ψ i ( u 2 ) + ⋯ + n N ¯ ∑ i = 1 i = N + 1 a i ψ i ( u N ) = 0.

3. Proof of the Main Theorem

Since the field E p has order p m , the group A u t ( E p ) has order m and π : E p → E p where π ( a ) = a p for all a ∈ E p , is the generating automorphism of A u t ( E p ) . So if ψ : D → E p is any homomorphism then so is π ⋅ ψ . Since ψ and π ⋅ ψ are two homomorphisms from D to E p there exist σ ∈ A u t ( E / ℚ ) such that π ⋅ ψ = ψ ⋅ σ or ψ − 1 ⋅ π ⋅ ψ = σ . This proves that the action on σ on { r 1 , r 2 , ⋯ , r n } is similar to the action of π on { r 1 ¯ , r 2 ¯ , ⋯ , r n ¯ } . Note: In the following diagram the mapping

D → σ D

is the restriction of σ ∈ A u t ( E / ℚ ) to D and we are only concerned with the effect of the mappings σ , ψ and π on { r 1 , r 2 , ⋯ , r n } and { r 1 ¯ , r 2 ¯ , ⋯ , r n ¯ } . Clearly

{ r 1 , r 2 , ⋯ , r n } → σ { r 1 , r 2 , ⋯ , r n } { r 1 ¯ , r 2 ¯ , ⋯ , r n ¯ } → π { r 1 ¯ , r 2 ¯ , ⋯ , r n ¯ } { r 1 , r 2 , ⋯ , r n } → ψ { r 1 ¯ , r 2 ¯ , ⋯ , r n ¯ }

As ψ − 1 ⋅ π ⋅ ψ = σ and ψ ⋅ σ ⋅ ψ − 1 = π the effect of σ on { r 1 , r 2 , ⋯ , r n } is similar to the effect of π on { r 1 ¯ , r 2 ¯ , ⋯ , r n ¯ } . This is further illustrated by the following:

σ ( r i ) = r j ⇒ π ( r i ¯ ) = r j ¯

r i ¯ → ψ − 1 r i → σ r j → ψ r j ¯

π ( r i ¯ ) = r j ¯ ⇒ σ ( r i ) = r j

r i → ψ r i ¯ → π r j ¯ → ψ − 1 r j

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

References Jacobson, N. (2014) Basic Algebra. 2nd Edition, Dover Publications, Inc., Mine-ola, New York.