Let K / ℚ be a Galois extension of ℚ of degree n. The question we ask ourselves is the following: Do such Galois extensions exist, which are free of inert primes? To our knowledge, there are no studies on such a property, although we have many theorems on ramification, which could allow in certain cases to seriously study this problem. The question is relevant, because apart from the arithmetic description of the prime ideals of ℤ K , this kind of results can facilitate the calculation of the number of classes of K.

All the reminders and definitions that follow can be found in [1].

- Let us recall, that it is well known that the ring of integers ℤ K of K, is a free ℤ -module of rank n = [ ℤ K : ℤ ] , moreover it is a Dedekind ring, and therefore, ∀ p a prime of ℕ , then:

n = e f g (1)

where: p ℤ K = P 1 e ⋯ P g e with P i / p being all prime ideals of ℤ K lying above p, and that they are in number g = g K / ℚ , where e = e K / ℚ is the ramification index of the P i , and f = f K / ℚ = [ ℤ K ∕ P i : F p ] , ∀ i = 1, ⋯ , g , is the residual degree of the P i / p .

Definition 1.1. The notations remaining the same, we have:

1) The prime number p ∈ ℕ ,is said to be inert in K ⇔ ∃ P a prime ideal of K such that p ℤ K = P ⇔ f = n , e = 1 and g = 1 .

2) The prime number p ∈ ℕ is said to be ramified in K ⇔ e ≥ 2 ⇔ p / d i s c r ( K ) .

Definition 1.2. The Galois number field K is said to be monogenic ⇔ ∃ θ ∈ ℤ K such that ℤ K = ℤ [ θ ] ⇔ { 1, θ 2 , ⋯ , θ n − 1 } forms a ℤ -basis (called power basis) of the ℤ -module ℤ K .

Notations 1.1. When K is monogenic, then in particular K = ℚ [ θ ] . Let denotes by f θ ( X ) or simply f ( X ) , the irreducible polynomial of θ , then:

f ( X ) = X n + a n − 1 X n − 1 + a n − 2 X n − 2 + ⋯ + a 1 X + a 0 ∈ ℤ [ X ] ;

And ∀ p a prime number of ℕ , we note

f ¯ p ( X ) = X n + a n − 1 ¯ X n − 1 + ⋯ + a 1 ¯ X + a 0 ¯ ∈ F p [ X ] , and by f ¯ p ( X ) = ∏ i = 1 g ( f ¯ i , p ( X ) ) e , its decomposition into irreducible factors product of F p [ X ] :

Where: f i , p ( X ) = X n i + b i , n i − 1 X n 1 − 1 + ⋯ + b i ,1 X + b i ,0 ∈ ℤ [ X ] ,and f ¯ i , p ( X ) = X n i + b i , n i − 1 ¯ X n 1 − 1 + b i , n i − 2 ¯ X n i − 2 + ⋯ + b i ,1 ¯ X + b i ,0 ¯ ∈ F p [ X ] ,are irreduciblesin F p [ X ] .

Definition 1.3. We denote by M = { K = ℚ ( d m , d n ) ∕ ∃ θ ∈ ℤ K   such   thay   ℤ K = ℤ [ θ ] } , the set of monogenic biquadratic number fields. (This set has infinite cardinality cf. [2]);

We have the following well-known important proposition, the notations being the same:

Proposition 1.1. Let K be a monogenic Galois number field and θ ∈ ℤ K ∕ ℤ K = ℤ [ θ ] , then ∀ p a prime number of ℕ , we have:

p ℤ K = p ℤ [ θ ] = ∏ i = 1 g ( p ℤ [ θ ] + f i , p ( θ ) ℤ [ θ ] ) e .

As a result we have the following corollary.

Corollary 1.1. Let K be a monogenic Galois number field. If for all prime numbers p ∈ ℕ , the polynomial f ¯ p ( X ) is decomposed in F p [ X ] . (i.e. is not irreducible), then there does not exist any prime number p ∈ ℕ , which is inert in K.

In this paper, from the introduction, we ask whether there exist number fields K, such that ∀ p , a prime integer, p is never inert in K. The answer is positive for small degree 4, since we have found an infinite family of such degree, for which all its fields satisfy this property.

From a general point of view, and as a relevant issue, one can look for other families of fields where this property remains true, those could be Galois or not.

The main goal of this article, is to show Theorem 3.1., which proves the existence of an infinite family of Galoisian numbers fields, namely M (cf. Definition 1.3.), having no inert prime integer p, when considering each field K ∈ M .

To proove this, we show that all fields belonging to the family M , are such that the irreducible polynomials associated with their monogenic element, satisfy Proposition 3.1.; then as a result, the Proposition 1.1. is true, but with necessarily g ≠ 1 , which established the seeked result.

Thus, in the family M , the equation: 4 = e f g , cf. (1), is always checked with necessarily f ≠ 4 .

Let’s take f ( X ) = X 4 + a 3 X 3 + a 2 X 2 + a 1 X + a 0 ∈ ℤ [ X ] an irreducible normalized integer polynomial, and for any prime number p, and let’s consider the map α p ¯ (cf. Lemma 2.2.) and denotes its restriction morphisme map by

α p ¯ | ℤ [ X ] : ℤ [ X ] → F p [ X ] , defined by: ( α p ¯ | ℤ [ X ] ( f ) ) ( X ) = ( α p ¯ ( f ) ) ( X ) = f ¯ p ( X )

= X 4 + a 3 ¯ X 3 + a 2 ¯ X 2 + a 1 ¯ X + a 0 ¯ ∈ F p [ X ] .

Let’s put Σ f for the splitting field of the polynomial f ( X ) . Then it is known that Σ f can be (cf. [3]): either a biquadratic field; or a cyclic quartic field (with a single sub-quadratic field); or a non-galoisian quartic field (with a single sub-quadratic field); or to finish a non-galoisian quartic field, without any subquadratic fields. In our case we will deal with Σ f when it belongs to a biquadratic field type.

Notations 2.1. For a square free integer a ≠ 0,1 , let’s denotes by ℤ ℚ ( a ) the integral ring of ℚ ( a ) . It is well known that:

ℤ ℚ ( a ) = ℤ [ a ∗ ] = { ℤ [ 1 + a 2 ]     if   a ≡ 1 ( mod 4 ) ℤ [ a ]     if   a ≡ − 1   or   2 ( mod 4 ) . ,

And that the conjugate of a ∗ is a ∗ ¯ = { 1 − a 2 = 1 − 1 + a 2 ,     if   a ≡ 1 ( mod 4 ) − a ,     if   a ≡ − 1   or   2 ( mod 4 ) . .

Let’s recall that for a prime ideal p of ℤ ℚ ( a ) over a prime p, we get

( ℤ ℚ ( a ) ) ∕ p = F p [ a ∗ ] , where the element a ∗ is a root of X 2 − X + 1 − a 4 ∈ ℤ [ X ] if a ≡ 1 ( mod 4 ) (resp. X 2 − a ∈ ℤ [ X ] , if a ≡ − 1 or 2 (mod 4)).

As a consequence:

Proposition 2.1. For a square free integer a ≠ 0,1 , and a rationnal prime integer p/a, if we take a look at F p [ a ∗ ] , we have:

F p ∋ a ∗ = { 2 − 1 ;     if   a ≡ 1 ( mod 4 ) ;   because   X 2 − X + 4 − 1 = ( X − 2 − 1 ) 2 ; 0 F p ;     if   a ≡ − 1   or   2   ( mod 4 ) .

Note that we have a canonical written form for ℚ ( d m , d n ) , which we will use in this paper (cf. [2]):

Remarks 2.1. In the following we’ll use the unique written form for a biquadratic field ℚ ( d m , d n ) , that is:

1) d , m , n ∈ ℤ ∗ ,square free and pairwise relatively prime,and d > 0 , m > n odds, d m , d n , m n ≠ 1 , d m ≡ d n ≡ − 1,1 or 2(mod 4),and in the case where d m ≡ d n ≡ 1 ( mod 4 ) we put d < I n f ( | m | , | n | ) .

In addition,in the other cases,we define δ = 0 or 1such that m n ≡ ( − 1 ) δ ( mod 4 ) ,when d m ≡ d n ≡ − 1 or 2(mod 4).That is δ = 0 when d m ≡ d n ≡ − 1 ( mod 4 ) ,and δ = 1 when d m ≡ d n ≡ 2 ( mod 4 ) .

2)When α 0 ∈ ℤ ℚ ( d m , d n ) (the ring of integer elements of the field

ℚ ( d m , d n ) ),we will consider its conjugates under the Galois group, G = { σ 0 , σ 1 , σ 2 , σ 3 } ,where the action on the basis of K: ε = { ε 0 = 1 , ε 1 = d m , ε 2 = d n , ε 3 = m n } is given by the matrix:

( a i j ) 0 ≤ i , j ≤ 3 = ( σ i ( ε j ) / ε j ) 0 ≤ i , j ≤ 3 = ( 1 1 1 1 1 1 − 1 − 1 1 − 1 1 − 1 1 − 1 − 1 1 ) .

There let’s make some other remarks and lemmas we’ll need further:

Under these notations we get the following main theorem:

Proposition 3.1. Let f ( X ) = X 4 + a 3 X 3 + a 2 X 2 + a 1 X + a 0 ∈ ℤ [ X ] be an irreducible polynomial, and Σ f its splitting fields. Then there is an implication from proposition (i) to proposition (ii):

(i) Σ f = ℚ ( d m , d n ) .

(ii) For every prime number p ∈ ℕ , the polynomial f p ¯ ( X ) = X 4 + a 3 ¯ X 3 + a 2 ¯ X 2 + a 1 ¯ X + a 0 ¯ , is always reducible in F p [ X ] .

Proof 3.1. • Let’s show that (i) ⇒ (ii).

The hypothesis (i)implies that f ( X ) = ∏ i = 0 3 ( X − α i ) ,with the α i ∈ ℤ ℚ ( d m , d n ) .

Let’s solve simultaneously,the both cases d m ≡ d n ≡ 1 ( mod 4 ) and d m ≡ d n ≡ − 1 or 2(mod 4).

So from Proposition 2.2.&2.3.,we know that there exists A 0 , A 1 , A 2 , A 3 ∈ ℚ such that:

α 0 = A 0 + A 1 d m + A 2 d n + A 3 m n ,

α 1 = A 0 + A 1 d m − A 2 d n − A 3 m n ,

α 2 = A 0 − A 1 d m + A 2 d n − A 3 m n ,

α 3 = A 0 − A 1 d m − A 2 d n + A 3 m n .

Let’s consider the following six polynomials,having noticed that:

( α 0 + α 1 , α 0 α 1 ) and ( α 2 + α 3 , α 2 α 3 ) ∈ ℤ ℚ ( d m ⋆ ) ,

( α 0 + α 2 , α 0 α 2 ) and ( α 1 + α 3 , α 1 α 3 ) ∈ ℤ ℚ ( d n ⋆ ) ,And

( α 0 + α 3 , α 0 α 3 ) and ( α 1 + α 2 , α 1 α 2 ) ∈ ℤ ℚ ( m n ⋆ ) :

• g 1 ( X ) = ( X − α 0 ) ( X − α 1 ) ∈ ℤ ℚ ( d m ⋆ ) [ X ] = X 2 − 2 ( A 2 + A 1 d m ) X + ( A 0 2 + A 1 2 d m − A 2 2 d n − A 3 2 m n     + 2 ( A 0 A 1 − A 2 A 3 n ) d m ) = X 2 − B 1 X + B 2 = X 2 − ( b 10 + b 11 d m ⋆ ) X + ( b 20 + b 21 d m ⋆ ) ,

with b i j ∈ ℤ ;

And:

h 1 ( X ) = ( X − α 2 ) ( X − α 3 ) ∈ ℤ ℚ ( d m ⋆ ) [ X ] = X 2 − 2 ( A 0 − A 1 d m ) X − ( A 0 2 + A 1 2 d m − A 2 2 d n − A 3 2 m n     − 2 ( A 0 A 1 − A 2 A 3 n ) d m ) , = X 2 − B 1 ¯ X + B 2 ¯ = X 2 − ( b 10 + b 11 d m ⋆ ¯ ) X + ( b 20 + b 21 d m ⋆ ¯ ) .

Similarly consider:

• g 2 ( X ) = ( X − α 0 ) ( X − α 2 ) ∈ ℤ ℚ ( d n ⋆ ) [ X ] = X 2 − 2 ( A 0 + A 2 d n ) X + ( A 0 2 + A 2 2 d n − A 1 2 d m − A 3 2 m n + 2 ( A 0 A 2 − A 1 A 3 m ) d n ) = X 2 − C 1 X + C 2 = X 2 − ( c 10 + c 11 d n ⋆ ) X + ( c 20 + c 21 d n ⋆ ) ,

with c i j ∈ ℤ ;

And:

h 2 ( X ) = ( X − α 1 ) ( X − α 3 ) ∈ ℤ ℚ ( d n ⋆ ) [ X ] = X 2 − 2 ( A 0 − A 2 d n ) X + ( A 0 2 + A 2 2 d n − A 1 2 d m − A 3 2 m n     − 2 ( A 0 A 2 − 2 A 1 A 3 m ) d n ) = X 2 − C 1 ¯ X + C 2 ¯ = X 2 − ( c 10 + c 11 d n ⋆ ¯ ) X + ( c 20 + c 21 d n ⋆ ¯ ) .

And,to finish similarly consider:

g 3 ( X ) = ( X − α 0 ) ( X − α 3 ) ∈ ℤ ℚ ( m n ⋆ ) [ X ] = X 2 − 2 ( A 0 + A 3 m n ) X + ( A 0 2 + A 3 2 m n − A 1 2 d m − A 2 2 d n     + 2 ( A 0 A 3 − A 1 A 2 d ) m n ) = X 2 − D 1 X + D 2 = X 2 − ( d 10 + d 11 m n ⋆ ) X + ( d 20 + d 21 m n ⋆ ) ,

with d i j ∈ ℤ ;

And:

h 3 ( X ) = ( X − α 1 ) ( X − α 2 ) ∈ ℤ ℚ ( m n ⋆ ) [ X ] = X 2 − 2 ( A 0 − A 3 m n ) X − ( A 0 2 + A 3 2 m n − A 1 2 d m − A 2 2 d n     − 2 ( A 0 A 3 − A 1 A 2 d ) m n ) = X 2 − D 1 ¯ X + D 2 ¯ = X 2 − ( d 10 + d 11 m n ⋆ ¯ ) X + ( d 20 + d 21 m n ⋆ ¯ ) .

We deduce the factorization:

g j × h j = f ,with deg ( g j ) = deg ( h j ) = 2 , ∀ j = 1 , 2 , 3 ;

And this:Respectively in ℤ [ d m ⋆ ] [ X ] , ℤ [ d n ⋆ ] [ X ] and ℤ [ m n ⋆ ] [ X ] .

These factorizations will lead to non-trivial factorizations of α p ¯ ( f ) = f p ¯ in F p [ a ∗ ] [ X ] = F p [ X ] ,for all prime number p.(cf.Remarks 2.2.).To prove it,it suffices to show that the propositions of Lemma 2.1.is holding.

Let’s prove it effectively:

A)Suppose that p ∕ d m n .

1) p ≠ 2 .

a)If p ∕ d then p ∕ d m (and p ∕ d n )then d m = 0 F p (and d n = 0 F p ).

We have d m ∗ = { 2 − 1     if   d m ≡ 1 ( mod 4 ) 0 F p ,     if   d m ≡ − 1   or   2 ( mod 4 ) ∈ F p .

(Resp.similarly for d n ∗ ).

We can choose to factorize in ℤ [ d m ∗ ] [ X ] :

f = g 1 × h 1 ,lies modulo p in F p [ X ] ,so we get the following factorization of f p ¯ :

For details:

- In the case: d m ≡ d n ≡ 1 ( mod ( 4 ) ) :

f p ¯ ( X ) = ( ( g 1 ) p ¯ ( X ) ) × ( ( h 1 ) p ¯ ( X ) ) = ( X 2 − ( b 10 + ( 2 ¯ − 1 ) b 11 ) X + b 20 + ( 2 ¯ ) − 1 b 21 )     × ( X 2 − ( b 10 − ( 2 ¯ ) − 1 b 11 ) X + b 20 − ( 2 ¯ ) − 1 b 21 ) .

- And in the case d m ≡ d n ≡ − 1 or 2 ( mod ( 4 ) ) we get:

f p ¯ ( X ) = ( ( g 1 ) p ¯ ( X ) ) × ( ( h 1 ) p ¯ ( X ) ) = ( X 2 − b 10 X + b 20 ) 2 .

b)If p ∕ m then d m = 0 F p ,and m n = 0 F p .We can choose to factorize in ℤ [ d m ∗ ] [ X ] ,we get for f p ¯ the same factorization.

c)If p ∕ n then d n = 0 F p ,and m n = 0 F p .We can choose to factorize in ℤ [ d n ∗ ] [ X ] ,we get for f p a factorization via ℤ [ d n ∗ ] [ X ] :

For details:

In the case: d m ≡ d n ≡ 1 ( mod ( 4 ) )

( ( f p ) ¯ ( X ) ) = ( ( g 2 ) p ¯ ( X ) ) × ( ( h 2 ) p ¯ ( X ) ) = ( X 2 − ( c 10 + ( 2 ¯ ) − 1 c 11 ) X + c 20 + ( 2 ¯ ) − 1 c 21 )     × ( X 2 − ( c 10 − ( 2 ¯ ) − 1 c 11 ) X + c 20 − ( 2 ¯ ) − 1 c 21 ) .

And in the case d m ≡ d n ≡ − 1 or 2 ( mod ( 4 ) ) we get:

f p ¯ ( X ) = ( ( g 2 ) p ¯ ( X ) ) × ( ( h 2 ) p ¯ ( X ) ) = ( X 2 − c 10 X + c 20 ) 2 .

2)If p = 2 ∕ d m n ,necessarily 2 ∕ d (cf.Remarks 2.2.1)),then d m = 0 F 2 and d n = 0 F 2 ⇒ X 2 − d m = X 2 and X 2 − d n = X 2 and the factorization of f,in ℤ [ d m ] [ X ] and in ℤ [ d n ] [ X ] given by: f = g 1 × h 1 and f = g 2 × h 2 are in fact modulo 2in F 2 [ X ] ,so we get two factorisations for f 2 ¯ ,that are f 2 ¯ = ( g 1 ) 2 ¯ × ( h 1 ) 2 ¯ and f 2 ¯ = ( g 2 ) p = 2 ¯ × ( h 2 ) p = 2 ¯ .

B)Let p prime such that p ∤ d m n .

1)First assume p ≠ 2 .

Then necessarily exactly one,or all,of the ( d m p ) , ( d n p ) , ( m n p ) are equal to 1.Otherwise: ( d 2 m 2 n 2 p ) = − 1 that is absurd.

Suppose that it is ( d m p ) = 1 (The other cases are done equivalently).

The conditions of the Lemma 2.1.(ii)are realized,so:

The factorization: f = g 1 × h 1 in ℤ [ d m ⋆ ] [ X ] .

Induces the factorization α p ( f ) = α p ¯ ( g 1 ) × α p ¯ ( h 1 ) ,in F p [ d m ⋆ ] [ X ] = F p [ X ] .

But in one hand,the left side gives α p ¯ ( f ) = f p ¯ ,and in the other hand,the right side give: α p ¯ ( g 1 ) = ( g 1 ) p ¯ and α p ¯ ( h 1 ) = ( h 1 ) p ¯ ,and the factorization f p ¯ = ( g 1 ) p ¯ × ( h 1 ) p ¯ lies to F p [ X ] ,so f is reducible mod ( p ) for any such prime p.

2)Assume now p = 2 (recall that 2 ∤ d m n ).

a)In case d m ≡ 3 ( mod 4 ) then X 2 − d m = ( X + 1 ) 2 in F 2 [ X ] ⇒ d m ∗ = d m = 1 ∈ F 2 ⇒ f 2 ¯ is reducible in F 2 [ X ] ,via the factorization f = g 1 × h 1 in ℤ [ d m ] [ X ] ,which implies: f 2 ¯ = ( g 1 ) 2 ¯ × ( h 1 ) 2 ¯ .

b)In case d m ≡ 1 ( mod 4 ) let’s note that we get also d n ≡ m n ≡ 1 ( mod ( 4 ) ) .

b₁)If d m ≡ 1 ( mod 8 ) or d n ≡ 1 ( mod 8 ) or m n ≡ 1 ( mod 8 ) then in F 2 [ X ] :

X 2 − X + 1 − d m 4 = X 2 − X ⇒ d m ∗ = 1 + d m 2 ∈ { 0 F p ,1 } ⊂ F 2 ,

or d n ∗ = 1 + d n 2 ∈ { 0 F p , 1 } ⊂ F 2 ,or m n ∗ = 1 + m n 2 ∈ { 0 F p ,1 } ⊂ F 2 .

So f 2 ¯ is reducible in F 2 [ X ] via the factorisations:

f = g 1 × h 1 in ℤ [ 1 + d m 2 ] [ X ] ;

(or f = g 2 × h 2 in ℤ [ 1 + d n 2 ] [ X ] or f = g 3 × h 3 in ℤ [ 1 + m n 2 ] [ X ] ).

Equalities that lead to:

f 2 ¯ = ( g 1 ) 2 ¯ × ( h 1 ) 2 ¯ ;

(Or f 2 ¯ = ( g 2 ) p = 2 ¯ × ( h 2 ) p = 2 ¯ ;or f 2 ¯ = ( g 3 ) 2 ¯ × ( h 3 ) 2 ¯ )

b₂)Now to close the proof,it remains the case d m ≡ d n ≡ m n ≡ 5 ( mod ( 8 ) ) .

But this case don’t exist for biquadratic fields ℚ ( d m , d n ) ,because:

d m ≡ d n ≡ 5 ( mod 8 ) ⇒ d m × d n ≡ m n ≡ 1 ( mod 8 ) which is absurd.

In conclusion,we have prooved that:

Once Σ f = ℚ ( d m , d n ) ,then f p ¯ ( X ) = X 4 + a 3 ¯ X 3 + a 2 ¯ X 2 + a 1 ¯ X + a 0 ¯ is reducible over F p [ X ] ,for any prime p.

We deduce the following theorem which is the goal of this paper:

The conclusive results of our study raise in a relevant way the question of the existence of algebraic fields without inert prime numbers, in relation to their monogeneity or not.

This encourages the search for other examples of families of such fields, possessing this property. We can think of well-known families in number theory.

I thank Prof. F. Pappalardi, from Roma 3 University, for suggesting this topic.

The author declares no conflicts of interest regarding the publication of this paper.

Tanoé, F.E. (2022) An Infinite Family of Number Fields with No Inert Primes. Advances in Pure Mathematics, 12, 744-756. https://doi.org/10.4236/apm.2022.1212057