_{1}

^{*}

The polynomial x
^{4}+1
is irreducible in Ζ[x]
but is locally reducible, that is, it factors modulo p for all primes p. In this paper we investigate this phenomenon and prove that
for any composite natural number N there are monic irreducible polynomials in Ζ[x]
which are reducible modulo every prime.

The polynomials of the title of this article have been discussed by Brandl [

We shall first enumerate the known results which we shall use in this article. Several of these results are true more generally but we shall state them as needed in this article.

1) Let f ( x ) ∈ ℚ [ x ] be a non-constant polynomial. Then the Galois group of f(x) over ℚ acts transitively on its roots if and only if f(x) is a power of an irreducible polynomial over ℚ .

2) Let K 1 / ℚ , K 2 / ℚ be finite normal extensions that is, splitting fields of some polynomial. Let K 1 K 2 denote the compositum of the fields K 1 , K 2 , that is, the smallest subfield of containing K 1 , K 2 . Then K is a normal extension of ℚ and if [ K 1 : ℚ ] and [ K 2 : ℚ ] are coprime. Then

A u t ( K / ℚ ) = A u t ( K 1 / ℚ ) × A u t ( K 2 / ℚ )

3) Every finite solvable group can be realized as a Galois group of some polynomials over ℚ . Same is true of the symmetric groups S n and alternating groups A n . We shall only need this result for cyclic groups, Frobenius groups and for the groups S n and A n [

4) Let f ( x ) ∈ ℚ [ x ] be an irreducible polynomials of degree n. Let α 1 , α 2 , ⋯ , α n be its roots. Let r be an integer, 1 < r < n − 1 and C n r = m . Let f r ( x ) denote the polynomial whose roots are all sums of r different α 1 . Then f r ( x ) ∈ ℚ [ x ] and f ( x ) and f r ( x ) have the same splitting field [

5) Let f ( x ) ∈ ℤ [ x ] be a monic irreducible polynomial of degree n and p a prime which does not divide the discriminant of f(x). Let G = G a l ( f ) be the Galois group of f ( x ) over ℚ . Suppose that modulo p the polynomial f ( x ) factors into irreducible polynomials of degrees n 1 , n 2 , ⋯ , n t so n 1 + n 2 + ⋯ + n t = n . Then there is σ ∈ G such that as a permutation on the n roots of f ( x ) , σ = σ 1 σ 2 ⋯ σ t , where σ i is acyclic permutation of length n i for 1 ≤ i ≤ n . See [

6) Let N be any composite natural number and f ( x ) ∈ ℤ [ x ] be a monic irreducible polynomial of degree N whose Galois group over ℚ does not have any element of order N. Then f ( x ) is reducible modulo every prime. This is an immediate consequence of (5) above.

Theorem: For every composite natural number N there is a monic irreducible polynomial f ( x ) ∈ ℤ [ x ] of degree N which is reducible modulo every prime.

Case I N is not square-free

We write N = p t m where t > 1 and p is a prime which does not divide m. Let G 1 be any non-cyclic group of order p t and G 2 a cyclic group of order m. Let f 1 ( x ) ∈ ℤ [ x ] be an irreducible polynomial of degree p t with Galois group isomorphic to G 1 and f 2 ( x ) ∈ ℤ [ x ] be an irreducible polynomial of degree m with Galois group isomorphic to G 2 . Let K 1 and K 2 be splitting fields of f 1 ( x ) and f 2 ( x ) respectively. Let K = K 1 K 2 be the compositum of the fields K 1 and K 2 . Then K is of degree N over ℚ and G = A u t ( K / ℚ ) is isomorphic to G 1 × G 2 and so it does not have any element of order N. Let α be any algebraic integer such that K = ℚ ( \ α ) . Let f ( x ) be the minimum polynomial of α. Then f ( x ) ∈ ℤ [ x ] is a monic irreducible polynomial of degree N and its Galois group does not have any element of order N and therefore f ( x ) has the desired property.

Case II N is square-free and gcd(N, φ(N)) > 1

In this case we can write N = p q m where p, q are primes, p divides q − 1 and gcd ( p q , m ) = 1 . Let G 1 be a non-abelian group of order pq and G 2 a cyclic group of order m. Just as in the previous case we get a monic irreducible polynomial in ℤ [ x ] of degree N whose Galois group does not contain an elementof order N.

Case III, N is square-free and gcd(N, φ(N)) = 1

In this case N is necessarily odd. First we assume that N is a product of just two primes. So let N = p q , where p and q are distinct primes, p < q and p does not divide q − 1 . Let t be the order of p modulo q. So t > 1 is the smallest integer such that p t ≡ 1 ( mod q ) . Let G 1 be an elementary Abelian p-group of order p t and G 2 be a group of order q. We note that A u t ( G 1 ) is isomorphic to G L ( t , p ) and so its order is divisible by q. Let G = G 1 × G 2 be the semi-direct product of G 1 by G 2 . Evidently G is not a direct product of G 1 and G 2 . Therefore G 2 is not a normal subgroup of G. We claim that G 2 is its own normalizer in G. For otherwisethe index of the normalizer of G 2 in G would be p r , for some r, 1 ≤ r < t which would contradict the fact that t is the smallest integer satisfying p t ≡ 1 ( mod q ) . Since G 2 has prime order q it is disjoint from its conjugates. Therefore G is a Frobenius group of order p t , q and every non-identity element of G 2 induces a fixed-point-free automorphism of G 1 .

Let K be a normal extension of ℚ with Galois group isomorphic to G. Then [ K : ℚ ] = p t q . Let H be a subgroup of G of order p t − 1 and let F ⊆ K be its fixed subfield.

Then by FTGT ({Fundamental Theorem of Galois Theory}) the field F is of degree pq over ℚ . We also note that as H is not a normal subgroup of G, F is not a normal extension of ℚ . Let α be an algebraic integer such that F = ℚ ( α ) and let f ( x ) be its minimal polynomial over ℚ . Then f ( x ) ∈ ℤ [ x ] is irreducible of degree pq.

We claim that K is the splitting field of f ( x ) (i.e. it is the normal closure of the field F ) and G is its Galois group over ℚ .

If the normal closure of F were a proper subfield of K then it would imply that G has a proper normal subgroup of order p r where r < t , but this is not possible, as G is a Frobenius group. So f ( x ) ∈ ℤ [ x ] is a monic irreducible polynomials of degree N = p q and its Galois group over ℚ does not have any element of order N = p q .

Finally assume that N and φ ( N ) are coprime and N is a product of more than two primes. We write N = p q m , where p, q are primes and gcd ( p q , m ) = 1 . Let t = order of p modulo q. As discussed in the previous case let f 1 ( x ) ∈ ℤ [ x ] be a monic irreducible polynomial of degree pq whose Galois group is the semi-direct product of an elementary group of order p t by a cyclic group of order q and is a Frobenius group.

Let G 1 denote this Frobenius group of order p t q and K 1 denote the splitting field of f 1 ( x ) . Let G 2 be a cyclic group of order m and f 2 ( x ) ∈ ℤ [ x ] be a monic irreducible polynomialof degree m whose splitting field is K 2 and Galois group over ℚ is G 2 .

Let K = K 1 K 2 be the compositum of the fields K 1 and K 2 . Let p q = n and

f 1 ( x ) = ∏ i = 1 n ( x − α i )

f 2 ( x ) = ∏ j = 1 m ( x − β j )

f ( x ) = ∏ j = 1 m ∏ i = 1 n ( x − α i β j )

We note the following:

1) [ K 1 : ℚ ] = p t q , [ K 2 : ℚ ] = m , [ K : ℚ ] = p t q m ;

2) K 1 = ℚ ( α 1 , α 2 , ⋯ , α n ) ;

3) K 2 = ℚ ( β 1 , β 2 , ⋯ , β m ) ;

4) K = ℚ ( α 1 , α 2 , ⋯ , α n , β 1 , β 2 , ⋯ , β m ) ;

5) G = A u t ( K / ℚ ) is a group of order p t q m isomorphic to the direct product of aFrobenius group of order p t q and a cyclic group of order . Therefore it does not have an element of order N = p q m . Note that this Frobenius groupdoes not have any subgroup of order pq.

6) The group G transitively permutes the nm algebraic numbers α i β j , 1 ≤ i ≤ n , 1 ≤ j ≤ m . So f ( x ) ∈ ℤ [ x ] is an irreducible polynomial of degree N = p q m , whose Galois group does not have any element of order N. This completes the proof of our theorem.

As we noticed the construction of irreducible polynomials in ℤ [ x ] of odd composite degree N where gcd ( N , φ ( N ) ) = 1 , and whose Galois group does not contain an element of order N is not so straight forward. In some case such as N = 15 or N = 35 there is another interesting method of construction of such polynomials. In fact it works for most N’s (with very few exceptions) which are such that C n r = N for some n and r such that 1 < r < n − 1 . The method we are about to describe fails in cases where C n r = N but the symmetric group S n does have an element of order N, as it happens when n = 15 , r = 2 and N = 105 .

As the symmetric group on 15 letters does have an element of order 105 = C 15 2 , namely a permutation which is a product of 3, 5 and a 7-cycle. Let f ( x ) ∈ ℤ [ x ] be a monic irreducible polynomial of degree n > 4 whose Galois group is isomorphic to either A n or S n . Let r be such that 1 < r < n − 1 and C n r = N . Further assume that S n does not have any element order N. We know that S n is n-transitive and A n is ( n − 2 ) -transitive on n letters. Let f r ( x ) denote a polynomial of degree N = C n r whose roots are sum of all r different roots of f ( x ) . Let the roots of f r ( x ) be β i , where 1 ≤ i ≤ N. The polynomials f ( x ) and let f r ( x ) have the same splitting field. Since both S n . and A n transitively permute the N, roots of Let f r ( x ) this polynomial is irreducible. So the polynomial let f r ( x ) is the required polynomial of degree N, whose Galois group does not have any element of order N.

1) The first interesting case is for N = 15 . Let f ( x ) ∈ ℤ [ x ] be an irreduciblemonic polynomial of degree six whose Galois group over ℚ is isomorphic to symmetric oralternating group on five or six letters. Then f 2 ( x ) ∈ ℤ [ x ] is an irreducible monic polynomial whose Galois group is the same as that of f ( x ) and so does not have any element of order 15. Therefore f 2 ( x ) is reducible modulo every prime. For instance let f ( x ) = x 6 + 24 x − 20 whose discriminant is 2 16 ⋅ 3 6 ⋅ 5 6 . We note that

f ( x ) ≡ ( x + 3 ) ( x 5 + 4 x 4 + 2 x 3 + x 2 + 4 x + 5 ) ( mod 7 )

f ( x ) ≡ ( x + 7 ) ( x + 12 ) ( x + 21 ) ( x 3 + 6 x 2 + 13 x + 16 ) ( mod 23 )

f ( x ) ≡ ( x 2 + 26 x + 10 ) ( x 4 + 3 x 3 + 28 x 2 + 25 x + 27 ) ( mod 29 )

It follows that f ( x ) is irreducible over

The Galois group of this polynomial is the same as that of

2) The second example is for

So

Suppose that the roots of

This polynomial is irreducible over

The polynomial

This polynomial is irreducible over

Note: The composite natural numbers N below 100 which are such that

Among these numbers the method described above works for N = 15, 35 and 91. This is so because

3) The method discussed in the previous examples above does not work for

This Frobenius group of order 363 does not have any subgroup of order 33. Let

Let H be a subgroup of G of order 11 and

It remains to be seen that given a degree n polynomial

be polynomials of degree

As this resultant is equal to the following determinant of order

In this determinant all the missing entries are zeros. The entries in the first m rows are the coefficients of

This observation and (and similar ones) will be used repeatedly in what follows. As before we let

We shall show how to find

Let

Then

Then

is a polynomial of degree

We first note that, as a polynomial in y,

Let

Then

We check that the total number adds up to the right degree, namely

We shall now find a polynomial with zeros

Here as before we have multiplied by

So the zeros of

So

is a polynomial of degree

is the required polynomial of degree

Bernard Dominique [

If the Galois group of any of these polynomials regarded as a permutation on its 33 roots had a 33-cycle then according to Cebotarev Density Theorem the density of primes p for which any of these polynomials is irreducible should be

In this paper we have shown that for any composite natural number N there are polynomials of degree N with integer coefficients which are irreducible in

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

Gupta, S. (2019) Irreducible Polynomials in ℤ[x] That Are Reducible Modulo All Primes. Open Journal of Discrete Mathematics, 9, 52-61. https://doi.org/10.4236/ojdm.2019.92006