Irreducible Polynomials in Z [ x ] That Are Reducible Modulo All Primes

The polynomial 4 1 x + is irreducible in [ ] 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 [ ] x  which are reducible modulo every prime.


Introduction
The polynomials of the title of this article have been discussed by Brandl [1], and Guralnick et al. [2].Brandl's paper excludes those N which are such that ( ) ( ) These are precisely the composite integers N for which there is only one abstract group of order N.The paper by Guralnick et al. does show the existence of such polynomials for all composite N's.Our proof of the same is different, more elementary, and in some cases even constructive.
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 ( ) [ ]
be a non-constant polynomial.Then the Galois group of f(x) over  acts transitively on its roots if and only if f(x) is a power of an irreducible polynomial over  .
2) Let 1 2 , / /     be finite normal extensions that is, splitting fields of some polynomial.Let 1 2   denote the compositum of the fields 1 2 ,   , that is, the smallest subfield of containing 1 2 ,   .Then  is a normal extension ( ) ( ) ( ) 3) Every finite solvable group can be realized as a Galois group of some polynomials over  .Same is true of the symmetric groups n S and alternating groups n A .We shall only need this result for cyclic groups, Frobenius groups and for the groups n S and n A [3] [4] and [5].
4) Let ( ) [ ] be an irreducible polynomials of degree n.Let be a monic irreducible polynomial of degree n and p a prime which does not divide the discriminant of f(x).Let ( ) Galois group of ( ) f x over  .Suppose that modulo p the polynomial ( ) Then there is G σ ∈ such that as a permutation on the n roots of ( ) See [7].
6) Let N be any composite natural number and ( ) [ ] be a monic irreducible polynomial of degree N whose Galois group over  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 and Proof
Theorem: For every composite natural number N there is a monic irreducible polynomial ( ) [ ]

G G
× and so it does not have any element of order N. Let α be any algebraic integer such that is a monic irreducible polynomial of degree N and its Galois group does not have any element of order N and therefore ( ) Case II N is square-free and gcd(N, φ(N)) > 1 In this case we can write N pqm = where p, q are primes, p divides 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 pq = , where p and q are distinct primes, p q < and p does not divide 1 q − .Let t be the order of p modulo q.So 1 t > is the smallest integer such that ( ) . Let 1 G be an elementary Abelian p-group of order t p and 2 G be a group of order q.We note that ( ) Aut G is isomor- phic to ( ) , GL t p and so its order is divisible by q.Let own normalizer in G.For otherwisethe index of the normalizer of 2 G in G would be r p , for some r, 1 ≤ r < t which would contradict the fact that t is the smallest integer satisfying ( ) . Since 2 G has prime order q it is disjoint from its conjugates.Therefore G is a Frobenius group of order , t p q and every non-identity element of 2 G induces a fixed-point-free automorphism of 1 G .
Let  be a normal extension of  with Galois group isomorphic to G.
Then [ ] . Let H be a subgroup of G of order Then by FTGT ({Fundamental Theorem of Galois Theory}) the field  is of degree pq over  .We also note that as H is not a normal subgroup of G,  is not a normal extension of  .Let α be an algebraic integer such that We claim that  is the splitting field of ( ) f x (i.e. it is the normal closure of the field  ) and G is its Galois group over  .
If the normal closure of  were a proper subfield of  then it would imply that G has a proper normal subgroup of order r p where r t < , but this is not possible, as G is a Frobenius group.So ( ) [ ] is a monic irreducible polynomials of degree N pq = and its Galois group over  does not have any element of order N pq = .
Finally assume that N and ( ) are coprime and N is a product of more than two primes.We write N pqm = , where p, q are primes and Let t = order of p modulo q.As discussed in the previous case let be a monic irreducible polynomial of degree pq whose Galois group is the semi-direct product of an elementary group of order t p by a cyclic group of order q and is a Frobenius group.

Let 1
G denote this Frobenius group of order t p q and 1  denote the split- ting field of ( ) be a monic irreducible polynomialof degree m whose splitting field is ( ) ( ) We note the following: : , : , : ( ) , , , , , , ,  is a group of order t p qm isomorphic to the direct product of aFrobenius group of order t p q and a cyclic group of order .Therefore it does not have an element of order N pqm

=
. Note that this Frobenius groupdoes not have any subgroup of order pq.
6) The group G transitively permutes the nm algebraic numbers ,1 ,1 is an irreducible polynomial of degree N pqm = , whose Galois group does not have any element of order N.This completes the proof of our theorem.

Alternate Methods
As we noticed the construction of irreducible polynomials in [ ]

Examples
1) The first interesting case is for 15  f x is an irreducible polynomial of degree 7 whose discriminant is a square and Galois group G has a 3-cycle.So G is isomorphic to 7 A [8].
Suppose that the roots of ( ) Among these numbers the method described above works for N = 15, 35 and 91.This is so because 2 A or 7 S , we con- structed an irreducible polynomial of degree 35 which is reducible modulo every prime, likewise starting with a polynomial of degree14 with Galois group isomorphic to 14 A or 14 S we can construct an irreducible polynomial of degree 91 which is reducible modulo every prime.
3) The method discussed in the previous examples above does not work for 33 N = .For this we proceed as in the proof of our theorem.As the order of 11 modulo 3 is 2 we construct a Frobenius group G of order This Frobenius group of order 363 does not have any subgroup of order 33.
Let /   be a normal extension whose Galois group is isomorphic to G. Let H be a subgroup of G of order 11 and  be its fixed subfield.By FTGT (Fundamental Theorem of Galois Theory) the field  has degree 33 over  .Let α be an algebraic integer such that , ,1 be the zeros of ( )  ( ) As this resultant is equal to the following determinant of order m n + , its value can be computed with the help of any symbolic computation package such as MATHEMATICA.
( ) In this determinant all the missing entries are zeros.The entries in the first m rows are the coefficients of ( ) f x and those in the last n rows are the coefficients of ( ) g x .If f and g are polynomials in two variables x and y then we can determine their resultant with respect to any of the variable.For the discussion of the calculation of ( ) r f x for a given polynomial ( ) f x it will be convenient to deal with monic polynomials.Let ( ) This observation and (and similar ones) will be used repeatedly in what follows.As before we let ( ) r f x denote the monic polynomial of degree r n C with zeros . The method discussed can be easily generalized to larger values of r.

Computation of f2(x)
A x is a monic polynomial of degree n with zeros 2 , Therefore, ( ) ( ) R x A x is a polynomial of degree 2 n n − , with zeros , i j i j α α + < , each zeroappearing twice.Therefore, ( ) ( ) ( ) is a polynomial of degree

Computation of f3(x)
We first note that, as a polynomial in y,  , i j k i j k α α α + + < < , each appearing three times.

, .
i j i j α α + ≠ We check that the total number adds up to the right degree, namely ( )( ) ( ) ( ) the polynomial whose roots are all sums of r different 1 the same splitting field.Since both n S .and n A transitively permute the N, roots of Let ( ) r f x this polynomial is irreduci- ble.So the polynomial let ( ) r f x is the required polynomial of degree N, whose Galois group does not have any element of order N.
As starting with a po- lynomial of degree 7 with Galois group isomorphic to 7

2 11 3 ⋅
which is a semi-direct product of 11 11 ×   by a group of order 3.More specifically we ex- the group ϕ of order three where ϕ is the automorphism of x y × easily seen that this automorphism has order three and is fixed-point-free.The resulting group, the semi-direct product of 11 11 ×   by ϕ is a Frobenius group of order 363 having the subgroup 11 11 ×   as its kernel and the group ϕ as its complement.
As proved in the theorem this polynomial has degree 33 and its Galois group is a Frobenius group of order 2 11 3 363 × = which does not have any subgroup of order 33 andtherefore the irreducible polynomials ( ) f x is reducible modulo every prime.It remains to be seen that given a degree n polynomial ( ) [ ] .This can be done with thehelp of the concept of the resultant of two polynomials.Let

for 1
i j n ≤ < ≤ .In other words we can write of degree six whose Galois group over  is isomorphic to symmetric oralternating group on five or six letters.Then x is irreducible over  and its Galois group G over  is 2-transitive on its roots and has a 3-cycle.Therefore G is isomorphic to 6 A Gupta DOI: 10.4236/ojdm.2019.9200656 Open Journal of Discrete Mathematics lemonic polynomial f x