# Irreducible polynomial in Q[x] but redicuble in Z[x]

k3k3

## Homework Statement

Give an example of a polynomial irreducible in Q[x], but reducible in Z[x]

## The Attempt at a Solution

I think there is no example of this. The coefficients of a reducible polynomial with integer coefficients will be rational, or am I mistaken?

Homework Helper
Gold Member
The question makes no sense. If a polynomial can be factored in Z[x], then it can be factored (with the same factors) in Q[x].

Also, I'm not sure what you mean by this question:
The coefficients of a reducible polynomial with integer coefficients will be rational, or am I mistaken?
Since integers are rational numbers, it's (trivially) true that integer coefficients are rational coefficients. But what does this buy you?

Staff Emeritus
Gold Member
There is an example.

Recall the definition of irreducible:
A non-unit x is irreducible if and only if it cannot be written as yz where y and z are non-units​

Now, as is very often a useful, we simply translate the question according to the definition:

We seek polynomials f,g,h with integer coefficients such that:

• f=gh, and none of f,g,h are units in Z[x]
• There do not exist non-units u,v in Q[x] such that f=uv

A more symmetric phrasing might be useful, to spot what you're missing:

We seek a polynomial f with integer coefficients such that

• There exist non-units u,v of Z[x] such that f=uv
• There do not exist non-units u,v of Q[x] such that f=uv

zukerm
The question is nonsense as written. If a polynomial f(x) in Q[x] is reducible (spelling!) in Z[x], then f(x) = g(x)h(x) were g(x) and h(x) are in Z[x]. We assume that the degrees of g(x) and h(x) are > 0. This means that g(x) and h(x) have integer coefficients. When multiplied, the product polynomial, f(x), must have integer coefficients in the first place.
g(x) and h(x) are also in Q[x], so f(x) IS reducible in Q[x].

The non-trivial result is that if a polynomial in Z[x] is irreducible, then it is also irreducible in Q[x]. This takes a bit of proof and is usually given in a first course in so-called abstract algebra.

Staff Emeritus
Gold Member
The question is nonsense as written.
Not only does the question make sense, the request can be satisfied.

We assume that the degrees of g(x) and h(x) are > 0. This means that g(x) and h(x) have integer coefficients. When multiplied, the product polynomial, f(x), must have integer coefficients in the first place.
g(x) and h(x) are also in Q[x], so f(x) IS reducible in Q[x].
And thus, we conclude that any example must violate the assumption...

k3k3
I am still stumped. Do you have another hint to try and make me see what I am missing?

Homework Helper
If a polynomial is reducible in the rational numbers, it can be written as a fraction (1 over the least common multiple of the denominators of the coefficients) times a polynomial that is reducible over the integers. That does NOT mean the original poynomial was reducible over the integers.

Staff Emeritus