# Irreducible polynomials

1. Apr 19, 2006

### mathwonk

im teeching algebra and had to prove that X^5 - 8 was irreducible over the rationals. so i did it using eisenstein.

then more generally i seem to have proved that X^n - a is irreducible over the rationals whenever it has no rational root.

but i used galois theory, and the course im teching does not have that in it.

is there an easier proof? im a rookie at this stuff.

thanks

2. Apr 19, 2006

### shmoe

That doesn't seem to be true in general, x^4-4=(x^2-2)(x^2+2), but it has no rational root.

3. Apr 19, 2006

### AKG

$$x^5-8 = \prod _{k=1}^5(x-\alpha\xi ^k)$$

where $\xi$ is one of the non-real fifth-roots of unity, and $\alpha$ is the real fifth-root of 8. It's clear that this polynomial isn't reducible over the rationals into a product of linear factors, so if it were reducible over the rationals, it would have an irreducible-over-Q factor which would be a product of 2, 3, or 4 of the $(x-\alpha\xi ^k)$. The constant term of this irreducible factor would have a constant term of the form $\pm\alpha ^j \xi ^{k_1 + \dots + k_j}$ where j is 2, 3, or 4, and the ki are in {1,2,3,4,5}. If $\alpha ^j$ is rational, then so is $\alpha ^{gcd(5,j)} = \alpha$. It's easy to prove this false, in a similar manner that we prove 21/2 is irrational. So $\alpha ^j$ is irrational, hence so is $\pm\alpha ^j\xi ^{k_1 + \dots + k_j}$, and so this factor which was supposedly an irreducible polynomial over Q is not a polynomial over Q at all, so there is no such factor, and so the original polynomial is indeed irreducible over Q.

Last edited: Apr 19, 2006
4. Apr 20, 2006

### mathwonk

sorry, i meant x^n -a where n is prime.

5. Apr 20, 2006

### mathwonk

x^5 -8 or x^p - n^k is easy since the field generated by a root is contained in Q[n^(1/p)] which has degree p by eisenstein if n is not a pth root. so Q[(n^k/p)] is a subfield of a field of prime degree, hence either the root is already in Q or the field is of degree p also.