Prove 1+8x-12x^3+2x^4 irreducible over Q[x]

[faradayslaw]

Homework Statement

Determine if 1+8x-12x^3+2x^4 irreducible over Q[x]

Homework Equations

Gauss's Lemma, Eisenstein Criterion

The Attempt at a Solution

If we multiply g(x)=1+8x-12x^3+2x^4 by 2^3, and then make the substitution y=(2*x), we recover a monic polynomial: h(y) = 8+32y-12y^3+y^4, but Eisenstein does not apply. h(y+/-1) doesn't help either. h(y) is of fourth degree, so it doesn't even suffice to check for absence of roots. But, I do know that if irreducibility in Z is shown, irreducibility over Q follows, for suppose such an f(x) is reducible over Q. Then, by Gauss's Lemma, there exists a nontrivial factorization of f(x) over Z into monic polynomials, but if f(x) is irreducible over Z, this is a contradiction. Thus, f(x) must be irreducible over Q.

Still, I have tried many things, so any help is appereciated.

 

[Dick]

Are you sure h(y+/-1) doesn't satisfy Eisenstein's criterion? It sure looks like they do to me.

 

[faradayslaw]

Thanks for the reply:

The constant term in h(y+1) is 29, a prime, and 29|\ all other coefficients. That of h(y-1) is 11, another prime which does not divide any other coefficeints. Thus, I don't think we can use Eisenstein Criterion.

However, if we consider b(x) = x^4*g(1/x) we obtain: b(x) = 2 -12x + 8x^3 + x^4, and we can use Eisenstein with p=2 here QED

 

[Dick]

Thanks for the reply:

The constant term in h(y+1) is 29, a prime, and 29|\ all other coefficients. That of h(y-1) is 11, another prime which does not divide any other coefficeints. Thus, I don't think we can use Eisenstein Criterion.

However, if we consider b(x) = x^4*g(1/x) we obtain: b(x) = 2 -12x + 8x^3 + x^4, and we can use Eisenstein with p=2 here QED

You're right. I misremembered the Eisenstein Criterion. And, yeah, that's a good idea!