Irreducibility of a polynomial over rationals. I am given the polynomial $x^4+1$ and I am asked to prove that it is irreducible in $\mathbb Q[x]$. I was just wondering if it is enough to show that $x^4+1$ does not contain a root in $\mathbb Q$ and therefore it is not irreducible in $\mathbb Q[x]$?
 A: No that is not enough.  That would only show that it has no linear factors.  As Daniel pointed out, it could factor as a product of two quadratic polynomials.  You could try a couple of approaches:
1) Use the factorization of this polynomial over $\mathbb{C}$ to conclude that it cannot factor over $\mathbb{Q}$.  Factor it completely over $\mathbb{C}$ and show that no product of these factors (other than the product of all four of them) lies in $\mathbb{Q}[x]$.
2) Make a change of variables and try to apply the Eisenstein Criterion.
A: No, showing that a polynomial does not have a root does not prove the irreducibility of the polynomial (both in general, and in the case you are considering). If the polynomial were of degree less than $4$, then showing it has no root does prove it is irreducible. 
A: It is not enough to show that it does not contain a root, because it could also factor as the product of two quadratics. Ab absurdo say $f\cdot g$ are quadratics that multiply to $x^4+1$. Then, you can assume that both $f$ and $g$ are monic (Why?), then consider $f(x)=x^2+ax+b$ and $g(x)=x^2+cx+d$, and if you multiply them you get a system of equations. If you try to solve this system you should get a contradiction. 
A: Note that pairing the complex conjugate roots of $x^4+1$ gives $$x^4+1=(x^2+\sqrt2x+1)(x-\sqrt2x+1)$$ but this is not a factorisation into polynomials with rational coefficients (and other pairings of the roots do not even give a factorisation over $\mathbb R$).
