How do I prove that this polynomial is irreducible? How do I prove that $x^4+1$ is an irreducible polynomial over $\mathbb Q$? I've already tried the Eisenstein criterion which gives to me any results to solve this question, I need help here.
Thanks
 A: Although the problem may be intended to be solved using a variant of the Eisenstein Criterion, one can also solve it using only elementary facts.
The polynomial $x^4+1$ has no real roots, so if we can reduce over $\mathbb{Q}$, it is as a product of quadratics. 
Without loss of generality these quadratics each have lead coefficient $1$. Since $x^4+1$ has no $x^3$ term, the two quadratics must have shape
$x^2-ax+b$ and $x^2+ax+c$.
The coefficient of $x$ in the product is $a(b-c)$. But it must be $0$. It is clear that we cannot have $a=0$. So $b=c$. That forces $b=c=1$ or $b=c=-1$.
But the coefficient of $x^2$ in the product is $b+c-a^2$. Thus $a^2=\pm 2$. This is not solvable in rationals. 
Remark: The polynomial does have the nice factorization
$x^4+1=\left(x^2-\sqrt{2}x+1\right)\left(x^2+\sqrt{2}x+1\right)$. This can be useful in some integration problems. The variant $x^4+4=(x^2-2x+2)(x^2+2x+2)$ has a habit of turning up in contest problems. 
A: Try Eisenstein's test on $(x+1)^4+1=x^4+4x^3+6x^2+4x+2$. Can you pick the prime number?
Do convince yourself that (ir)reducibility is preserved by translations in the variable. Nifty trick.
