# Factoring $x^4+1$ into irreducibles over $\mathbb{Q}$ and over $\mathbb{F}_2$

I've been struggling with factoring polynomials into irreducibles. I found an example and would like to know how to tackle it. If our polynomial is $x^4+1$, what are the irreducible polynomial factors in $\mathbb{Q}[x]$? What about in $\mathbb{F}_2[x]$?

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For $\mathbb{F}_2[x]$, maybe factorization is obvious. Or the slow way, note that $1$ is a root. So $x+1$ divides $x^4+1$. Do the division. See what happens. Remember that in general $-a=a$. –  André Nicolas Nov 30 '11 at 6:51

A factorization of $x^4+1$ over $\mathbb{Q}$ is also a factorization over $\mathbb{R}$. In $\mathbb{R}[x]$ we have $$x^4+ 1 = \bigl( x^2 + \sqrt{2}x + 1\bigr)\bigl(x^2 - \sqrt{2}x+1\bigr),$$ and this is a factorization into irreducibles. Since neither of these factors is in $\mathbb{Q}[x]$, we must have that $x^4+1$ is irreducible in $\mathbb{Q}[x]$ (it cannot factor as a product of two squares, because they would have to be the ones in $\mathbb{R}[x]$ above; and it cannot have linear factors because it has none in $\mathbb{R}[x]$).
For $\mathbb{F}_2[x]$, remember that $(a+b)^2 = a^2+b^2$ in characteristic $2$, so $$x^4 + 1 = x^4+1^4 = (x^2+1^2)^2 = ((x+1)^2)^2 = (x+1)^4.$$