Is $x^8+1$ irreducible?

Question

How to decide if $f(x)=x^8+1$ is irreducible or not over the following fields:

1. $F=\mathbb R$
2. $F=\mathbb Q$.

I can't use Eisenstein's criterion. So the only possibility is computing the complex 8th-roots of -1?

Answer 1

$X^{8}+1$ is indeed irreducible in $\mathbb Q[X]$: it is the $16$-th cyclotomic polynomial. A possible elementary way to prove this is to consider $(X+1)^{8}+1$, which is irreducible iff $X^{8}+1$ is. All the coefficients of $(X+1)^{8}+1$ are even except the first one, and the last coefficient is $2$, so Eisenstein criterion applies.

For $\mathbb{R}[X]$ Jihad already gave you an answer.

Answer 2

Hints:

1. Try seeing if you can find a factorization of $x^4 + 1$ of the form $(x^2 + ax + b)(x^2 + cx + d)$ for suitable real constants $a, b, c, d$. Can you then use this to find a factorization of $x^8 + 1$?

Answer 3

Irreducible polynomials over $\mathbb{R}$ have degree $\leq 2$ since you can find roots over $\mathbb{C}$ and then multiply conjugated terms.

Answer 4

Hint: Completing the square, we have $x^8+1=\Big(x^4\Big)^2+1^2=\Big(x^4+1\Big)^2-2x^4.~$ Can you take it from here ? :-)

Answer 5

We can start with something similar.

Factor $x^4 + 4$:

$x^4 + 4x^2 + 4 - 4x^2$

$(x^2+2)^2 - (2x)^2$

$(x^2+2-2x)(x^2+2+2x)$

$\boxed{x^4 + 4 = (x^2-2x+2)(x^2+2x+2)}$

We could use the same method to factor $x^4+1$:

$x^4 + 2x^2 + 1 - 2x^2$

$(x^2+1)^2 - (\sqrt{2}x)^2$

$(x^2+1-\sqrt{2}x)(x^2+1+\sqrt{2}x)$

$\boxed{x^4 + 1 = (x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}$

Here's what I discovered: $x^4 + y^4 = (x^2+y^2-\sqrt{2}xy)(x^2+y^2+\sqrt{2}xy)$

This shows that every sum of fourth powers can be factored. Now let's factor $x^8+1$:

$(x^2)^4 + 1^4$

$((x^2)^2+1-\sqrt{2}x^2)((x^2)^2+1+\sqrt{2}x^2)$

$\boxed{(x^4-\sqrt{2}x^2+1)(x^4+\sqrt{2}x^2+1)}$

Hope it helps!