Is the ideal generated by $X^4+1$ maximal in $\mathbb{R}[X]$?

Question

I've tried to check whether $\mathbb R[X]/(X^{4}+1)$ is a field, and because $\mathbb R[X]$ is an euclidian ring I just have to prove that $X^{4}+1$ is irreducible over $\mathbb{R}$. I think this is quite hard because I've only learned theorems such as:

Lemma of Gauss.

Eisenstein.

Root condition.

But these theorems only work on $\mathbb{Q}$ an $\mathbb{Z}$. I could assume that they are reducible an try to get a contradiction, but most of the time this is very troublesome. Are there any theorems/tricks/hints I don't know at this question?

Thanks.

Answer 1

$X^4+1=(X^2-\sqrt{2}X+1)(X^2+\sqrt{2}X+1)$ in $\Bbb R[X]$, so $(X^4+1)$ is not a (prime) maximal ideal. In fact, by CRT we have $$\Bbb R[X]/(X^4+1)\cong \Bbb C\times\Bbb C.$$

Answer 2

Hint 1 By the Fundamental Theorem of Algebra, the irreducible Polynomials over $\mathbb R$ have degree $1$ or $2$.

Alternate Hint

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