Is $x^2+1$ irreducible over a cyclotomic field?

Question

Let $K=\mathbb{Q}[\omega]$, where $1+\omega+\omega^2=0$, let $f(X)=X^2+1$. How can i prove irreducibility of $f$ over $K$?

Answer 1

This polynomial is irreducible if and only if it has no roots in the field. That is true for any polynomial of degree $2$ or $3$.

Showing that $x^2+1=0$ has no root should be essentially a calculation. Check whether the square of $s+\omega t$ can be equal to $-1$ for rational $s$ and $t$. Equivalently, check whether we can have $(a+b\sqrt{-3})^2=-1$ for some rational $a$ and $b$.

Added: We have $(a+b\sqrt{-3})^2=a^2-3b^2+2ab\sqrt{-3}$. If this is $-1$, then $ab=0$. But $b=0$ is impossible, since we cannot have $a^2=-1$. And $a=0$ is impossible, else we would have $3b^2=1$. This cannot happen, since $\sqrt{3}$ is irrational. A proof of irrationality can, if necessary, be given along the same lines as the familiar proof of the irrationality of $\sqrt{2}$.

Answer 2

Alternatively, prove first the possibly easier fact that $g(x) = x^{2} + x + 1$ is irreducible over $\Bbb{Q}(i)$. This is clear, as the roots of $g(x)$ are $$ \omega, \omega^{2} = \frac{-1 \pm \sqrt{-3}}{2} \notin \Bbb{Q}(i). $$

So we have $$ \lvert \Bbb{Q}(i, \omega) : \Bbb{Q} \rvert = \lvert (\Bbb{Q}(i)) (\omega) : \Bbb{Q}(i) \rvert \cdot \lvert \Bbb{Q}(i) : \Bbb{Q} \rvert = 2 \cdot 2 = 4. $$ It follows that $$ \lvert (\Bbb{Q}(\omega))(i) : \Bbb{Q}(\omega) \rvert = \frac{\lvert \Bbb{Q}(i, \omega): \Bbb{Q} \rvert}{\lvert \Bbb{Q}(\omega) : \Bbb{Q} \rvert} = \frac{4}{2} = 2, $$ which means $x^{2} + 1$ is irreducible over $\Bbb{Q}(\omega)$.