Splitting Field: Let $K$ be a field and let $f(x) = a_0+a_1x+ a_2x^2+\cdots+a_nx^n$ be a polynomial in $K[X]$ of degree $n \gt 0$. An extension field $F$ of $K$ is called a splitting field for $f(x)$ over $K$ if there exist elements $r_1, r_2,\ldots, r_n$ in $F$ such that
(i) $f(x) = a_n(x-r_1)(x-r_2)\ldots(x-r_n)$, and
(ii) $F = K(r_1,r_2,\ldots,r_n)$.
There is another concept: for $f(x) \in F[X]$, which is a polynomial of degree $n \ge 1$, there exist an extension $E$ of $F$ of degree at most $n!$ in which $f(x)$ has $n$ roots.
Let's take an example as $x^4+1$, then
\begin{align}
x^4+1 & = (x^4+2x^4+1)-2x^4 \\
& = (x^2+1)^2 – (\sqrt{2} x)^2 \\
& = (x^2+1+\sqrt{2} x)(x^2+1-\sqrt{2}x) \\
\end{align}
Let $r$ be the root of $(x^2+1-\sqrt{2}x)$, then $$r = \frac{-\sqrt{2} \pm \sqrt{2-4}}{2} = \frac{\sqrt{2}(i-1)}{2}$$
So, ${\Bbb{Q}(√2,i)}$ is the smallest field containing $\frac{\sqrt{2}(i-1)}{2}$. So,
\begin{align}
[\Bbb{Q}(\sqrt{2},i):\Bbb{Q}] & = [\Bbb{Q}(\sqrt{2},i):\Bbb{Q}(\sqrt{2})][\Bbb{Q}(\sqrt{2}):\Bbb{Q}] \\
& = 2×2 = 4
\end{align}
( Because $\sqrt{2}$ satisfies an irreducible polynomial $x^2-2$ over $\Bbb{Q}$ hence $[\Bbb{Q}(\sqrt{2}):\Bbb{Q}] = 2$. Similarly $i$ satisfies an irreducible polynomial $x^2+1$ over $\Bbb{Q}$, therefore it satisfies $x^2+1$ over $\Bbb{Q}(\sqrt{2})$ also hence $[\Bbb{Q}(\sqrt{2},i):\Bbb{Q}(\sqrt{2})]=2 )$ Is my explaination correct?
So, first we need to find the factors of $x^5 – 1$.
This equation can be simplified to $(x^4+x^3+x^2+x+1)(x-1)$
But now i am stuck!
Any more insight in this question would be appreciated. Thanks
Best Answer
I'll answer this question more generally by considering $x^n-1$ instead.
If we let $\zeta_n = e^{2 \pi i /n}$, then the roots of $x^n-1$ are $1,\zeta_n, \zeta_n^2,...,\zeta_n^{n-1}$. Therefore the splitting field of $x^n-1$ over $\mathbb{Q}$ is $\mathbb{Q}(\zeta_n)$.
One can show that $[\mathbb{Q}(\zeta_n) \ : \ \mathbb{Q}]= \varphi(n)$ where $\varphi$ is Euler's totient function (This is not totally obvious). Therefore, in your case, the splitting field of $x^5-1$ is $\mathbb{Q}(e^{2 \pi i /5})$ and its degree over $\mathbb{Q}$ is $4$.
Also when you show that $[\mathbb{Q}(\sqrt{2}, i) \ : \ \mathbb{Q}]= 4$, you should actually compute $[\mathbb{Q}(\sqrt{2}, i) \ : \ \mathbb{Q}(\sqrt{2})]$ and $[\mathbb{Q}(\sqrt{2}) \ : \ \mathbb{Q}]$ by showing that $x^2+1$ is irreducible over $\mathbb{Q}(\sqrt{2})$ (You already mentioned that $x^2-2$ is irreducible over $\mathbb{Q}$).