I was wondering if anyone could check to see if i have done the following problem correct.
I want to find the splitting Field for $f(x)$ over $\mathbb Q$, $E$ say, and also evaluate $[E:\mathbb Q]$
Let $f(x)=x^{7}+3$
EDIT:
The roots of $x^{7}+3$ are the seven, $7$-th roots of $-3$. Now since $7$ is odd, $\exists$ a real $7$-th root of $-3$, namely $-(\sqrt[7]{3})$. Let $\alpha=-(\sqrt[7]{3})$. Let $\omega=e^{\frac{2\pi i}{7}}$ (a primitive root $7$-th root of unity. Then the seven,$7$-th roots of $-3$ are $\alpha,\alpha\omega,\alpha\omega^{2},\alpha\omega^{3},\alpha\omega^{4}...\alpha\omega^{6}$. Thus $\mathbb E=\mathbb Q(\alpha,\omega)$
Now to evaluate $[\mathbb E:\mathbb Q]=[\mathbb E:\mathbb Q(\omega)][\mathbb Q(\omega):\mathbb Q]$
So first calculating $[\mathbb Q(\omega):\mathbb Q]$.
First note that $f(x)$ is monic and also irreducible using E.I.T with prime $3$.This shows that $[\mathbb Q(\omega):\mathbb Q]=\deg f(x)=7$
Now calculating $[\mathbb E:\mathbb Q(\omega)]$.
So, now note that $\mathbb E=\mathbb Q(\alpha,\omega)$ and $x^{7}-1$ is the minimum polynomial of $\omega$ over $\mathbb Q(\alpha)$. This is because $\mathbb Q(\alpha)$ contains only real numbers and $x^{7}-1$ has only non-real solutions. Thus $[\mathbb E:\mathbb Q(\omega)]=\deg(x^{7}-1)=\deg \Phi_{7}(x)=6$. Thus $[\mathbb E:\mathbb Q]=6*7=42$.
You need the $\;7\,-$ th roots of $\;-3=3e^{\pi i+2k\pi i},\,\,k\in\Bbb Z\;$ , and these are the seven numbers
$$z_k=\sqrt[7]3\,e^{\frac{\pi i}7\left(1+2k\right)}\;,\;\;k=0,1,2,...,6$$
Observe that $\;\zeta_7:=e^{\frac{\pi i}7}\;$ is a primitive unit root of order $\;7\;$ , so your polynomial's roots are
$$z_k:=\sqrt[7]{-3}\,\zeta_7^k\;,\;\;k=0,1,...,6$$
and from here you get a rasonably easy way to express the splitting field:
$$\Bbb E=\Bbb Q\left(\sqrt[7]{-3}\,,\,\,\zeta_7\right)$$
and its $\;\Bbb Q\,-$ dimension is $\;[\Bbb E:\Bbb Q(\zeta_7)][\Bbb Q(\zeta_7):\Bbb Q]=7\cdot6=42\;$
Your calculations are fine up to the point when you try to determine the splitting field since
$$\;\color{red}{(-i\sqrt[7]3)^7}=-\left(i\sqrt[7]3\right)^7=-i^7\cdot3=-3i\color{red}{\neq-3}\,\ldots\;$$
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