I was wondering if anyone could explain how i can find the splitting field for $f(x)$ over $\mathbb{Q}$, $E$ say, and also evaluate $[E:\mathbb{Q}]$
for a basic case of $f(x)=x^{4}-15$
I know that i would need to find the $4$ $4th$ roots of $15$, would this then just be $E=\mathbb{Q}(^{4}\sqrt{15},-^{4}\sqrt{15},^{4}\sqrt{15}i,-^{4}\sqrt{15}i)$ $\implies E=\mathbb{Q}(^{4}\sqrt{15},i)$
now $^{4}\sqrt{15}$ is a root to the poly $f(x)$ and $i$ is a root of $x^{2}+1$ thus $\mathbb{Q}(^{4}\sqrt{15},i)$ is finitely generated algebraic extension of $\mathbb{Q}$, thus $\mathbb{Q}(^{4}\sqrt{15},i)$ is a finite extension of $\mathbb{Q}$. I don't understand on where to go from here, i think my roots may be incorrect?
As you said, the polynomial splits if you add the roots $15^{1/4}$ and $i$. Therefore, $\mathbb{Q}(15^{1/4},i)$ is a splitting field.
On the other hand, if $E$ is a splitting field, then $15^{1/4}\in E$, and therefore $\mathbb{Q}(15^{1/4})\subset E$. But since $15^{1/4}i$ is a root of $f$, then $15^{1/4}i\in E$. Therefore $i=15^{1/4}i/15^{1/4}\in E$. Therefore $\mathbb{Q}(15^{1/4},i)\subset E$.
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To evaluate $[E:\mathbb{Q}]=[E:\mathbb{Q}(15^{1/4})][\mathbb{Q}(15^{1/4}):\mathbb{Q}]$.
To compute the factor $[\mathbb{Q}(15^{1/4}):\mathbb{Q}]$ you just need to show that $f$ is irreducible. You can use Eisenstein with the prime $3$ (or $5$). This shows that $[\mathbb{Q}(15^{1/4}):\mathbb{Q}]=\deg(f)=4$.
The factor $[E:\mathbb{Q}(15^{1/4})]=2$, because $E=\mathbb{Q}(15^{1/4})(i)$, and $x^2+1$ is the minimal polynomial of $i$ over $\mathbb{Q}(15^{1/4})$. This is because $\mathbb{Q}(15^{1/4})$ consists of real numbers and $x^2+1$ has only non-real solutions.