Here's the situation. I am in this algebra class, and so far we have defined splitting fields and proved their existence and uniqueness. We have not yet decided on any rigorous definition of complex numbers, by the way. For a homework question (and yes, we are allowed to use any internet resources we want),
I have to find a splitting field for $x^8-3$ over $\mathbb{Q}$ and find its degree of extension.
I don't really know how to go about it. If I can use that the complex numbers are algebraically closed, can't I just adjoin all of the roots, or is there something more explicit that I can do in order to find the degree of extension?
Best Answer
Denote by $K$ the splitting field of $x^8-1$. Clearly $x^8-1$ has $8$ complex roots, namely $\zeta^i\cdot\sqrt[8]{3}$ for $i=0,\dots,7$ and $\zeta$ is a primitive $8^{th}$ root of unity. Since $K$ contains $\zeta\cdot\sqrt[8]{3}$ and $\sqrt[8]{3}$, it must contain their quotient $\zeta$. It also contains $\sqrt[8]{3}$, and we can conclude $K\supseteq\mathbb{Q}(\zeta,\sqrt[8]{3})$. On the other hand, any field containing these two elements must contain the splitting field for $x^8-1$, hence $K=\mathbb{Q}(\zeta,\sqrt[8]{3})$.
To determine the dimension of this extension, recall that field automorphisms permute the roots of minimal polynomials. $\sqrt[8]{3}$ has minimal polynomial $x^8-3$ over $\mathbb{Q}$, ($x^8-3$ is 3-Eisenstein, hence irreducible). $\zeta$ has minimal polynomial $\Phi_8(x)=x^4+1$ (see below), hence $\mathbb{Q}(\zeta)$ is an extension of dimension 4 over $\mathbb{Q}$. It is not difficult to see that $\zeta\notin\mathbb{Q}(\sqrt[8]{3})$, hence $K$ is of dimension $4\cdot8=32$ over $\mathbb{Q}$.
Last, recall (see Dummit & Foote 13.6, for instance) that the $n^{th}$ roots of unity satisfy
$x^n-1=\prod\Phi_d(x)$
where $d$ goes over all the divisors of $n$. In our case, $n=8$,
$x^8-1 = (x^4-1)\cdot(x^4+1)=(x-1)(x+1)(x^2+1)\cdot(x^4+1)$
where the first three multipliers at the last equality above are $\Phi_1(x)=x-1$, $\Phi_2(x)=x+1$ and $\Phi_4(x)=x^2+1$.
As a final remark, we could have taken $\zeta=(1+i)\sqrt{2}/2$. From a similar argument to before, $\mathbb{Q}(\sqrt{2},i)\supseteq\mathbb{Q}(\zeta)$. However, both sides are a 4-dimensional extension over $\mathbb{Q}$, hence they must be equal. This can also be proven directly, by writing $i$ and $\sqrt{2}$ only using powers of $\zeta$.
To summarize, the splitting field of $x^8-3$ is $\mathbb{Q}(\sqrt{2},i,\sqrt[8]{3})=\mathbb{Q}(\zeta,\sqrt[8]{3})$.