# How do I find a splitting field $x^8-3$ over $\mathbb{Q}$?

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?

• I would first calculate the 8th roots of $3$. These are simply $\sqrt{3}\omega^i$ for $0\leq i\leq 7$, where $\omega=e^{\pi i/4}$. You can read about them here. Actually calculating $\omega$ in terms of values on the unit circle helps in seeing which elements you need to adjoin to $\mathbb{Q}$. – yunone Nov 13 '11 at 6:45
• @yunone: I admire your restraint in not linking this to a certain recent question. – Jyrki Lahtonen Nov 13 '11 at 8:47

## 2 Answers

Just saying "adjoin all roots" isn't really going to help in figuring out the degree of the extension.

So you really want to be more explicit.

One root is clearly $\sqrt{3}$. Another is $-\sqrt{3}$; but these are the only two real roots, and the splitting field is supposed to have 8 roots, so they are not it; you'll certainly need to deal with complex numbers. But you should know what the degree of $\mathbb{Q}[\sqrt{3}]$ over $\mathbb{Q}$ is, which will also get you started with the degrees.

How much do you know about "complex roots of unity"? You want some complex number $\zeta$ with the property that $\zeta^8 = 1$. Look up "primitive roots of unity". That should help you continue.

• But what's bothering me, though, is that in class the professor told us that we aren't going to prove that the complex numbers are algebraically closed until next week. But don't I need to use that to use primitive roots of unity? – MathTeacher Nov 13 '11 at 6:50
• @MathMastersStudent: You don't need to know that the complex numbers are algebraically closed (i.e., that every nonconstant polynomial with complex coefficients has a root in $\mathbb{C}$). You only need to exhibit the roots of this polynomials, and they can be explicitly written down. You also don't need $\mathbb{C}$ to be algebraically closed to show that a particular complex number $\zeta$ has the property that $\zeta,\zeta^2,\ldots,\zeta^7$, and $1$ are all the roots of $x^8-1=0$ in $\mathbb{C}$. These are not arbitrary polynomials. – Arturo Magidin Nov 13 '11 at 6:53
• How do we know that these specific roots actually exist, then? (other than just saying that there exist roots by existence of splitting fields theorem) – MathTeacher Nov 13 '11 at 7:11
• @MathMastersStudent: You exhibit them. You write them down explicitly, and you verify that they are roots. – Arturo Magidin Nov 13 '11 at 7:15
• ok. I'll give that a shot. Thanks! – MathTeacher Nov 13 '11 at 7:58

Denote by $K$ the splitting field of $x^8-1$. Clearly $x^8-1$ has $8$ complex roots, namely $\zeta^i\cdot\sqrt{3}$ for $i=0,\dots,7$ and $\zeta$ is a primitive $8^{th}$ root of unity. Since $K$ contains $\zeta\cdot\sqrt{3}$ and $\sqrt{3}$, it must contain their quotient $\zeta$. It also contains $\sqrt{3}$, and we can conclude $K\supseteq\mathbb{Q}(\zeta,\sqrt{3})$. On the other hand, any field containing these two elements must contain the splitting field for $x^8-1$, hence $K=\mathbb{Q}(\zeta,\sqrt{3})$.

To determine the dimension of this extension, recall that field automorphisms permute the roots of minimal polynomials. $\sqrt{3}$ 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 $\zeta\notin\mathbb{Q}(\sqrt{3})$, 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 $\mathbb{Q}(\sqrt{2},i,\sqrt{3})=\mathbb{Q}(\zeta,\sqrt{3})$.

• It is not enough that $\zeta\notin\mathbb{Q}(\sqrt{3})$ for $K$ to be of dimension $4\cdot8=32$ over $\mathbb{Q}$. For example, the splitting field of $x^8-2$ over $\Bbb{Q}$ has degree $16$ even though $\zeta\notin\mathbb{Q}(\sqrt{2})$. – Sarastro Dec 14 '15 at 22:11