Degree of field extension over the rationals

Question

So I started with $x^4+1\\$

$\begin{eqnarray} x^4+1 &=& (x^2+1)^2-2x^2\\ &=& (x^2 + x\sqrt{2} + 1)(x^2 - x\sqrt{2} +1) \end{eqnarray}$

So $x=\pm \frac{\sqrt{2}}{2}(1 \pm i)$ and the splitting field is $K=\mathbf{Q}[\sqrt{2},i]$.

Is $[K:\mathbf{Q}]=4$ because $x^4+1$ has degree 4 and is irreducible over the rationals? So it is minimal over the rationals, but not the splitting field?

Answer 1

When $f$ is monic and $R$ a ring, the degree of the ring extension $R[x] / f(x)$ over $R$ is always well-defined and equal to the degree of $f$.

When $F$ is a field and $f$ irreducible with a root $\alpha$, then $F(\alpha)$ over $F$ is isomorphic to an extension of that form.

So, setting $\alpha = (1+\mathbf{i})/\sqrt{2}$, you have $[\mathbf{Q}(\alpha) : \mathbf{Q}] = 4$.

You're maybe missing this idea:

Exercise: Prove $K = \mathbf{Q}(\alpha)$.

This is an unusual feature of this polynomial: that adjoining a single root happens to give you the entire splitting field.

As an aside, here are two proof ideas:

Determine the degree of the extension $K / \mathbf{Q}(\alpha)$

and

Directly show that $\mathbf{i}$ and $\sqrt{2}$ are both in $\mathbf{Q}(\alpha)$. Possibly by observing $\alpha$ is an eighth root of unity.

Answer 2

You're right in that, if a degree $d$ polynomial $f(x)$ is irreducible over $\mathbb Q$, and $\alpha$ is a single root of $f(x)$, then $[\mathbb Q(\alpha) : \mathbb Q] = d$.

To get the splitting field of $f(x)$ over $\mathbb Q$, you need to adjoin all of the roots of $f(x)$ to $\mathbb Q$! In other words, if $\alpha_1, \dots , \alpha_d$ are the roots of $f(x)$, then the splitting field of $f(x)$ over $\mathbb Q$ is $\mathbb Q(\alpha_1, \dots, \alpha_d)$. Of course, $\mathbb Q(\alpha_1, \dots, \alpha_d)$ may well be bigger than $\mathbb Q(\alpha_i)$ for any single $\alpha_i$, and therefore, $[\mathbb Q(\alpha_1, \dots, \alpha_d) : \mathbb Q]$ may well be bigger than $d$.

However, your example is quite special. The four roots of $f(x)$ are $\zeta_8, \zeta_8^3, \zeta_8^5$ and $\zeta_8^7$, where $\zeta_8 = \exp (2\pi i /8)$. The reason it is special is that the splitting field $\mathbb Q(\zeta_8, \zeta_8^3, \zeta_8^5, \zeta_8^7)$ happens to be equal to $ \mathbb Q(\zeta_8)$! Can you prove this?

Since $[\mathbb Q(\zeta_8) : \mathbb Q] = 4$, and since $\mathbb Q(\zeta_8, \zeta_8^3, \zeta_8^5, \zeta_8^7) = \mathbb Q(\zeta_8)$, it follows that $[\mathbb Q(\zeta_8, \zeta_8^3, \zeta_8^5, \zeta_8^7) : \mathbb Q] = 4$.

I encourage you find the splitting field of $f(x) = x^3 - 2$ over $\mathbb Q$. This is a irreducible cubic, but you'll discover that the splitting field is a degree six extension over $\mathbb Q$.