Calculate the degree of the field extensions.

Question

I have been staring at this question for a while. I'm sure there is a little trick I am missing...anyway, it is the following:

$ f = x^3 + x + 3 $

a) Show $f$ is irreducible over $\mathbb{Q}[x]$: Did this fine, using the following theorem: "If $f$ is primitive (i.e. the GCD of the coefficients is 1), then $f$ is irreducible in $\mathbb{Z}[x] \iff f$ is irreducible in $\mathbb{Q}[x]$, then showed by contradiction that $f$ was irreducible in $\mathbb{Z}[x]$.

b) Show that $f$ has exactly one real root: Pretty straightforward with Intermediate Value Theorem and showing that $~f' > 0. $

c) Let $\theta$ be the real root of $f$. Let $\phi, \phi'$ be the two other roots. Compute:

[$\mathbb{Q}(\theta)$ : $\mathbb{Q}$], [$\mathbb{Q}(\theta, \phi)$ : $\mathbb{Q}$], [$\mathbb{Q}(\theta, \phi, \phi')$ : $\mathbb{Q}$].

Managed to compute the first one as 3, as $f$ is the minimal polynomial for $\theta$ and is of degree 3.

As for the second one, using the tower law, I can get

$\displaystyle [\mathbb{Q}(\theta, \phi) : \mathbb{Q}] = [\mathbb{Q}(\theta) : \mathbb{Q}][\mathbb{Q}(\theta, \phi) : \mathbb{Q}(\theta)] = 3[\mathbb{Q}(\theta, \phi) : \mathbb{Q}(\theta)]. $

My question is, how do I compute $[\mathbb{Q}(\theta, \phi) : \mathbb{Q}(\theta)]$? If I know this, i'm sure I can extend the method to get the final part. Would I need to find the minimal polynomial of $\phi$ over $\mathbb{Q}(\theta)$? If so, how do I show the polynomial is in $\mathbb{Q}(\theta)$ and not just $\mathbb{Q}$ ?

Answer 1

Over $\mathbb Q(\theta)$, you can divide out the factor $x-\theta$ from $f$ to obtain a quadratic polynomial with $\phi,\phi'$ as roots. But is it minimal? Could possibly $\mathbb Q(\theta,\phi)=\mathbb Q(\theta)$? (Use that $\theta\in\mathbb R$)

Answer 2

The extension $\Bbb{Q}(\theta)$ is indeed degree $3$ and has a basis $1,\theta, \theta^2$. Now in $\Bbb{Q}(\theta, \phi,\phi')$ the polynomial $f$ can be written as $f(x)=(x-\theta)(x^2+\theta x+\theta^2+1)$. Since $\phi+\phi'+\theta=0$ we have $\mathbb{Q}(\theta, \phi, \phi')=\mathbb{Q}(\theta, \phi)$. Since $\phi$ is complex $\phi\notin \Bbb{Q}(\theta)$ its minimal polynomial over $\mathbb{Q}(\theta)$ is quadratic so $[\mathbb{Q}(\theta) : \mathbb{Q}]=2$. A basis for $\mathbb{Q}(\theta, \phi)$ is $1,f,f^2,\phi,\phi f, \phi f^2$.