Degree of field extension

[mathmari]

Hey! I want to calculate the degree $[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}]$.

It holds that $$[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\sqrt{3}][\mathbb{Q}[\sqrt{3}] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\sqrt{3}]]\cdot 2$$ and $$[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]][\mathbb{Q}[\rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]]\cdot 4$$ since we know that the polynomial $f=x^4-2x^2-1\in \mathbb{Q}[x]$ is irreducible, and $\rho\in \mathbb{C}$ is a root of $f$.

When we divide these two relations we get $[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\sqrt{3}]]=2[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]]$.

From that we get that $[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\sqrt{3}]]\geq 2$ and it is even, right? (Wondering)

It holds that $[\mathbb{Q}[\sqrt{3}, \rho]:\mathbb{Q}] \leq [\mathbb{Q}[\sqrt{3}] :\mathbb{Q}]\cdot [\mathbb{Q}[\rho] :\mathbb{Q}]$, or not? (Wondering)

Therefore, we get that
$[\mathbb{Q}[\sqrt{3}, \rho]:\mathbb{Q}]\leq 8$.

From $$[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]][\mathbb{Q}[\rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]]\cdot 4$$
we get that $[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]]$ is either $1$ or $2$. So $[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\sqrt{3}]]$ is either $2$ or $4$.

How could we continue? (Wondering)

 

[jvicens]

Hi there! It seems like you are trying to calculate the degree of the field extension $\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}$. This is an interesting problem that involves some abstract algebra and number theory.

First, let's define some terms. The degree of a field extension $L:K$ is denoted by $[L:K]$ and it represents the dimension of $L$ as a vector space over $K$. In simpler terms, it is the number of elements in $L$ that can be expressed as linear combinations of elements in $K$.

Now, let's take a closer look at your calculations. You correctly used the tower property of field extensions to break down the degree into smaller parts. This is a useful tool in determining the degree of a field extension. However, there are a few things to consider.

First, you mentioned that $f=x^4-2x^2-1$ is irreducible in $\mathbb{Q}[x]$. This is true, but it is also important to note that $\rho$ is a root of $f$ in $\mathbb{C}$, not in $\mathbb{Q}[\sqrt{3}]$. This means that $\rho$ does not necessarily belong to $\mathbb{Q}[\sqrt{3}]$ and we cannot simply use $[\mathbb{Q}[\rho] :\mathbb{Q}] = 4$. We need to find the minimal polynomial of $\rho$ over $\mathbb{Q}[\sqrt{3}]$ and use that to calculate $[\mathbb{Q}[\rho] :\mathbb{Q}[\sqrt{3}]]$.

Second, you mentioned that $[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]]$ is either $1$ or $2$. This is not necessarily true. It could also be $4$ since $\rho$ is a root of a degree $4$ polynomial.

To continue, you can try finding the minimal polynomial of $\rho$ over $\mathbb{Q}[\sqrt{3}]$ and use that to calculate $[\mathbb{Q}[\rho] :\mathbb{Q}[\sqrt{3}]]$. Then, use the tower property again to calculate $[\mathbb{Q}[\sqrt{3