Dummit and Foote page 526

Question

I'm having trouble with a line of example 2 on page 526.

Consider the field $\mathbb{Q}(\sqrt{2},\sqrt{3})$. generated over $\mathbb{Q}$ by $\sqrt{2}$ and $\sqrt{3}$. Since $\sqrt{3}$ is of degree $2$ over $\mathbb{Q}$, the degree of the extension $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}(\sqrt{2})$ is at most $2$ and is precisely $2$ iff $ x^2 - 3$ is irreducible over $\mathbb{Q}(\sqrt{2})$.

Why is it at most two? From the previous page we know that $[\mathbb{Q}(\sqrt 2 , \sqrt 3) : \mathbb{Q} \sqrt 2][\mathbb{Q}(\sqrt 2 ) : \mathbb{Q}]=[\mathbb{Q}(\sqrt 2 , \sqrt 3) : \mathbb{Q}]$

Can I somehow use this?

Answer 1

Because $\{1,\sqrt 3\}$ spans $K(\sqrt{3})/K$ for any field $K$, so as a vector space, $K(\sqrt{3})/K$ has dimension at most $2$.

Here $K=\mathbb Q(\sqrt{2})$ since $\mathbb Q(\sqrt{2})(\sqrt{3})=\mathbb Q(\sqrt 2,\sqrt 3)$.

Answer 2

It is at most $2$ because the quadratic polynomial $f(x) = x^2 - 3 \in \mathbb{Q}[\sqrt{2}][x]$ has a root in the field extension. Worst case scenario, $f(x)$ is irreducible, which would mean that the field extension is degree $2$. Otherwise, if the polynomial was reducible, then it would be a (trivial) degree $1$ extension.

The important think to note is that, given a field extension $F[\alpha]$ over $F$, the degree of that field extension will be exactly the degree of the minimal (irreducible) polynomial in $F[x]$ with $\alpha$ as a root.

Answer 3

When you add an algebraic element $r$ to a field $F$, the degree of the extension equals the degree of the minimum polynomial of $r$ over $F$. The minimum polynomial is a factor of any polynomial $f(X)\in F[X]$ such that $f(r)=0$.

In this case $F=\mathbb{Q}(\sqrt{2})$, $r=\sqrt{3}$ and $f(X)=X^2-3$.