The key lemma for proving the primitive Element Theorem (for finite extension of a field $F$ with characteristic $0$) in Artin's Algebra (2nd edition) is the following:
Suppose $char F=0$ and $K=F(\alpha,\beta)$ is a finite extension of $F$. Then $F(\alpha,\beta)=F(\alpha+c\beta)$ for all but finitely many $c$ in $F$.
This does give the existence of primitive elements but not all of them.
Consider $F=\Bbb{Q}$ and $\alpha=\sqrt{2}$ and $\beta=\sqrt{3}$. Suppose $Q(\sqrt{2},\sqrt{3})=\Bbb{Q}(\gamma)$. Then $\gamma$ must be of the form
$$
\gamma=c_1+c_2\sqrt{2}+c_3\sqrt{3}+c_4\sqrt{6},\quad c_i\in{\Bbb Q}.
$$
Using the method in the proof of the lemma quoted above, one can find all the primitive elements of the form $\sqrt{2}+c\sqrt{3}$ where $c\in\Bbb{Q}$.
A quick search on Google returns the article of Primitive elements theorem, which says that
Generally, the set of all primitive elements for a finite separable extension $L / K$ is the complement of a finite collection of proper $K$-subspaces of $L$, namely the intermediate fields.
This looks something related to the Galois theory about which I don't know much yet.
Here is my question:
How can one find all the $\gamma$ such that $\Bbb{Q}(\gamma)=\Bbb{Q}(\sqrt{2},\sqrt{3})$ [EDITED: without using Galois theory]?
Best Answer
Easy. There are exactly five subfields of ${\mathbb Q}(\sqrt{2},\sqrt{3})$ : $K_1={\mathbb Q},K_2={\mathbb Q}(\sqrt{2}), K_3={\mathbb Q}(\sqrt{3}), K_4={\mathbb Q}(\sqrt{6}), K_5={\mathbb Q}(\sqrt{2},\sqrt{3})$ and ${\mathbb Q}(\gamma)$ is one of them.
It follows that $\gamma$ is a primtive element iff $\gamma\not\in K_2,\gamma\not\in K_3,$ and $\gamma\not\in K_4$.
Writing $\gamma=c_1+c_2\sqrt{2}+c_3\sqrt{3}+c_4\sqrt{6}$, this is equivalent to $(c_3,c_4)\neq(0,0),(c_2,c_3)\neq(0,0),(c_2,c_4)\neq(0,0)$, in other words at most one of $c_2,c_3,c_4$ is zero.