Primitive elements for $K=\Bbb{Q}(\sqrt{2},\sqrt{3})$ 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]?
 A: 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.
A: This is an exercise in a section (Ch 15.8 Primitive Elements) before Galois Theory in Artin's Algebra (2nd edition).
While the fundamental theorem of the Galois theory is implicitly used in Ewan's nice answer, here is a tedious one without Galois theory, which is recorded mostly for my own benefit. 
Let $\alpha=\sqrt{2}$ and $\beta=\sqrt{3}$. 
Then $K$ has the following five subfields (without using Galois theory, one cannot conclude immediately that these are "exactly" the intermediate fields so far.)
$$
K_1=\Bbb{Q}, \ \ K_2=\Bbb{Q}(\alpha),\ \ K_3=\Bbb{Q}(\beta), \ \ K_4=\Bbb{Q}(\alpha\beta),\ \ K.
$$
Let $\gamma\in K$ be such that $K=\Bbb{Q}(\gamma)$. Then $\gamma$ must be of the form 
$$
\gamma=a+b\alpha+c\beta+d\alpha\beta
$$
for some $a,b,c,d\in \Bbb{Q}$, since $(1,\alpha,\beta,\alpha\beta)$ is a basis of the $\Bbb{Q}$-vector space $K$. Since $[K:\Bbb{Q}]=4$, $\gamma$ must be of degree $4$. We show that this would happen if and only if at most one of $b,c,d$ is zero. 
Suppose that $\gamma$ has degree $4$ and that more than one of $b,c,d$ are zero. Then
$$
\gamma\in K_i
$$
for some $i\in\{1,2,3,4\}$, which contradicts the degree of $\gamma$. 
Conversely, suppose at most one of $b,c,d$ is zero. We want to show that $\gamma$ is of degree $4$. Note that $\Bbb{Q}\subset\Bbb{Q}(\gamma)\subset K$. Thus, the only possible degrees of $\gamma$ are $1,2,4$. It thus suffices to show that the degree can neither be $1$ nor $2$. Since "at most one of $b,c,d$" is zero implies that one of $b,c,d$ must be non zero, we have $\gamma\not\in \Bbb{Q}$ and thus its degree cannot be $1$. Supose $\gamma$ has degree $2$. Then $\gamma^2$ can be written as a $\Bbb{Q}$-linear combination of $\{1,\gamma\}$:
$$
\gamma^2=k\cdot 1+l\cdot\gamma,\quad l,k\in\Bbb{Q}.
$$
Note that
$$
\gamma^2=(a^2+2b^2+3c^2+6d^2)+(2ab+6cd)\alpha+(2ac+4bd)\beta+(2ad+2bc)\alpha\beta.
$$
One the other hand
$$
k+l\gamma=(k+la)+(lb)\alpha+(lc)\beta+(ld)\alpha\beta.
$$
Comparing the coefficients of $\alpha,\beta,\alpha\beta$ gives
$$
\begin{align}
&l=(2ab+6cd)/b=2a+6cd/b,\\
&2ac+4bd=lc=2ac+6c^2d/b, \\
&l=(2ac+4bd)/c=2a+4bd/c,\\
&2ad+2bc=ld=2ad+4bd^2/c, 
\end{align}
$$
which implies that
$$
\begin{align} 
2d(3c^2-2b^2)=0
\end{align}
$$
and 
$$
\begin{align} 2b(2d^2-c^2)=0,
\end{align}
$$
If $3c^2-2b^2=0$, then $c=b=0$ since otherwise both $c$ and $b$ cannot be zero, which implies that $\sqrt{3/2}\in\Bbb{Q}$ which is impossible. But this contradicts our assumption that "at most one of $b,c,d$" is zero. Hence, $3c^2-2b^2\not=0$. It follows that $d=0$. Similarly, $b=0$. Therefore, $d=b=0$, which is again a contradiction. 
To sum up, $\gamma$ is such that $\Bbb{Q}(\gamma)=K$ if and only if it is of the form
$$
\gamma=a+b\alpha+c\beta+d\alpha\beta
$$
where at most one of ${b,c,d}$ is zero.
