Basis of Field F over $\mathbb{Q}$ I am self studying Field theory and got struck on this problem.

If $\;F=\mathbb{Q} \left(\sqrt{2}, \sqrt{3}\right),\;$ find
$\;\left[F:\mathbb{Q}\right]\;$ and a basis of $F$ over $\mathbb{Q}$.

I have proved that $\left[F:\mathbb{Q}\right] =4$, but there is a problem in basis elements.
My basis set is $\left\{a, b\sqrt{2}, c\sqrt{3}\right\}$ such that $a, b, c$ belongs to $\mathbb{Q}$. But what should be fourth element and why ?
I am not able to see .
Kindly help.
 A: A basis of $F$ over $\mathbb{Q}$ is $\;\left\{1,\sqrt{2},\sqrt{3},\sqrt{6}\right\},$ indeed for any element $\;x\in F\;$ there exist $\;r_1,\;r_2,\;r_3,\;r_4,\;r_5,\;r_6,\;r_7,\;r_8\in\mathbb{Q}\;$ such that
\begin{align}
x&=\dfrac{r_1+r_2\sqrt{2}+r_3\sqrt{3}+r_4\sqrt{6}}{r_5+r_6\sqrt{2}+r_7\sqrt{3}+r_8\sqrt{6}}\\
&=\dfrac{r_1+r_2\sqrt{2}+r_3\sqrt{3}+r_4\sqrt{6}}{r_5+r_6\sqrt{2}+r_7\sqrt{3}+r_8\sqrt{6}}\cdot\dfrac{r_5+r_6\sqrt{2}-r_7\sqrt{3}-r_8\sqrt{6}}{r_5+r_6\sqrt{2}-r_7\sqrt{3}-r_8\sqrt{6}}\\
&=\dfrac{\left(r_1+r_2\sqrt{2}+r_3\sqrt{3}+r_4\sqrt{6}\right)\left(r_5+r_6\sqrt{2}-r_7\sqrt{3}-r_8\sqrt{6}\right)}{\left(r_5+r_6\sqrt{2}\right)^2-\left(r_7\sqrt{3}+r_8\sqrt{6}\right)^2}\\
&=\dfrac{s_1+s_2\sqrt{2}+s_3\sqrt{3}+s_4\sqrt{6}}{s_5+s_6\sqrt{2}}\\
&=\dfrac{s_1+s_2\sqrt{2}+s_3\sqrt{3}+s_4\sqrt{6}}{s_5+s_6\sqrt{2}}\cdot\dfrac{s_5-s_6\sqrt{2}}{s_5-s_6\sqrt{2}}\\
&=\dfrac{\left(s_1+s_2\sqrt{2}+s_3\sqrt{3}+s_4\sqrt{6}\right)\left(s_5-s_6\sqrt{2}\right)}{\left(s_5+s_6\sqrt{2}\right)\left(s_5-s_6\sqrt{2}\right)}\\
&=q_11+q_2\sqrt{2}+q_3\sqrt{3}+q_4\sqrt{6}
\end{align}
where $\;q_1,\;q_2,\;q_3,\;q_4\in\mathbb{Q}\;$.
