Show that this is an integral basis of the ring of integers of $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Show that $1,\sqrt{3},\sqrt{2},\frac{\sqrt{2}+\sqrt{6}}{2}$ is an integral basis for $\mathcal{O}_K$ where $K=\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{6})$. 
Clearly, the rank of $\mathcal{O}_K$ is indeed $4$, so it suffices to show that this set spans. My objective is to take an arbitrary $\alpha\in \mathcal{O}_K$ and show that it can be written in terms of my claimed basis with integer coefficients, but I am not sure how one would achieve this. I know there is the exercise treating the general case in Marcus's $\textit{Number Fields}$, but I am interested in keeping the computations straightforward and limited to this concrete example. (In any case, I do not quite see how the exercise in Marcus is done either.)
Any and all help would be appreciated.
 A: Here’s another way of doing it, using the (relatively easy) fact that a basis of $K$ is an integral basis (i.e. a $\Bbb Z$-basis of $\mathcal O$) if and only if it‘s a basis locally at every prime of $\Bbb Z$.
I checked the discriminant of your basis, and found $2^8\cdot3^2$, so that your basis is good at all primes not dividing $6$ — the discriminant of any true basis would divide your $2^8\cdot3^2$. Now to check $2$ and $3$:
I’m going to make typing easier for myself by writing your fourth basis element $\beta=\frac{\sqrt2+\sqrt6}2=\frac{1+\sqrt3}{\sqrt2}$.
Now above $2$, there’s only one prime $\mathfrak p_2$ of $K$, with ramification index $4$, and I checked (just by computing the norms down to $\Bbb Q$) that $\beta-1=\pi$ is locally a prime, that is $v_{\mathfrak p_2}(\pi)=1$, furthermore that $v_{\mathfrak p_2}(\sqrt2\,)=2$ (of course), and equally obviously that $v_{\mathfrak p_2}(1+\sqrt3-\sqrt2)=3$, since it’s $\pi\sqrt2$. Thus, locally at $2$, your basis is good, since together with $1$, the numbers I checked must be a basis.
Now, at $3$, the situation is a bit different. There’s still just one prime above $3$, call it $\mathfrak p_3$, but the ramification index and the residue field degree are both $2$. Now $1$ and $\sqrt2$ form a basis of the unramified part, i.e. give independent elements of the residue field (over $\Bbb F_3$), and $v_{\mathfrak p _3}(\sqrt3\,)=1$, so all we need to do is find something else of valuation $1$, but which, as element of $\mathfrak p_3/\mathfrak p_3^2$ is not an $\Bbb F_3$-multiple of $\sqrt3$, this “something” to be a $\Bbb Z$-linear combination of elements of your proposed basis. But we get:
$$
\beta+\sqrt2=\frac{1+\sqrt3+2}{\sqrt2}=\frac{3+\sqrt3}{\sqrt2}\equiv\frac{\sqrt2\cdot\sqrt3}{2}\pmod{\mathfrak p_3^2}\,,
$$
which fills the bill.
So, without dependence on any further discriminant calculation, your basis turns out to be an integral basis.
