Finding a basis for $\Bbb{Q}(\sqrt{2}+\sqrt{3})$ over $\Bbb{Q}$. I have to find a basis for $\Bbb{Q}(\sqrt{2}+\sqrt{3})$ over $\Bbb{Q}$. 
I determined that $\sqrt{2}+\sqrt{3}$ satisfies the equation $(x^2-5)^2-24$ in $\Bbb{Q}$. 
Hence, the basis should be $1,(\sqrt{2}+\sqrt{3}),(\sqrt{2}+\sqrt{3})^2$ and $(\sqrt{2}+\sqrt{3})^3$.
However, this is not rigorous. How can I be certain that $(x^2-5)^2-24$ is the minimal polynomial that $\sqrt{2}+\sqrt{3}$ satisfies in $\Bbb{Q}$? What if the situation was more complicated? In general, how can we ascertain thta a given polynomial is irreducible in a field?
Moreover, checking for linear independence of the basis elements may also prove to be a hassle. Is there a more convenient way of doing this?
Thanks. 

A basis of $\mathbb Q\big[\sqrt{2},\sqrt{3}\big]$ consists of the elements $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$, and hence its dimension over $\mathbb Q$ is equal to $4$.
Clearly, all the above elements belong to $\mathbb Q\big[\sqrt{2},\sqrt{3}\big]$, and hence it remains to show that they are independent over $\mathbb Q$.
There is a rather elegant way to show this, and in fact something more general:
If $n_1,n_2,\ldots,n_k$ are distinct square-free positive integers, then $\sqrt{n_1},\sqrt{n_2},\ldots,\sqrt{n_k}$ are linearly independent over $\mathbb Q$.
(A number $n$ is said to be square free if $k^2\mid n$, implies that $k=1$.)
