I'll give a proof of the following.
THEOREM: Let $x = p+\sqrt{q}$, with $p,q \in \mathbb{Q}$, and $m$ an integer. Then $x^m = a+b\sqrt{q}$ with $a,b \in \mathbb{Q}$.
PROOF
Using the Binomial Theorem
${x^m} = C_0^m{p^m} + C_1^m{p^{m - 1}}{q^{\frac{1}{2}}} + C_2^m{p^{m - 2}}{q^{\frac{2}{2}}} + \cdots + C_{m - 2}^m{p^2}{q^{\frac{{m - 2}}{2}}} + C_{m - 1}^mp{q^{\frac{{m - 1}}{2}}} + C_m^m{q^{\frac{m}{2}}}$
Let m= 2·j then
${x^{2j}} = C_0^{2j}{p^{2j}} + C_1^{2j}{p^{2j - 1}}{q^{\frac{1}{2}}} + C_2^{2j}{p^{2j - 2}}{q^{\frac{2}{2}}} + \cdots + C_{2j - 2}^{2j}{p^2}{q^{\frac{{2j - 2}}{2}}} + C_{2j - 1}^{2j}p{q^{\frac{{2j - 1}}{2}}} + C_{2j}^{2j}{q^{\frac{{2j}}{2}}}$
Grouping produces
${x^{2j}} = \sum\limits_{k = 0}^j {C_{2k}^{2j}{p^{2j - 2k}}{q^k}} + \sum\limits_{k = 1}^j {C_{2k - 1}^{2j}{p^{2j - 2k + 1}}{q^{k - 1}}\sqrt q } $
But since every binomial coefficient is integer, and every power of $p$ and $q$ is rational then one has
${x^{2j}} = a+b\sqrt{q} \text{ ; and } a,b \in \mathbb{Q}$ where $b$ and $a$ are the sums.
If $m = 2j+1$ then
$x^{2j+1} =(a+b\sqrt{q}) (p+\sqrt{q}) = c+d\sqrt{q}$
which is also in our set.
(I don't know if the ring-theory tag is appropiate.)