On the basis of field extension Ok, so, $\mathbb{Q} (\sqrt{3} + \sqrt{5})$ means all the numbers $ a + b (\sqrt{3} + \sqrt{5}) $ , with $a,b \in \mathbb{Q}$ .
So know that $ \sqrt{3} + \sqrt{5} $ is algebraic of order 4 in $ \mathbb{Q} $ , so the basis of $\mathbb{Q} (\sqrt{3} + \sqrt{5})$ over $\mathbb{Q}$ must be $\{1, \sqrt{3} + \sqrt{5}, (\sqrt{3} + \sqrt{5})^2 , (\sqrt{3} + \sqrt{5})^3 \}$.
But shouldn't it be just $\{1, \sqrt{3} + \sqrt{5} \}$ ? Because any element should be a linear combination of only these two, right? What am I missing?
 A: $\mathbb{Q}(\sqrt{3}+\sqrt{5})$ is not equal to what you wrote, but it is equal to the smallest field that contains $\mathbb{Q}$ and $\sqrt{3}+\sqrt{5}$ (and that is equal to $\{a+b(\sqrt{3}+\sqrt{5})+c(\sqrt{3}+\sqrt{5})^2+d(\sqrt{3}+\sqrt{5})^3, a,b,c,d\in\mathbb{Q}\}$). 
In general $\mathbb{Q}(a)$ denotes the smallest field extension of $\mathbb{Q}$ that contains $a$ and that is not always equal to $\{q_1+q_2a, q_1,q_2\in\mathbb{Q}\}$ (it is for example when $a=\sqrt{n}$, for some $n$.. but not in most cases. In fact it is what you wrote if and only if $a$ is algebraic of order 2). 
A: Take$\def\Q{\mathbb Q}$ $(\sqrt3+\sqrt5)^2-8=2\sqrt{15}$ that is not an element of $\Q+(\sqrt3+\sqrt5)\Q$.

all the numbers $ a + b (\sqrt{3} + \sqrt{5}) $ , with $a,b \in \mathbb{Q}$

You would denote that $\Q$-module as $\Q + (\sqrt{3} + \sqrt{5})\Q$.
Notation $\Q(\zeta)$ denotes the smallest field containing $\Q$ and $\zeta$, a finite field extension of $\Q$ if $\zeta$ is algebraic. $\Q(\zeta)$ can also be regarded as $\Q$-module, but in general it's not the same as $\zeta\Q$ or $\Q+\zeta\Q$, and it's basis is not that obvious.
A: $\mathbb Q(\alpha)$ does not mean "all the numbers $a + b\alpha$" it means the smallest field containing $\mathbb Q$ and $\alpha$ which means all rational functions in $\alpha$. For instance $\frac{1 + 2\alpha^3 - \alpha^5}{3\alpha - \alpha^4}$. It is a theorem that when $\alpha$ is algebraic, you can always clear denominators so that $\mathbb Q(\alpha) = \mathbb Q[\alpha] = $ the smallest ring containing $\mathbb{Q}$ and $\alpha = $ all polynomials in $\alpha$.
Then if $\alpha$ satisfies a polynomial like $\alpha^n + a_{n - 1}\alpha^{n - 1} + \dots + a_0=0$, you can always reduce larger powers of $\alpha$ by replacing $\alpha^n$ with $-(a_{n - 1}\alpha^{n - 1} + \dots + a_0)$.
