Show that $\mathbf{Q}(\alpha + \beta) = \mathbf{Q}(\alpha, \beta)$ I am looking for help in proving that:
$\mathbf{Q}(\alpha + \beta) = \mathbf{Q}(\alpha, \beta)$,
where $\mathbf{Q}(X)$ denotes the field of rational numbers adjoin $X$. Is it enough to quote the Primitive Element Theorem?
$\alpha, \beta$ are elements of distinct extension fields of $\mathbf{Q}$. That is to say, the extensions are such that $\mathbf{Q}(\alpha)\cap\mathbf{Q}(\beta) = \mathbf{Q}$. Apologies for any confusing regarding the use of 'distinct', I hope the clarification helps.
 A: The primitive element theorem tells you that there is a primitive element. It does not say that $\alpha+\beta$ is a primitive element, and this is what you are trying to prove.
There is a reason that then primitive element theorem does not go via $\alpha+\beta$, which is that it is not that simple - as even the simple counter-examples to your unedited question show, some care needs to be taken.
If you wanted to ask a related question, you might ask for necessary and sufficient criteria for $\alpha+\beta$ to be a primitive element - which is what you have been working towards with your edits.
But to answer your primary question again, it is insufficient to quote the primitive element theorem to achieve what you want.
A: This is not true in general (even with the assumption of trivial intersection in place). 
Consider the case of $\alpha=e^{2\pi i/3}\root3\of2$, one of the complex zeros of $p(x)=x^3-2$. Let $\beta=\overline{\alpha}$. Then $\Bbb{Q}(\alpha)$ and $\Bbb{Q}(\beta)$ are two distinct cubic extensions, and hence they intersect trivially. Here $\Bbb{Q}(\alpha,\beta)$ is the splitting field of $p(x)$, a degree six extension. 
On the other hand $\alpha+\beta=-\root3\of2$ generates only a cubic extension.
