How can I show that $\mathbb{Q}(\sqrt{p},\sqrt{q}) \subseteq \mathbb{Q}(\sqrt{p}+\sqrt{q})$, for distinct primes $p,q?$ The other inclusion is trivial.
I tried saying $$(\sqrt{p}+\sqrt{q})^{-1} = \frac{1}{\sqrt{p}+\sqrt{q}} = \frac{\sqrt{p}-\sqrt{q}}{p-q},$$ and since $p-q = -(q-p) \in \mathbb{Z},$ note that $(p-q)(\sqrt{p}+\sqrt{q})^{-1} + \sqrt{q}$ and $(p-q)(\sqrt{p}+\sqrt{q})^{-1} + \sqrt{p}$ are in $\mathbb{Q}(\sqrt{p}+\sqrt{q}).$
I'm almost there.
We find $$\sqrt{pq}=\frac12\bigl((\sqrt p+\sqrt q)^2-p-q\bigr)\in\mathbb Q(\sqrt p+\sqrt q)$$ and then $$p\sqrt q+q\sqrt p= (\sqrt p+\sqrt q)\sqrt{pq}\in\mathbb Q(\sqrt p+\sqrt q)$$ so finally $$\sqrt p= \frac{p(\sqrt p+\sqrt q)-(p\sqrt q+q\sqrt p)}{p-q}\in\mathbb Q(\sqrt p+\sqrt q)$$ and$$\sqrt q= \frac{q(\sqrt p+\sqrt q)-(p\sqrt q+q\sqrt p)}{q-p}\in\mathbb Q(\sqrt p+\sqrt q).$$
Let $\alpha=\sqrt p+\sqrt q$.
Find $\alpha ^3=(p+3q)\sqrt{p}+(3p+q)\sqrt q$.
These equalities can be rewritten as $$\begin{bmatrix} 1 & 1\\ p+3q & 3p+q\end{bmatrix}\begin{bmatrix}\sqrt p\\ \sqrt q \end{bmatrix}=\begin{bmatrix} \alpha \\ \alpha ^3\end{bmatrix}_.$$
The square matrix is clearly invertible and its entries are in $\mathbb Q(\alpha)$ and so are the entries of the matrix on the RHS, thus $$\begin{bmatrix}\sqrt p\\ \sqrt q \end{bmatrix}=\begin{bmatrix} 1 & 1\\ p+3q & 3p+q\end{bmatrix}^{-1}\begin{bmatrix} \alpha \\ \alpha ^3\end{bmatrix}\in \mathcal M_{2\times 2}(\mathbb Q(\alpha)).$$
Therefore $\sqrt p,\sqrt q\in \mathbb Q(\alpha)$.