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.

  • $\begingroup$ you tried something? $\endgroup$
    – user87543
    May 3, 2015 at 14:38
  • 2
    $\begingroup$ You can see that $\sqrt{p}\sqrt{q}\in\mathbb{Q}(\sqrt{p}+\sqrt{q})$ by squaring $\sqrt{p}+\sqrt{q}$. $\endgroup$
    – Alamos
    May 3, 2015 at 14:43
  • $\begingroup$ @PraphullaKoushik See edit. $\endgroup$
    – user85362
    May 3, 2015 at 14:45
  • 3
    $\begingroup$ Hey, your attempt is even faster! Try linear combining $\sqrt p$ and $\sqrt q$ from $\sqrt p+\sqrt q$ and $\sqrt p-\sqrt q$ $\endgroup$ May 3, 2015 at 14:45
  • 2
    $\begingroup$ to finish off in your argument you really want to say $\frac12 (p-q)(\sqrt{p}+\sqrt{q})^{-1} + (\sqrt{p}+\sqrt{q})$ and $\frac12(p-q)(\sqrt{p}+\sqrt{q})^{-1} - (\sqrt{p}+\sqrt{q})$ are in $\mathbb{Q}(\sqrt{p}+\sqrt{q})$ $\endgroup$ May 3, 2015 at 15:20

2 Answers 2


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 $\sqrt p,\sqrt q\in \mathbb Q(\alpha)$.


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$$\sqrt q= \frac{q(\sqrt p+\sqrt q)-(p\sqrt q+q\sqrt p)}{q-p}\in\mathbb Q(\sqrt p+\sqrt q).$$


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