# Show that $\mathbb Q(\sqrt m,\sqrt n)=\mathbb Q(\sqrt {m}+\sqrt {n})$.

Show that $$\mathbb Q(\sqrt m,\sqrt n)=\mathbb Q(\sqrt {m}+\sqrt {n})$$

My attempt: It is obvious that $$\mathbb Q(\sqrt {m}+\sqrt {n}) \subset \mathbb Q(\sqrt m,\sqrt n)$$ .

Is this proof is correct?

• Compare your answer with this post. The proof there also works here. Your fourth line is no equation. Commented Dec 14, 2019 at 20:04
• The only problem I can see is that we aren't guaranteed $4m + 2n \neq 0$ and $2m - 2n \neq 0$. I suppose these two situations could be treated as special cases. I'm trying to write a proof that doesn't need case management. @DietrichBurde is right though. The fourth line is not clear. I know what you mean by it, but it should be an equation or a statement of set membership or something like that. Commented Dec 14, 2019 at 20:21
• @DietrichBurde, I think the OP simply omitted a "$\in\mathbb{Q}(\sqrt m+\sqrt n)$" at the end of the fourth line. (There should really also be one at the end of the third line as well). Commented Dec 14, 2019 at 20:23
• @CharlesHudgins, I agree with respect to $4m+2n$, but it seems clear enough that $\mathbb{Q}(\sqrt m,\sqrt n)=\mathbb{Q}(\sqrt m+\sqrt n)$ if $m=n$. However, it's enough to show that $\sqrt n\in\mathbb{Q}(\sqrt m+\sqrt n)$, since $\sqrt m=(\sqrt m+\sqrt n)-\sqrt n$. Commented Dec 14, 2019 at 20:27
• @BarryCipra Seems simple enough. I was hoping for a proof that did it all in one fluid motion, but that seems like it should work. I wonder why the "no perfect squares" assumption was included in the problem statement. Commented Dec 14, 2019 at 20:29

It is good, but you can shorten it to just deduce that $$2(m-n)\sqrt{n}\in\mathbb{Q}(\sqrt{m}+\sqrt{n})$$ If $$m=n$$ the statement $$\mathbb{Q}(\sqrt{m}+\sqrt{n})=\mathbb{Q}(\sqrt{m},\sqrt{n})$$ is obvious, so we can assume $$m\ne n$$. Thus $$\sqrt{n}\in\mathbb{Q}(\sqrt{m}+\sqrt{n})$$ and so also $$\sqrt{m}=(\sqrt{m}+\sqrt{n})-\sqrt{n}\in\mathbb{Q}(\sqrt{m}+\sqrt{n})$$