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?

  • 1
    $\begingroup$ Compare your answer with this post. The proof there also works here. Your fourth line is no equation. $\endgroup$ Dec 14, 2019 at 20:04
  • $\begingroup$ 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. $\endgroup$ Dec 14, 2019 at 20:21
  • $\begingroup$ @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). $\endgroup$ Dec 14, 2019 at 20:23
  • $\begingroup$ @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 $\sqrt m=(\sqrt m+\sqrt n)-\sqrt n$. $\endgroup$ Dec 14, 2019 at 20:27
  • $\begingroup$ @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. $\endgroup$ Dec 14, 2019 at 20:29

1 Answer 1


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 $$ \sqrt{m}=(\sqrt{m}+\sqrt{n})-\sqrt{n}\in\mathbb{Q}(\sqrt{m}+\sqrt{n}) $$


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