2

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?

user786
  • 109
  • 1
    Compare your answer with this post. The proof there also works here. Your fourth line is no equation. – Dietrich Burde Dec 14 '19 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. – Charles Hudgins Dec 14 '19 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). – Barry Cipra Dec 14 '19 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 $\sqrt m=(\sqrt m+\sqrt n)-\sqrt n$. – Barry Cipra Dec 14 '19 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. – Charles Hudgins Dec 14 '19 at 20:29
  • @user786, where is it given that ${m,n,mn}$ contains no perfect square? The OP makes no mention of any such assumption. (Nor is it necessary, as my other comment indicates.) – Barry Cipra Dec 14 '19 at 20:30
  • This question caught my eye in part because of a question from earlier today, https://math.stackexchange.com/questions/3475953/mathbb-q21-331-3-mathbb-q21-3-31-3 – Barry Cipra Dec 14 '19 at 20:32
  • @user786, ah, that makes sense. But as I said, it's an unnecessary assumption (and your proof makes no use of it). Your proof is basically OK. I think the only thing it lacks is mention that $4m+2n$ and $2m-2n$ cannot both be $0$ (unless, of course $m=n=0$, for which equality of the fields is trivial!), so at least one of $\sqrt m$ and $\sqrt n$ is in $\mathbb{Q}(\sqrt m+\sqrt n)$, and as soon as one of them is, then so is the other. – Barry Cipra Dec 14 '19 at 20:41
  • @BarryCipra, I believe this condition works for distinct and non zero m and n. – user786 Dec 14 '19 at 20:43

1 Answers1

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

egreg
  • 238,574