Square roots in arbitrary fields I'm a little confused about a certain argument concerning square roots. The problem is Dummit and Foote, 13.2.9., detailed here with a solution also given.
Specifically my problem is as follows: suppose I have shown that $(\sqrt{m}+\sqrt{n})^2=a+\sqrt{b}$. How should I then conclude that $\sqrt{m}+\sqrt{n}=\sqrt{a+\sqrt{b}}$?
I think I'm just a little confused about what the symbol $\sqrt{a}$ actually means in an arbitrary field. How do we make a choice between the two algebraically indistinguishable roots of $x^2-a$?
Thank you.
 A: In my opinion, formulations such as

Suppose $a^2-b$ is a square where $a,b\in F$ and $b$ is not a square. Show that $\sqrt{a+\sqrt b}=\sqrt m+\sqrt n$ for some $m,n\in F$

are to some degree an abuse of notation when the field is not $\mathbb R$ (or a subfield thereof). There are also problems if the field is $\mathbb R$ and roots of negative numbers are to be taken (as is the case here as $b$ is explicitly not a square).  A more adequate formulation might be

Suppose $a^2-b$ is a square where $a,b\in F$ and $b$ is not a square. Show that there exist $m,n\in F$ such that $F[X,Y]/(X^2-m,Y^2-n)$ has an element $\gamma$ such that $(\gamma^2-a)^2=b$.

But this is probably less appealing to read.
Just keep in mind that oneshould, in this context, take $\sqrt a$ to mean "for all $\alpha$ with $\alpha^2=a$" or "there exists $\alpha$ such that $\alpha^2=a$", depending on where in the statement the square root occurs. Thus:

For all $\beta$ with $\beta^2=b$ and all $\gamma$ with $\gamma^2=a+\beta$ there exist $m,n\in F$ and $\mu,\nu$ with $\mu^2=m$ and $\nu^2=n$ and $\gamma=\mu+\nu$. 

or

For all $\beta$ with $\beta^2=b$  there exist $m,n\in F$ and $\mu,\nu$ with $\mu^2=m$ and $\nu^2=n$ and $(\mu+\nu)^2=a+\beta$.

In both cases, "in a suitable extension field of $F$" or "in $\overline F$" should be added for all those greek letters. Thus by the end of the day, the "abusive" notation as originally given is nicely concise and is fine as long as one watches out for sign problems.
A: Usually, you would get two answers: $\sqrt{(\sqrt{m}+\sqrt{n})^2} = \pm \sqrt{a + \sqrt{b}}$. However, the negative solution you can drop here because $\sqrt{m}$ and $\sqrt{n}$ are already positive, by definition of the symbol $\sqrt{~}$, so you cannot get the negative root $-\sqrt{a + \sqrt{b}}$ as solution if you want $\sqrt{m}+\sqrt{n}$ and not $-(\sqrt{m}+\sqrt{n})$.
