Working through Dummit and Foote as a hobby. I'm on Section 13.2, problem 9, and it's a doozy. Please note, the goal here is to solve this problem without any machinery beyond the very basics of field extensions and minimal polynomials. I also have already solved a problem in which I prove that $ F(\sqrt{D_1}, \sqrt{D_2}) $ is degree 4 provided $ D_1, D_2, $ and $ D_1D_2 $ are not perfect squares. With these ideas in mind, I am stuck on:
Problem Let $ F $ be a field (not characteristic 2), $ a,b \in F $ with $ b $ not a square in $ F $. Prove that a necessary and sufficient condition for $ \sqrt{a+\sqrt{b}} = \sqrt{m} + \sqrt{n} $ for some $ m, n \in F $ is that $ a^2 - b $ is a square in $ F $. Use this to determine when the field $ \mathbb{Q}(\sqrt{a+\sqrt{b}}) $ ($ a,b\in\mathbb{Q} $) is biquadratic over $ \mathbb{Q} $.
My Solution So Far Throughout this, denote $ \alpha = \sqrt{a + \sqrt{b}} $. We observe that $ \alpha $ is a root of the fourth-degree polynomial $ f(x) = (x^2 - a)^2 - b $ (although this is not necessarily minimal). We also note that, for any $ m, n \in F $, the element $ \beta = \sqrt{m} + \sqrt{n} $ is a root of the fourth degree polynomial $ g(x) = (x^2 - (m+n))^2 - 4mn $. With these in mind, we proceed.
$ \Longrightarrow $: Suppose $ \alpha = \beta $ . We need to show that $ a^2 - b $ is a perfect square. I am having some difficulty with this. If $\alpha$ is degree $ 4 $ over $ F $, then I know that, since $ f(\alpha) = 0 = g(\beta) = g(\alpha) $, we must have $ f(x) = g(x) $ since they are both minimal polynomials of $ \alpha $ over $ F $. Expanding and equating coefficients, we immediately get $ a^2 - b = (m - n)^2 $ from comparing the constant terms. I am stuck on the case where $ \alpha $ is not degree 4 over $ F $.
$ \Longleftarrow $: Suppose $ a^2 - b = c^2 $ for some $ c \in F $. Then we can select (using the fact that $ ch(F) \neq 2 $) $ m = \frac{1}{2}(a - c) $ and $ n = \frac{1}{2}(a + c) $. One then finds (by direct computation, say) that $ \beta = \sqrt{m} + \sqrt{n} $ is a root of $ f(x) $. This is precisely what it means for $ \beta $ to equal $ \sqrt{a+\sqrt{b}} $.
This is all the progress I made on the "if and only if" portion of the problem. My proof is almost done, but I don't yet see how to treat the case where $ \alpha $ is of degree 2, so that the polynomials are not minimal.
Next I turn my attention to $ \mathbb{Q}(\alpha) $ where $ \alpha $ is of the form $ \sqrt{a+\sqrt{b}} $, determining when this extension of $ \mathbb{Q} $ is biquadratic. If $ a^2 - b $ is a perfect square, $ a^2 - b = c^2 $, then we've shown that $ \alpha = \sqrt{m} + \sqrt{n} $, where (as we showed), $ m = \frac{a-c}{2} $ and $ n = \frac{a+c}{2} $. If, in turn, neither $ m $ nor $ n $ is a perfect square, then neither can be $ mn $ (or else we could show that $ b $ is a perfect square, a contradiction), and so $ F(\alpha) = F(\sqrt{m} + \sqrt{n}) = F(\sqrt{n}, \sqrt{m}) $, showing that $ F(\alpha) $ is a biquadratic extension of $ F $ (I think).
On the other hand, if $ a^2 - b = c^2 $, it can happen that $ m = (a-c)/2 $ or $ n = (a+c)/2 $ happens to be a perfect square. For example, $ a = 4, b = 12, c = 2 $. In this case we still have $ \alpha = \sqrt{4 + \sqrt{12}} = \sqrt{m} + \sqrt{n} = 1 + \sqrt{3} $, but $ F(\alpha) $ is merely a quadratic extension, as $ \alpha $ has minimal polynomial $ (x - \sqrt{m})^2 - n $.
I believe the last two paragraphs allow us to conclude that if $ a^2 - b = c^2 $ then $ \mathbb{Q}(\alpha) $ is biquadratic if neither $ m=(a+c)/2 $ nor $ n=(a-c)/2 $ is a perfect square or quadratic otherwise.
I have made little progress on showing that this implication works the other way. That is, given that $ F(\alpha) = F(\sqrt{m}, \sqrt{n}) $ is a biquadratic extension, I have not so far been able to show that $ a^2 - b $ is a perfect square. Nor am I even sure that this is true yet. This is the second place where I need help.
In your first case, $\alpha=\beta$ entails $$a+\sqrt b=(\sqrt m+\sqrt n)^2=m+n+2\sqrt{mn}.$$ One must have $a=m+n$ and $b=4mn$. Why is this? If I write $c=m+n$ and $d=4mn$ then $$a-c+\sqrt b=\sqrt d$$ and squaring gives $$(a-c)^2+b+2(a-c)\sqrt b=d.$$ But as $\sqrt b\notin F$ we must have $2(a-c)=0$. It follows that $a=c$ and $b=d$.
Then $$a=m+n\qquad\text{and}\qquad b=4mn$$ and so $$a^2-b=(m-n)^2.$$
This argument does not need to separate the degree $2$ case (where one of $\sqrt m$ or $\sqrt n$ is an element of $F$) from the degree $4$ case.