TITLE: Is a biquadratic ring uniquely determined by two intermediate quadratic rings?
QUESTION [6 upvotes]: Given positive squarefree integers $a$ and $b$, neither equal to $1$, but not necessarily coprime, we see that $\sqrt{a}$ and $\sqrt{b}$ are irrational algebraic integers of degree $2$, and the fields $\mathbb{Q}(\sqrt{a})$ and $\mathbb{Q}(\sqrt{b})$ are fields of algebraic numbers, also of degree $2$. The intersection of $\mathbb{Q}(\sqrt{a})$ and $\mathbb{Q}(\sqrt{b})$ is $\mathbb{Q}$.
Then $\mathbb{Q}(\sqrt{a} + \sqrt{b})$ is of degree $4$ and has $\mathbb{Q}(\sqrt{a})$ and $\mathbb{Q}(\sqrt{b})$ as intermediate fields. If $a$ and $b$ are indeed coprime, then $\mathbb{Q}(\sqrt{ab})$ is also an intermediate field.
But then wouldn't $\mathbb{Q}(\sqrt{a} + \sqrt{ab})$ and $\mathbb{Q}(\sqrt{ab} + \sqrt{b})$ also constitute biquadratic fields? If I have calculated these things correctly, they would all be the same field.
For example, $\mathbb{Q}(\sqrt{2} + \sqrt{3})$ is not uniquely determined by $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$. It can also be obtained from $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{6})$, or $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}(\sqrt{6})$, since for example $(\sqrt{2} + \sqrt{6})^2 = 8 + 4 \sqrt{3}$ and $(\sqrt{3} + \sqrt{6})^2 = 9 + 6 \sqrt{2}$.
This proves that in each case, there are the same three intermediate fields, and therefore $\mathbb{Q}(\sqrt{a} + \sqrt{b}) = \mathbb{Q}(\sqrt{a} + \sqrt{ab}) = \mathbb{Q}(\sqrt{ab} + \sqrt{b})$.
Have I calculated these correctly? And if so, have I drawn the right conclusions?

REPLY [6 votes]: I'm going to take a stab at this question.
The following stipulations remain in effect: $a, b \in \mathbb{Z}$, $\gcd(a, b) = 1$, and either $a > b > 1$ or $b > a > 1$.
As you have already established, $(\sqrt{a} + \sqrt{b})^2 = a + b + 2 \sqrt{ab}$. From this it's not too difficult to show that $\sqrt{ab} \in \mathbb{Q}(\sqrt{a} + \sqrt{b})$.
I think that your apprehension comes from the possibility that maybe $\mathbb{Q}(\sqrt{a} + \sqrt{ab})$ or $\mathbb{Q}(\sqrt{b} + \sqrt{ab})$ could be a different biquadratic field.
However, $(\sqrt{a} + \sqrt{ab})^2 = ab + a + 2a \sqrt{b}$. Then we merely need to prove that $\sqrt{b} \in \mathbb{Q}(\sqrt{a} + \sqrt{ab})$. And then the case of $\mathbb{Q}(\sqrt{b} + \sqrt{ab})$ becomes plainly obvious.