Is a biquadratic ring uniquely determined by two intermediate quadratic rings?

Question

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?

Answer 1

The answer to your question (where you should have said quadratic "fields" instead of "rings") is yes, but the developments in your post are somewhat muddled. So let me first try to put some order. You start with two distinct quadratic fields $\mathbf Q(\sqrt a)$ and $\mathbf Q(\sqrt b)$, which you view as subfields of the quartic field $K=\mathbf Q(\sqrt a + \sqrt b)$, and you ask for different primitive elements generating $K$.

A systematic approach would be to notice that, since $\mathbf Q(\sqrt a)$ and $\mathbf Q(\sqrt b)$ are linearly disjoint, $\mathbf Q(\sqrt a,\sqrt b)$ is biquadratic, and the common degree 4 implies $K=\mathbf Q(\sqrt a,\sqrt b)$, with Galois group $G=C_2\times C_2$, which has exactly 3 subgroups of order 2. It follows that $K$ has exactly 3 quadratic subfields, which are obviously $\mathbf Q(\sqrt a),\mathbf Q(\sqrt b)$ and $\mathbf Q(\sqrt {ab})$. NB: this is a particular case of Kummer theory, which can be found in any textbook on Galois theory.

To get primitive generators of $K$, it suffices to combine primitive generators of its quadratic subfields in order to find an element which does not belong to any of these subfields. Your starting example was $x=\sqrt a + \sqrt b$, which does not belong to any of the subfields $\mathbf Q(\sqrt a),\mathbf Q(\sqrt b),\mathbf Q(\sqrt {ab})$ (if $\mathbf Q(\sqrt a)$ contained $x$, it would contain $\sqrt b$, etc.), hence generates $K$. Exactly the same argument gives the primitive elements $y=\sqrt a + \sqrt {ab}, z=\sqrt b +\sqrt {ab}$ which you mention.

Answer 2

No, but it is uniquely determined by three intermediate fields. No. As Gerry Myerson explained in a comment, a biquadratic field can be determined by any two of its three intermediate quadratic fields.

For now it may be best for you to limit yourself to $a$ and $b$ positive. Without loss of generality, let's say that indeed $\gcd(a, b) = 1$ does hold for these squarefree integers. Then $\Bbb{Q}(\sqrt{a})$, $\Bbb{Q}(\sqrt{b})$, $\Bbb{Q}(\sqrt{ab})$ are intermediate to $\Bbb{Q}(\sqrt{a} + \sqrt{b})$.

The LMFDB does have plenty of examples, but you have to know where to look. The most effective way I've found is to compute (or have Wolfram Mathematica compute for you) the discriminant of the field. For example, the discriminant of $\Bbb{Q}(\sqrt{2} + \sqrt{3})$ is $2304$. In global number field search enter that number by itself in "discriminant range." Be sure to put in $4$ for degree. The results are $\Bbb{Q}(\sqrt{2}, \sqrt{3})$, $\Bbb{Q}(\sqrt{-2}, \sqrt{3})$, $\Bbb{Q}(i, \sqrt{6})$ (link 1, link 2, link 3 respectively).

Answer 3

Yes, they are all the same field. If $a$ and $b$ are indeed coprime, $\textbf Q(\sqrt{ab})$ is an intermediate field of $\textbf Q(\sqrt a + \sqrt b)$. But if $a$ and $b$ share a nontrivial divisor, then there is a number $c$ which is coprime to $a$ or to $b$ (but not to both) such that $\textbf Q(\sqrt c)$ is also an intermediate field of $\textbf Q(\sqrt a + \sqrt b)$.

Since all these numbers ($a$, $b$ and $c$) are stipulated to be squarefree and greater than $1$, the existence of exactly one trio of numbers sharing these relationships of common divisors is guaranteed, and therefore, the field $\textbf Q(\sqrt a + \sqrt b)$ is determined by any combination of two out of the three numbers.

Answer 4

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.