6
$\begingroup$

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?

$\endgroup$
4
  • 2
    $\begingroup$ What exactly is the question? It seems like you have a bunch of true things written down that answer all the questions you pose. $\endgroup$ Commented Aug 16, 2019 at 0:00
  • $\begingroup$ Did you look at en.wikipedia.org/wiki/… it is quite the general method to understand $K(a+b),K(a+ab),K(ab)\subset K(a,b)$ $\endgroup$
    – reuns
    Commented Aug 16, 2019 at 1:06
  • 5
    $\begingroup$ @Milo It seems to me that the question is the last line. Maybe he should have offset it in some way. $\endgroup$ Commented Aug 16, 2019 at 3:35
  • 3
    $\begingroup$ I think, Mr.B, that you are using the term, "uniquely determined", in a non-standard way. When someone says, A is uniquely determined by B and C, that means once you have B and C you have A, and only A. When you write A is uniquely determined by B and C, it seems that you mean once you have B and C you have A, whereas having D and E doesn't get you A. The standard interpretation holds for these fields; two quadratic fields uniquely determine a biquadratic field. Your interpretation doesn't; a biquadratic field has three quadratic subfields, any two of which determine it. $\endgroup$ Commented Aug 23, 2019 at 0:35

4 Answers 4

9
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Good thing I didn't include my thoughts on what happens with $a$ or $b$ negative, or you would have thought I have gone completely senile. I still have a few lucid days left in me... $\endgroup$
    – Mr. Brooks
    Commented Aug 19, 2019 at 21:39
6
+250
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ Since one half is rational, and so is half $a + b$, can't you just do $$\frac{1}{2}(\sqrt{a} + \sqrt{b})^2 - \frac{a + b}{2} = \sqrt{ab} ?$$ $\endgroup$
    – Bob Happ
    Commented Aug 29, 2019 at 21:34
  • 1
    $\begingroup$ I don't recall ever writing $a + b + 2 \sqrt{ab}$ on this website, but it is in my notes, for what that's worth. $\endgroup$
    – Mr. Brooks
    Commented Aug 30, 2019 at 21:30
5
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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).

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .