Let $d_1,d_2,d_3,d_4$ be squarefree integers such that $d_1 \neq d_2$ and $d_3 \neq d_4$. Then I have an impression of believing that the biquadratic extension $\mathbb{Q}(\sqrt{d_1},\sqrt{d_2})$ is isomorphic to $\mathbb{Q}(\sqrt{d_3},\sqrt{d_4})$ iff $\{d_3,d_4\}$ is a permutation of $\{d_1,d_2\}$. Is my intuition right? Does similar statement hold for the compositum of three quadratic extensions?

  • $\begingroup$ It boils down to when $1,\sqrt{d_1},\sqrt{d_2},\sqrt{d_3}$ are linearly independent. $\endgroup$ – lhf Jul 28 '16 at 13:08
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    $\begingroup$ $\Bbb{Q}(\sqrt2,\sqrt3)=\Bbb{Q}(\sqrt2,\sqrt6)$ and generalizations. $\endgroup$ – Jyrki Lahtonen Jul 28 '16 at 13:39
  • $\begingroup$ @lhf : Could you explain the proof for biquadratic extension? You have no condition on $d_4$? $\endgroup$ – MathStudent Jul 28 '16 at 15:01
  • $\begingroup$ @Lahtoonen: So if $\mathbb{Q}(\sqrt{d_1},\sqrt{d_2})\cong \mathbb{Q}(\sqrt{d_3},\sqrt{d_4})$ then is it true that, upto permutation. $d_3=d_1$ and $d_4=d_1d_2$? $\endgroup$ – MathStudent Jul 28 '16 at 15:27
  • $\begingroup$ Pretty much, $d_1d_2$ need not be square-free, and that adds a twist. For example with $d_1=2,d_2=6$ you get $d_1d_2=12$, but $\sqrt{12}$ and $\sqrt{3}$ give the same extension field. A clean proof comes from Galois theory, when you learn that ALL the quadratic fields contained in $\Bbb{Q}(\sqrt{d_1},\sqrt{d_2})$ are $\Bbb{Q}(\sqrt{d_1})$, $\Bbb{Q}(\sqrt{d_2})$ and $\Bbb{Q}(\sqrt{d_1d_2})$, so $d_3,d_4$ need to be one of $d_1,d_2,d_1d_2$ apart from a square factor. $\endgroup$ – Jyrki Lahtonen Jul 28 '16 at 17:13

Jyrki Lahtonen’s comment says it all, but your later comments suggest that you’re perhaps thinking in terms that are too close-in to the situation, when stepping back a bit will help you see the forest and not the trees.

I hope that I’m neither insulting you by being too basic, nor overwhelming you with new theory. And I’m going to commit the mathematical solecism of calling $-1$ a prime, since for my discussion, it behaves just like the proper primes.

What, really, is a square-free integer? It’s a product of finitely many of our primes, $\{-1,2,3,5,7,11,\cdots\}$; and you can, if you wish, combine two such things by multiplying and dropping off all the squares. I’ll just call this operation multiplication again, so that $2\cdot3\cdot5*3\cdot5\cdot11=2\cdot11$, in other words $30*165=22$.

I hope you recognize the significance of this, because $\sqrt{30}\sqrt{165}=15\sqrt{22}$ — in other words, in $\Bbb Q(\sqrt{30},\sqrt{165}\,)$, you’ll also find the subfield $\Bbb Q(\sqrt{22}\,)$.

If you’ve seen Abstract Algebra, you’ll recognize that I’m talking about the multiplicative group of nonzero rationals, modulo squares, denoted $\Bbb Q^\times/{\Bbb Q^\times}^2$. As I said, you can multiply two squarefrees to get another, and you can multiply any squarefree to itself to get the trivial squarefree $1$.

If it pleases you, you can think of $\Bbb Q^\times/{\Bbb Q^\times}^2$ as an infinite-dimensional vector space over the field with two elements $\Bbb F_2$ (you may have seen it denoted $\Bbb Z_2$). If it does so please you, all I’m doing is talking about finding bases for finite-dimensional subspaces of our space of squarefrees.

In case it doesn’t so please you, I’ll talk instead of proper generating sets of a mulltiquadratic extension $\Bbb Q(\sqrt a,\sqrt b,\cdots)$. Such a pgs is a set of squarefrees with the property that no one of them can be written as a product of some of the others. So $\{6,15,10\}$ is not proper, but $\{6,15,11\}$ is proper. But any time you have a pgs, say with $n$ elements, you can take any subset of them, multiply them together, and get precisely $2^n$ results. This is the subgroup (or subspace) of $\Bbb Q^\times/{\Bbb Q^\times}^2$ “generated” by your pgs. For example, if your pgs is $\{2,3,-10\}$, the eight products are $1,2,3,-10,6,-5,-30$, and $-15$. These form a “group” in the sensse that when you multiply any two, you’ll get an element from this set of eight.

Now, finally, you know that to describe a multiquadratic extension of $\Bbb Q$ efficiently, you need to specify a pgs of the proper cardinality. For instance $\Bbb Q(\sqrt2,\sqrt3,\sqrt{-5}\,)$ has been described by the pgs $\{2,3,-5\}$. But now, once we’ve gotten beyond the biquadratic extensions, to a triquadratic extension like this last, you see that there are other pgs’s that describe the same field. Here, you could have used $\{6,-10,-30\}$. I’m just saying that $\Bbb Q(\sqrt2,\sqrt3,\sqrt{-5}\,)=\Bbb Q(\sqrt6,\sqrt{-10},\sqrt{-30}\,)$, as you will easily verify.

Too much talk on my part, but I have to say that this is the simplest part of the Kummer Theory of finite extensions generatable by adjunction of $n$-th roots of elements of the base field.


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