Let $R[x]$ be the ring of polynomials with coefficients in $R$. Let $p_k$ be the $k$-th prime number with $p_0 = 2$.

Now consider

$\mathbb{Q}^0_0 := \{ P(\sqrt{2})\ |\ P \in \mathbb{Q}[x]\} =\mathbb{Q}[\sqrt{2}]$

$\mathbb{Q}^2_0 := \{ a + b\sqrt{2}\ |\ a, b \in \mathbb{Q}\}$

$\mathbb{Q}^4_0 := \{ \frac{a}{b} \ |\ a, b \in \mathbb{Q}[\sqrt{2}]\} = \mathbb{Q}(\sqrt{2})$

It's easy to show that $\mathbb{Q}^0_0 = \mathbb{Q}^2_0 = \mathbb{Q}^4_0$.

Now consider

$\mathbb{Q}^0_1 := \{ P(\sqrt{3})\ |\ P \in \mathbb{Q}[\sqrt{2}][x]\} =\mathbb{Q}[\sqrt{2},\sqrt{3}]$

$\mathbb{Q}^1_1 := \{ P(\sqrt{2} + \sqrt{3})\ |\ P \in \mathbb{Q}[x]\} = \mathbb{Q}[\sqrt{2} + \sqrt{3}]$

$\mathbb{Q}^2_1 := \{ a + b\sqrt{3}\ |\ a, b \in \mathbb{Q}[\sqrt{2}]\}$

$\mathbb{Q}^3_1 := \{ a + b\sqrt{2} + c\sqrt{3} + d\sqrt{2}\sqrt{3}\ |\ a, b, c, d \in \mathbb{Q}\}$

$\mathbb{Q}^4_1 := \{ \frac{a}{b} \ |\ a, b \in \mathbb{Q}[\sqrt{2},\sqrt{3}]\} = \mathbb{Q}(\sqrt{2},\sqrt{3})$

$\mathbb{Q}^5_1 := \{ \frac{a}{b} \ |\ a, b \in \mathbb{Q}[\sqrt{2} + \sqrt{3}]\} = \mathbb{Q}(\sqrt{2} + \sqrt{3})$

It takes some more work to show that $\mathbb{Q}^0_1 = \mathbb{Q}^1_1 = \mathbb{Q}^2_1 = \mathbb{Q}^3_1 = \mathbb{Q}^4_1 = \mathbb{Q}^5_1$.

The generalizations for $\mathbb{Q}^0_k, \mathbb{Q}^2_k, \mathbb{Q}^3_k, \mathbb{Q}^4_k$ seem obvious:

$\mathbb{Q}^0_k := \{ P(\sqrt{p_k})\ |\ P \in \mathbb{Q}_{k-1}[x]\} $

$\mathbb{Q}^2_k := \{ a + b\sqrt{p_k}\ |\ a, b \in \mathbb{Q}_{k-1}\}$

$\mathbb{Q}^3_k := \{ \sum_{S \subset \{0,\dots k-1\}} a_S \prod_{i \in S}\sqrt{p_i} \ |\ a_S \in \mathbb{Q}\}$

$\mathbb{Q}^4_k := \{ \frac{a}{b} \ |\ a, b \in \mathbb{Q}_k\} $

and it seems straightforward to show that $\mathbb{Q}^0_k = \mathbb{Q}^2_k = \mathbb{Q}^4_k$ and possibly $\mathbb{Q}^0_k = \mathbb{Q}^3_k$.

But how to generalize $\mathbb{Q}^1_k$ and $\mathbb{Q}^5_k$ and to prove $\mathbb{Q}^0_k =\mathbb{Q}^1_k$ and/or $\mathbb{Q}^1_k = \mathbb{Q}^5_k$? Which combinations of square roots $\sqrt{p_i}$ should be taken into account?

Finally consider

$$\mathbb{Q}^0_\omega = \lim_{k\rightarrow \infty} \mathbb{Q}^0_k = \mathbb{Q}[\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\dots]$$

$$\mathbb{Q}^4_\omega = \lim_{k\rightarrow \infty} \mathbb{Q}^4_k = \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\dots)$$

i.e. the extensions of $\mathbb{Q}$ by the square roots of all prime numbers. I assume these extensions are well-defined, and I assume they are equal.

With which otherwise defined extension of $\mathbb{Q}$ are these extensions identical?

  • $\begingroup$ You only need the sum to generate $\mathbb{Q}^1_k$. To see this, show $\mathbb{Q}^0_k$ is Galois, and show the Galois group is elementary abelian of order $2^k$. Show that you have elements in the Galois group fixing all but one square root. Now look at the orbit of $\sum\sqrt{p_i}$. This is from problems 18.12-18.14 in Isaacs's Algebra. $\endgroup$ – Steve D Sep 24 '18 at 13:50
  • $\begingroup$ "The sum": But which one? BTW: Looking at the orbit of $\sum\sqrt{p_i}$ is rather advanced, isn't it? $\endgroup$ – Hans-Peter Stricker Sep 24 '18 at 13:55
  • $\begingroup$ The sum is the one I mention in my comment: the sum of the square roots of the $k$ different primes. And I guess if you consider Galois theory "advanced", then yes it is. I'm not sure the orbit of a finite group action would be considered "advanced" by anyone who knows what a field extension is. $\endgroup$ – Steve D Sep 24 '18 at 13:57
  • $\begingroup$ This depends on what "knowing what a field extension is" means. Knowing only the definition doesn't seem to suffice. $\endgroup$ – Hans-Peter Stricker Sep 24 '18 at 13:59
  • $\begingroup$ "Knowing what a field extension is" cannot mean "being aware of everything that can be derived from its definition". (You may say: "but at least the most important facts". I would agree, and in this case I have to admit: I don't "know" what a field extension is, i.e. which important facts the pure definition - which I believe to have understood - implies.) $\endgroup$ – Hans-Peter Stricker Sep 24 '18 at 14:03

I dare to give by myself an answer to the first part of my question which is suggested by Steve's comment above and spelled out in a little more detail here – in an answer to a very closely related question:

It holds that


So one can choose

$$\mathbb{Q}^1_k = \mathbb{Q}[\sqrt{p_0}+\sqrt{p_1}+\dots+\sqrt{p_k}]$$

i.e. $\mathbb{Q}^0_k = \mathbb{Q}^1_k$.

From this it follows immediately that $\mathbb{Q}^4_k = \mathbb{Q}^5_k$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.