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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?

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  • $\begingroup$ 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
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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

$$\mathbb{Q}[\sqrt{2},\sqrt{3},\dots,\sqrt{n}]=\mathbb{Q}[\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}]$$

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 $\mathbb{Q}^4_k = \mathbb{Q}^5_k$.

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