# What is $\mathbb{Q}[\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\dots]$ and related questions

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?

• 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. – Steve D Sep 24 '18 at 13:50
• "The sum": But which one? BTW: Looking at the orbit of $\sum\sqrt{p_i}$ is rather advanced, isn't it? – Hans-Peter Stricker Sep 24 '18 at 13:55
• 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. – Steve D Sep 24 '18 at 13:57
• This depends on what "knowing what a field extension is" means. Knowing only the definition doesn't seem to suffice. – Hans-Peter Stricker Sep 24 '18 at 13:59
• "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.) – 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

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

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