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Let $K$ be a number field, prove: For the fields $K_i$, we have $K_i\neq K_{i+1}$ for all $i\geq 0$.

$K_i$ is defined as: $K_0=K$, and $K_i=K_{i-1}(\sqrt{K_{i-1}})$ for $i\geq 1$. Also, $K(\sqrt{K})$ is defined as the subfield $K(S)\subset\overline{K}$ generated by the set $S=\{w\in\overline{K}|w^2\in K\}$ (the set of square roots of elements of $K$).

cut
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  • I'm not sure this works, but the first possibility that occured to me was to show that, given a natural number $m$, $K$ has an extension field $M_m$ such that $M_m/K$ is Galois with a 2-group $G=Gal(M_m/K)$ such that there exists an element $\sigma_m\in G$ of order $2^m$. Then anything in $K_i\cap M_m$ is a fixed point of $\sigma_m^{2^i}$, allowing us to conclude that the fields $K_i\cap M_m$ are distinct for $i=1,2,\ldots,m$. – Jyrki Lahtonen May 15 '23 at 11:52
  • What is the definition of 2-group? – cut May 15 '23 at 13:00
  • Any finite group of order $p^n$, $p$ a prime, $n$ a natural number, is called a $p$-group. I have this fond hope that you can find $M_m$ by first adjoining the required roots of unity to $K$ and then taking the splitting field of a suitable polynomial of the form $x^{2^m}-a$, $a\in K$. – Jyrki Lahtonen May 15 '23 at 13:07
  • Okay, thanks. And why is there such a $M_m$ for every natural number $m$? – cut May 15 '23 at 13:08
  • I would try and select $a$ such that these criteria are met. I don't if it can be made to work. If $\sqrt{K}\subseteq K$, it surely won't. So that would be the first problem. – Jyrki Lahtonen May 15 '23 at 13:11
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    Note that the roots of unity in $K$ form a finite cyclic group, generated by some $n$'th root of unity $\zeta_n$. Since $\sqrt{k}$ for $k\in K_i$ is a root of unity if and only if $k$ is already, it should be easy to see that the roots of unity in $K_1$ are generated by $\zeta_{2n}$, in $K_2$ by $\zeta_{4n}$, and in general $K_j$ has roots of unity generated by $\zeta_{jn}$. As this is a cyclic group of order $jn$, we see that $K_i\neq K_j$ for $i\neq j$. – SomeCallMeTim May 15 '23 at 14:17
  • I see that $\sqrt{\zeta_n}=\zeta_{2n}$, $\sqrt{\zeta_{2n}}=\zeta_{4n}$, etc. But I don't really see the argument why every root of unity in $K_1$ is generated by $\zeta_{2n}$. Is it obvious? – cut May 15 '23 at 14:57
  • There is something wrong with your statement. I think it should be that $K_j$ has roots of unity generated by $\zeta_{2^jn}$. Also, why does that mean $K_j\neq K_{j+1}$?. Isn't it possible that $K_{j-1}$ also contains the cyclic group generated by $\zeta_{2^jn}$? – cut May 16 '23 at 15:11
  • @SomeCallMeTim If $K=\mathbb{Q}$, then the roots of unity in $K_0$ are generated by $\zeta_2$, but $\zeta_3\in K_1$ since $2\zeta_3 +1=\sqrt{-3}\in K_1$. – Pace Nielsen May 18 '23 at 22:23
  • Both of you are correct, I forgot that we consider the field generated by $\sqrt{k}$ and not just the set of $\sqrt{k}$'s. Thus, my claim is incorrect. – SomeCallMeTim May 18 '23 at 22:41

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I personally found this question fascinating, and am surprised it didn't get more upvotes.

Here is a solution. Let $K$ be a number field, and define $K_n$ as in the question.

We can characterize membership in $K_n$ in terms of group theory. Suppose we have an element $a\in K_n$. This means that there is a sequence of algebraic field extensions $K=L_0\subseteq L_1\subseteq \ldots \subseteq L_n$, where $a\in L_n$, and each extension $L_{i+1}/L_i$ is formed by adjoining square-roots of elements of $L_i$ (and all their Galois conjugates over $K$, to keep each of these new number fields Galois over $K$). So (the Galois closure of) $K(a)$ belongs to a Galois extension $L_n/K$, whose Galois group $G$ has a normal series $G=G_0\trianglerighteq G_1\trianglerighteq \ldots \trianglerighteq G_n=\{1\}$, where each quotient $G_i/G_{i+1}$ is isomorphic to an elementary abelian $2$-group. (The converse is also true.)

Let $H\trianglelefteq G$ be the fixing field for the Galois closure of $K(a)$. We then have the normal series $$ G=G_0H\trianglerighteq G_1H\trianglerighteq \ldots \trianglerighteq G_nH=H. $$ This gives rise to the normal series $$ G/H =G_0H/H \trianglerighteq G_1H/H\trianglerighteq\ldots \trianglerighteq G_nH/H = H/H. $$ Each quotient $(G_iH/H)/(G_{i+1}H/H)$ is isomorphic to a homomorphic image of $G_i/G_{i+1}$, and hence is an elementary abelian $2$-group.

In other words, if $a\in K_n$, then the Galois group of the Galois closure of $K(a)$ over $K$ must have a normal series of length $n$ whose invariant factors are elementary abelian $2$-groups.

We know that since $K$ is a number field, there is some cyclic Galois extension $M/K$ of degree $2^{n+1}$. (Use Shafarevich's theorem on the inverse Galois group problem for solvable groups over global fields, although I'm sure a simpler argument would suffice here.) By the primitive element theorem, write $M=K(b)$ for some $b\in M$.

Since $M/K$ is Galois, with Galois group cyclic of order $2^{n+1}$, we know that $M$ can be obtained via a sequence of $n+1$ quadratic extensions. Thus, $b\in K_{n+1}$. On the other hand, such a cyclic group does not have a normal series of length $n$ whose invariant factors are elementary abelian $2$-groups. So $b\notin K_n$.

Pace Nielsen
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  • @ThePhoenix I thought about that option, but $K(2^{1/2^n},\zeta_{2^n})$ is not a cyclic extension of $K$ of order $2^n$. There may be a simple fix, but I didn't see it right away. – Pace Nielsen May 19 '23 at 21:42
  • @ThePhoenix I think I understand your point. Do you have an easy argument that as $n\to \infty$, then ${\rm Gal}(\mathbb{Q}(2^{1/2^n},\zeta_{2^n})/\mathbb{Q})$ has elements of $2$-power order getting arbitrarily large? Again, I might just be missing something obvious. – Pace Nielsen May 20 '23 at 16:03