# Fertile fields for roots of unity

Is there a field $K$, an odd prime $p$, and a positive integer $n$, such that $K[ζ] = K[ζ^p]$ where $ζ = ζ_{p^n}$ is a primitive $p^n$th root of unity not contained in $K$?

In other words, can a base field $K$ be chosen such that adjoining a "small" root of unity automatically adjoins some "large" roots of unity?

This came up while understanding the answer to my Sylow p-subgroups of PGL2 question. It does not affect the proof particularly, but it definitely affects my understanding of it.

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I like the title as much as the solution! – rschwieb Jun 11 '12 at 0:57

I think you will find that $K = \mathbb{R}$ fits your requirements in a fairly dramatic way.

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Perfect thanks! – Jack Schmidt Jun 8 '11 at 14:14
Very nice Pete! It's the obvious examples that are easiest to forget about... – Zev Chonoles Jun 8 '11 at 14:19

Here is an example for those of you who don't wish to leave the land of algebraic extensions of $\mathbb{Q}$.

Let $p$ be a prime number greater than 2 and $E = \mathbb{Q}(\zeta_{p^n}).$ Then $Gal(E/\mathbb{Q}) \cong (\mathbb{Z}/p^n)^{\times} \cong \mathbb{Z}/(p-1) \oplus \mathbb{Z}/(p^{n-1}).$ Let $H$ be the subgroup of $Gal(E/\mathbb{Q})$ isomorphic to $\mathbb{Z}/(p-1)$ and set $K_n = E^H.$ Then $K_n(\zeta_p) = E.$ But note $[K_n:\mathbb{Q}] = p^{n-1},$ and thus as $[\mathbb{Q}(\zeta_p):\mathbb{Q}] = p-1,$ it must be the case that $\zeta_p\not\in K_n.$ It follows $K_n$ does not contain a $p^z$-th root of unity for any $z\in \mathbb{Z}_{+}.$

Hence, $K_n$ is a field containing no $p$-power roots of unity, that when a $p$-th root of unity is adjoined yields the field $\mathbb{Q}(\zeta_{p^n}).$

In fact, for each $n\in \mathbb{Z}_{+},$ the field $K_n \subset K_{n+1}.$ Thus, setting $K = \cup_{n\in\mathbb{Z}_{+}} K_n$, one obtains a field that contains no $p$-power roots of unity, which when a $p$-th root of unity is adjoined yields a field containing every $p$-power root of unity! For reference, the field $K$ is called the cyclotomic $\mathbb{Z}_p$ - extension of $\mathbb{Q}$.

A similar construction is possible for $p = 2,$ which yields a field containing no $2$-power roots of unity, save $-1,$ which when $i$ is adjoined yields a field containing all $2$-power roots of unity.

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Thanks! This is a great example for me, as I actually use cyclotomic fields. – Jack Schmidt Jun 8 '11 at 14:43
(Yeah, I think the p=2 case of the PGL2 proof is going to go very similarly, but there is going to be an honest-to-goodness matrix involved after handling the field. So far though, the only problem is order 2 elements, like −1.) – Jack Schmidt Jun 8 '11 at 14:45
Another similar example to both yours and Pete's is to let F be any field containing a bunch of roots of unity (say, a full cyclotomic field), and let K be the real subfield. Then [F:K]=2, and this K contains your Kn with index (p−1)/2. – Jack Schmidt Jun 8 '11 at 16:22