Kummer extension over cyclotomic fields Let $K=\mathbb{Q}(\mu_p)$, where $\mu_p$ is a primitive  $p^{th}$ roots of unity. Let $\Sigma_p$ be the set of primes of $K$ over $p$. Let $L$ be the composite of cyclic extensions of degree $p$ over $K$ which are  unramified outside of $\Sigma_p$. Let $\eta_1,..., \eta_r$ be the fundamental units of $K$.
I want to show that $L=K(p^{\frac{1}{p}}, \mu_p^{\frac{1}{p}}, \eta_1^{\frac{1}{p}},...\eta_r^{\frac{1}{p}})$.
Try: Kummer theory shows that the fields  $K(p^{\frac{1}{p}})$,$K(\mu_p^{\frac{1}{p}})$, $K(\eta_i^{\frac{1}{p}})$ are in $L$. Then I am stuck. How to show that $L$ is the composite of them? Thanks for help.
 A: (This is a strongly edited version of my original answer, since I was at first weak on the subject of irregular primes.)
I believe that your conjecture is false. Let’s call $\mathfrak p$ the unique prime of $K$ above $p$. You’re hoping that in the group $K^\times/{K^\times}^p$, the only things that give extensions unramified outside $p$ are units and the prime element $\pi=\zeta_p-1$ at $\mathfrak p$. (Since $\pi^{p-1}/p$ is a unit, you don’t need to worry about having $p$ on your list.)
Now I’m going to outline a strategy for finding a $\lambda$ that’s not in the subgroup of $K^\times/{K^\times}^p$ generated by the units and $\pi$, and whose $p$-th root induces an extension unramified above $p$ that’s not among those you know about. This strategy works because of the existence of “irregular” primes—these are the primes $p$ such that $p$ divides the class number of the field of $p$-th roots of unity. And I am going to suppose that $p$ is such a prime, like $37$.
Let $S$ be the set consisting of the $(p-1)/2$ archimedean primes of $K$, all of them complex of course, plus $\mathfrak p$. The $S$-ideal class group is the group of fractional ideals not involving $\mathfrak p$, modulo the principal ideals. The group is finite, of course, and in fact it is isomorphic to the ordinary class group because $\mathfrak p$ is principal.
Thus I start with an $S$-ideal $\mathfrak A$ that is nontrivial in the $S$-ideal class group, with $\mathfrak A^p$ trivial in that group. Now let $\lambda\in K$ such that $\mathfrak A^p=(\lambda)$. I say that adjoining the $p$-th root of $\lambda$ to $K$ gives an extension of $K$ that’s unramified outside $p$. At any prime $\mathfrak q\ne\mathfrak p$ of (the integers of) $K$, $v_{\mathfrak q}(\lambda)$ is divisible by $p$, and an argument local to $\mathfrak q$ shows that adjoining a $p$-th root of $\lambda$ gives an extension of $K$ unramified at $\mathfrak q$. There you are: this $\lambda$ is not in the subgroup of $K^\times/{K^\times}^p$ generated by the units and $\pi$, because anything in that group has a divisor of form $\mathfrak p^m$ for some $m$.
A: One can use stronger results from algebraic number theory to show quickly that your "guess" is wrong. For any number field K, let S be a finite set of primes of K containing the primes above p. Let X_S be the Galois group over K of the maximal abelian pro-p-extension of K which is unramified outside S. Class field theory shows that X_S is a noetherian Z_p-module , where Z_p denotes the ring of p-adic integers) of rank x = 1 + c + d, where c is the number of pairs of complex embeddings of K and d is a natural integer, conjectured to be null (this is the celebrated Leopoldt's conjecture, known to be true for abelian extensions of Q). Thus X_S, as a group, is isomorphic to the product of x copies of Z_p and of its (finite) p-torsion subgroup T. In your case, you are studying the quotient X_S/pX_S for K = Q(mu_p), so that Leopoldt's conjecture holds and consequently the dimension of X_S/pX_S over the prime field Z/pZ is equal to (p + 1)/2 + the dimension t of T/pT. The set of generators that you propose has cardinal (p + 1)/2, so t is missing. It can be shown that t is non null if and only if the prime p is irregular, as in Lubin's answer. For all this stuff, you can read for example L. Washington's book "Introduction to cyclotomic fields".   ¤
