Totally ramified cyclic extensions of degree $p^a$ of $\mathbb{Q}_p$

It's quite easy to show that the totally ramified extension $\mathbb{Q}_p(\zeta_{p^{a+1}})/\mathbb{Q}_p$ contains a unique subextension $E$ s.t. $E/\mathbb{Q}_p$ is a cyclic extension of degree $p^a$ and this is given by the fixed field of the unique subgroup of order $p-1$ of the Galois group. My questions is: can we explicitly write down what this field is?

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Yes, you can. Let $r$ be a primitive root modulo $p^{a+1}$, so that $\sigma: \zeta\mapsto \zeta^r$ is a generator of the Galois group of $\mathbb{Q}_p(\zeta)/\mathbb{Q}_p$ (I will drop the subscripts under the $\zeta$). For each $i=0,\ldots,p^a-1$, set $$\eta_i = \sum_{j=1}^{p-1}\zeta^{\displaystyle r^{jp^a+i}}.$$ It is easy to see that all the $\eta_i$ are fixed by the cyclic subgroup $\langle\sigma^{p^a}\rangle$ of order $p-1$, so $\eta_i\in E$. Moreover, $\sigma(\eta_i) = \eta_{i+1}$ for $0\leq i < p^a-1$, so each $\eta_i$ has exactly $p^a$ conjugates under $\text{Gal}(\mathbb{Q}_p(\zeta)/\mathbb{Q}_p)$, and so any $\eta_i$ generates $E$.
Shouldn't $\eta_i$ be the sum from $j=0$ to $p-1$? – CruiskeenLawn Nov 14 '12 at 5:57
@CruiskeenLawn: $j=0$ is the same as $j=p-1$, since $\phi(p^{a+1})=p^{a+1}-p^a$, and so $r^{p^{a+1}-p^a}\equiv r^0\equiv 1\pmod{p^{a+1}}$. Does that address your question or should I say more? – Alex B. Nov 15 '12 at 16:43