Subfield of degree $p$ of $\mathbf{Q}_p(\zeta_{p^2})$ In this answer, Keith Conrad claims that a generator of the subfield of degree $p$ over $\mathbf{Q}_p$ of $\mathbf{Q}_p(\zeta_{p^2})$ is given by $$\sum_{a^{p-1}\equiv 1\bmod{p^2}} \zeta_{p^2}^a.$$
I don't see why this is true. 
In any case, it is not true that the subfield of degree $p-1$, which is obviously $\zeta_{p}$, is generated by $\sum_{a^{p}\equiv 1\bmod{p^2}} \zeta_{p^2}^a$, which comes out to be $0$ .
Am I missing something obvious?
 A: Write $\zeta = \zeta_{p^2}$, $K=\mathbb{Q}_p(\zeta)$. Note that for $p\nmid a$, $\text{Tr}_{K/\mathbb{Q}_p}(\zeta^a)=0$, this follows from the appearance of cyclotomic polynomial $\Phi_{p^2}$. 
Let $$x = \sum_{a^{p-1}\equiv 1\bmod{p^2}} \zeta^a$$
we already know $x$ is in the degree-$p$ subfield of $\mathbb{Q}_p(\zeta)$. Assume $x\in \mathbb{Q}_p$, then taking trace both sides, gives $\text{Tr}_{K/\mathbb{Q}_p}(x)=p(p-1)x=0$, so $x=0$. Therefore
$$p-1 = \sum_{a^{p-1}\equiv 1\bmod{p^2}} (1-\zeta^a)$$
each $v_p(1-\zeta^a) > 0$, but $v_p(p-1) = 0$, contradiction. Hence $x\notin \mathbb{Q}_p$. 
A: Hint: To show something is in the ground field, you have to check if it is fixed by every element of the Galois group.
Added: Here's an argument similar but a slightly different from the other answer. Call $$x:= \sum_{a^{p-1}\equiv 1\bmod{p^2}} \zeta_{p^2}^a.$$ Via $(\mathbb Z/p^2)^\times \simeq Gal(\mathbb Q_p(\zeta_{p^2}) \vert \mathbb Q_p)$, $b \mapsto \sigma_b:=[\zeta_{p^2} \mapsto \zeta_{p^2}^a]$ we have 
$$\sigma_b(x) = \sum_{a^{p-1}\equiv 1\bmod{p^2}} \zeta_{p^2}^{ba}.$$
Now since $(\mathbb Z/p^2)^\times = \langle 1+p \rangle \times \{a \in (\mathbb Z/p^2)^\times: a^{p-1}=1\}$ (the first being the subgroup of order $p$, the second the one of order $p-1$), we get
$$\sum_{l=1}^p \sigma_{1+p}^l(x) = \sum_{a \in (\mathbb Z/p^2)^\times} \zeta_{p^2}^{a}$$
which is
$$=Tr_{\mathbb Q_p(\zeta_{p^2})\vert \mathbb Q_p}(\zeta_{p^2})=0.$$
This contradicts $\sigma_{p+1}(x)\stackrel{?}=x$ unless $x =0$ (which is the case for "the other Gauß sum", where a completely analogous argument goes through until here.)
So we need to show $x \neq 0$. The other answer does that with $p$-adic valuation; surely there's other ways.
