Quadratic extensions of p-adic rationals

Question

In Alain Robert's A Course in p-adic Analysis, the author uses Hensel's Lemma to analyze quadratic extensions of $\mathbb{Q}_p$. He wants to calculate the index of $(\mathbb{Q}^*_p)^2$ in $\mathbb{Q}^*_p$, as multiplicative groups; and to do that, he computes (page 50, $p$ is assumed odd):

$\mathbb{Q}^*_p/(\mathbb{Q}^*_p)^2 \cong (p^\mathbb{Z}/p^{2\mathbb{Z}}) \times (\mathbb{Z}^*_p/(\mathbb{Z}^*_p)^2) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$

I could use help understanding this argument - it doesn't seem to refer to anything earlier in the book, and I'm not even sure what $p^\mathbb{Z}$ and $p^{2\mathbb{Z}}$ are here.

The only part of this I seem to understand is why $\mathbb{Z}^*_p/(\mathbb{Z}^*_p)^2 \cong \mathbb{Z}/2\mathbb{Z}$. Please tell me in case this is wrong: this is the part that uses Hensel's Lemma. By the lemma, a p-adic integer is a square $\iff$ its $0$-th digit is a square in $\mathbb{F}^*_p$; but squares form a subgroup of index $2$ in the cyclic group $\mathbb{F}^*_p$ because they're precisely the even powers of (any) generator.

Answer 1

$p^{\mathbb{Z}}$ means the subgroup of $\mathbb{Q}_p^{\times}$ generated by $p$, which is abstractly isomorphic to $\mathbb{Z}$, but this notation names the isomorphism: it sends $n \in \mathbb{Z}$ to $p^n$. Similarly $p^{2\mathbb{Z}}$ means the subgroup generated by $p^2$.

Every element of $\mathbb{Q}_p^{\times}$ factors uniquely as an integer power of $p$ times a $p$-adic unit in $\mathbb{Z}_p^{\times}$. That is, we have a direct product decomposition

$$\mathbb{Q}_p^{\times} \cong p^{\mathbb{Z}} \times \mathbb{Z}_p^{\times}$$

and that's where the first isomorphism comes from. As you say, the second isomorphism comes from Hensel's lemma.