I have a question to squares in $\mathbb{Q}_p^\ast$. I know that $\mathbb{Q}_p^\ast\cong \mathbb{Z}\times V\times U_1$. I do not think that it is necessary what $V$ or $U_1$ are here, but I can give more context if needed.
We write $x\in\mathbb{Q}_p^\ast$ uniquely as $x=p^nuv$ where $n\in\mathbb{Z}, u\in U_1$ and $v\in V$. It is stated that $x$ is a square iff $n$ is even and $u$ is a square in $U_1$ as well as $v$ is a square in $V$.
But why is that? I see that n has to be even. But why has u and v be squares as well? Can it not happen that u and v are not squares but the product is?
Thanks in advance.
Asking ten questions won't change the answer : $$V= \{ a \in \mathbb{Z}_p^\times, a^{p-1} = 1\} \cong \mathbb{Z}_p^\times / (1+p\mathbb{Z})\cong (\mathbb{Z}_p/p\mathbb{Z}_p)^\times \cong \mathbf{F}_p^\times$$
Here you only need that $V$ is isomorphic to a subgroup of $(\mathbb{Z}_p/p\mathbb{Z}_p)^\times$ which is obvious from the ring homomorphism $\mathbb{Z}_p \to \mathbf{F}_p, a \mapsto a+p\mathbb{Z}_p$ which is also a group homomorphism $\mathbb{Z}_p^\times \to \mathbf{F}_p^\times$.
$V \to \mathbf{F}_p^\times$ is injective because the polynomial $X^{p-1} -1$ has at most $p-1$ roots in the field $\mathbb{Q}_p$. That it is surjective is a consequence of Hensel lemma, or more directly that $\lim_{n \to \infty} a^{p^n} \in \mathbb{Z}_p$ is a primitive $r$-th root of unity, with $r= \text{ord}(a\bmod p)$.
Finally for $p \ne 2$, $a \in 1+p \mathbb{Z}_p$ is always a square because $(1+pz)^{1/2} = \sum_{k=0}^\infty {1/2 \choose k} p^k z^k$ converges for $|z|_p \le 1$, with the $p$-adic absolute value $|z|_p = p^{-v_p(z)}$.