Show that i is an element of the p-adic integers if and only if p congruent to 1 mod 4

Question

This exercise was given in a graduate course on Local Class Field Theory. We want to prove that $i\in \mathbb{Z}_p$ (the $p-$adic integers) if and only if $p\equiv 1 \mod 4$. For $\Rightarrow$, we know that $$i^2 = (-1) = \left( \sum a_jp^j \right) = a_0^2 + p(\cdots)$$ where $(\cdots)$ denotes some expression. Thus $a_0^2 \equiv -1 \mod p$ has a solution, which implies that $p \equiv 1 \mod 4$. I'm having some difficulty with the other direction. I've tried to write out a $p-$adic expansion for $i$ when $p \equiv 1 \mod 4$, but haven't seemed to have any luck. I would greatly appreciate any hints or suggestions to solving the other direction of this problem.

Answer 1

I post solution based on Robert's textbook because the homework is probably already due and I dislike unanswered question.

All roots of unity in $\mathbb Q_p$ lie in $\mathbb Z_p^\times$. One has to use the polynomial $P(x) = x^{p-1} - 1$ with derivative $P'(x) = (p-1)x^{p-2}$. The reduction of $P(x)$ to $\mathbb F_p$ has $p-1$ distinct roots, namely all elements of $\mathbb F_p^\times$ and thus with help of Hensel's lemma we can deduce existence of $p-1$ distinct roots in $\mathbb Z_p^\times$.

The reduction homomorphism is $\mu(\mathbb Q_p) \to \mathbb F_p^\times$ is bijective: surjectivity follows from Hensel's lemma, injectivity from direct calculations starting with $\zeta^n = (1 + pt)^n = 1$ and the binomial theorem.

When $p$ is odd, $p-1$ is even and $-1$ belongs to $\mu_{p-1}$. The number $-1$ will have a square root in $\mathbb Q_p$ precisely when $\frac 12 (p-1)$ is still even, namely when $p \equiv 1 \mod 4$.