Squares in $\mathbb Z_p$

Question

Let $p\neq 2$, then I want to understand that every element in $\mathbb Z_p$ (p-adic integers) is a square.

For the prove one must see that $2$ is invertible in $\mathbb Z_p$. But $2$ is the element $(...,2 \mod p^2,2\mod p)=(...,2,2,2)$

What is the inverse of this in $\mathbb Z_p$? We are talking about the inverse in terms of multiplication, right?

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Edit: From the comments and the linked material it became clear that the question is about why the elements of the multiplicative group $1+p\Bbb{Z}_p$ are all squares. That group is isomorphic to the additive group $\Bbb{Z}_p$ and this may have created some confusion. JL

Answer 1

The $p$-adic integers $\mathbb Z_p$ form a ring, which is both an additive group and a monoid under multiplication. When the textbook says "$U_1$ is isomorphic to $\mathbb Z_p$", they mean the group $U_1$ under multiplication is isomorphic to the group $\mathbb Z_p$ under addition. So the analogue of a "square" in the relevant sense is an element of the form $y=x+x=2x$. This equation has the solution $x=2^{-1}y$, where yes, $2^{-1}$ is the multiplicative inverse of $2$.

In general, $2^{-1}$ is an element of the field $\mathbb Q_p$ of $p$-adic numbers. If $p\neq2$ then $|2|_p=1$, so we also have $|2^{-1}|_p=1$ and $2^{-1}\in\mathbb Z_p$. So if $y\in\mathbb Z_p$, then $2^{-1}y\in\mathbb Z_p$.

Answer 2

You don’t need to use the isomorphism between $U_1$ and $\Bbb Z_p$ (when $p\ne2$) to show that every element of $U_1$ is a square. Two methods: (1) your element of $U_1$ has form $1+pz$ for some $z\in\Bbb Z_p$. Then the binomial expansion for $(1+pz)^{1/2}$ converges because the only denominators in the series are powers of $2$ (units in $\Bbb Z_p$) and the powers of $pz$ go to zero.

(2) the equation $X^2-(1+pz)$ looks like $(X-1)(X+1)$ modulo the maximal ideal, and by (the strong form of) Hensel’s Lemma, these two factors are relatively prime in $\Bbb F_p[X]$, so the factorization lifts from characteristic $p$ to characteristic zero.