${\mathbb{Q}_p^*}^2$ is open in $\mathbb{Q}_p^*$

Question

Show that the set of squares in $\mathbb{Q}_p^*$ is open in $\mathbb{Q}_p^*$.

Here $\mathbb{Q}_p$ is the $p$-adic numbers and $\mathbb{Q}_p^*$ is the set of units in $\mathbb{Q}_p$.

I know that $\mathbb{Q}_p^*$ is isomorphic to $\mathbb{Z}\times \mathbb{Z}_p\times \mathbb{Z}/(p-1)\mathbb{Z}$ if $p$ is odd prime and $\mathbb{Q}_2^*$ is isomorphic to $\mathbb{Z}\times \mathbb{Z}_2\times \mathbb{Z}/2\mathbb{Z}$. But I am unable to solve the problems using these. I need some help...

Answer 1

It should be clear that it’s enough to show that there’s a neighborhood of $1$ consisting entirely of squares. For $p>2$, this neighborhood is $1+p\Bbb Z_p$: for $z\in\Bbb Z_p$, the equation $X^2-(1+pz)$ factors modulo $p$ as $(X-1)(X+1)$, product of two linears, to which you apply the strong Hensel’s Lemma: the original polynomial also factors as a product of two linears.

For $p=2$, the appropriate neighborhood is $1+8\Bbb Z_2$. The polynomial $X^2-(1+8z)$ clearly factors into linears if and only if $(X+1)^2-(1+8z)$ does; this is $X^2+2X-8z$. And this factors if and only if (dividing the roots by $2$) $X^2+X+2z$ factors. But this is $X(X+1)$ modulo $2$, and Hensel applies.

If you don’t like the appeal to the strong form of Hensel’s Lemma, you may like the observation that the Binomial expansion of $(1+4x)^{1/2}$ has all integer coefficients. This applies to give the result for all $p$: just plug in any element of $p\Bbb Z_p$.

Answer 2

I will show directly that ${\mathbb{Q}_p^\times}^2$ is open in $\mathbb{Q}_p^\times$. That is, around every non-zero square $\beta$ in $\mathbb{Q}_p^\times$ there exists an open ball around $\beta$ such that it contains only squares. For this I will make use of a result found in Koblitz's book on $p$-adic numbers. It is an exercise problem which I'll quote for your convenience:

Let $F(x)$ be a polynomial with coefficients in $\mathbb Z_p$. If $a_0 \in \mathbb Z_p$ satisfies $F'(a_0) \equiv 0 \mod p^M$ but $F'(a_0) \not\equiv 0 \mod p^{1+M}$, and if $F(a_0) \equiv 0 \mod p^{1+2M}$, then there is a unique $a \in \mathbb Z_p$ such that $F(a)=0$ and $a\equiv a_0 \mod p^{1+m}$.

Let $\beta \in {\mathbb Q_p^\times}^2$ and $\beta=\alpha^2$. First assume that $\alpha \in \mathbb Z_p$. (The case $\alpha \in \mathbb Q_p-\mathbb Z_p$ follows from this case.)

Let $\alpha=p^Mu$, $M \geq 0$ and $u\in \mathbb Z_p^\times$. Consider the polynomial $F_\epsilon(x)=x^2-\beta-\epsilon \in \mathbb Z_p[x]$. If $p^{1+2M} \mid \epsilon$ then $F_\epsilon$ satisfies the hypotheses of the above theorem with $a_0=\alpha$. That means for every $\epsilon$ such that $\mid \epsilon \mid \leq \tfrac{1}{p^{1+2M}}$ the number $\beta+\epsilon \in {\mathbb Q_p^\times}^2$.

Let us now consider the case when $\alpha=u/p^k$, $k>0$. Apply the above proved result to $u^2$ to get an $\epsilon>0$ such that whenever $\mid z \mid<p^{2k}\epsilon$, $u^2+zp^{2k}$ is a square which happens if and only if $\tfrac{u^2}{p^{2k}}+z=\beta+z$ is a square.

Q.E.D.

P.S. After finding the suitable open ball around $\beta$ if necessary we should decrease the radius suitably to make sure $0$ is not contained in it.