Square classes p-adic numbers isomorphism

Question

I was reading about the Hilbert symbol and the Hasse-Minkowski theorem and found this statement in the book I was reading :

$ |\frac{Q_p^{*}}{(Q_p^{*})^{2}} |=2^r$; with $r=2$ for $p \neq 2$ and $r=3$ for $p=2$

I was thinking about proving it directly by showing an isomorpishm, but I got a bit stuck. I'd apreciate any advice or hint you could give me in order to prove it.

Thank you so much in advance.

Answer 1

It may be possible to prove this via Hensel's lemma, but I think a more persuasive argument (perhaps the most standard one) is by p-adic versions of exponential and logarithm (yes, taking p-adic inputs and producing p-adic outputs). The $n!$ in the denominator of the exponential is no longer large, p-adically, indeed, so this causes some trouble, rather than helping convergence. The specifics about the convergence (in most textbooks...) show, qualitatively, that for given $n$, anything sufficiently near $1$ has an $n$-th root. The details work out to give what you are trying to prove in the square-root/square case. (Among other sources, my notes on alg no. th. cover such things: http://www-users.math.umn.edu/~garrett/m/number_theory/.)

Answer 2

The first step is certainly to note that $\Bbb Q_p^\times=p^\Bbb Z\times\Bbb Z_p^\times$. The first factor is infinite cyclic, so all you need to do is show that $\bigl|\Bbb Z_p^\times/(\Bbb Z_p^\times)^2\bigr|$ is $4$ or $2$ according as $p=2$ or $p>2$. For that, go to Paul Garrett’s answer.

Answer 3

Rather than using logarithms and exponentials you can use directly the binomial formula

$(1+x)^\frac 12 = 1 + \frac 12 x + \frac 12 \frac {-1}{2}\frac 1 {2!}x^2 + \frac 12 \frac {-1}2 \frac {-3}2\frac 1 {3!} x^3 + \ldots$

Using $v_p(n!) \le n$, you easily get that this converges $p$-adically if $v_p(x) \ge 1$ for $p \neq 2$, and if $v_p(x) \ge 3$ for $p=2$.

This shows that you have surjections $(\Bbb Z/8\Bbb Z)^* \to \Bbb Z_2 / (\Bbb Z_2^*)^2 $ and $(\Bbb Z/p\Bbb Z)^* \to \Bbb Z_p^* / (\Bbb Z_p^*)^2$ for $p \neq 2$.

If $x$ is a square in $\Bbb Z_p$ then it's a square mod $p^n$ for any $n$.
Conversely, if $x$ is a square $y^2$ mod $8$ or mod $p$, then $x/y^2$ is close enough to $1$ to be a square in $\Bbb Z_p$, and thus $x$ is a square in $\Bbb Z_p$.

Therefore the kernels of those surjections are exactly $((\Bbb Z/8 \Bbb Z)^*)^2$ and $((\Bbb Z/p \Bbb Z)^*)^2$ respectively.

The first one is trivial and so $\Bbb Z_2 / (\Bbb Z_2^*)^2$ is isomorphic to $(\Bbb Z/8\Bbb Z)^* \approx (\Bbb Z/2\Bbb Z)^2$. The second ones have size $(p-1)/2$, and so $\Bbb Z_p / (\Bbb Z_p^*)^2 \approx \Bbb Z/2\Bbb Z$ and the corresponding quotient map is the Legendre symbol.