$\mathbb Q^*_p/(\mathbb Q^*_p)^2$ is a group of order $8$ for $p=2$ and $4$ for $p\neq 2$

Question

I'm new to this topic and I'm currently reading Cassel's Rational quadratic forms and Serre's a Course in arithmetic and they're both stating $\mathbb Q^*_p/(\mathbb Q^*_p)^2$ is a group of order $8$ for $p=2$ and $4$ for $p\neq 2$.

I've had some troubles understanding their proofs, so please keep your answers as simple as possible.

Lemma 1.6 in Cassel's is saying that for $\alpha\in \mathbb Q^*_p$, if $|\alpha-1|\leq1/p$, for $p\neq2$, or $\leq1/8$ for $p=2$, then $\alpha\in\mathbb (Q^*_p)^2$.
As a corollary, one gets the aforementioned claim.

My first question is: in the case $p=2$, using the p-adic expension $a_0+a_1p+a_2p^2+a_3p^3+...$, are the groups' elements all the elements that have $a_0\neq0 \wedge a_1\neq0 \wedge a_2\neq0$? if so, how come we get $8$ elements?
Same question for the case $p\neq2$.

My second question is more like a request: Where can I find a simpler source for these proofs?

Answer 1

• For $p$ odd $$\Bbb{Q}_p^\times = p^\Bbb{Z} \times \Bbb{Z}_p^\times=p^\Bbb{Z} \times \mu_{p-1}\times (1+p\Bbb{Z}_p)$$ where $\mu_{p-1}$ is the cyclic group generated by $p-1$ roots of unity (if $g\in \Bbb{Z}$ is of order $p-1$ modulo $p$ then $\lim_{n\to \infty}g^{p^n}\in \Bbb{Z}_p$ is a primitive $p-1$ root of unity)

The binomial series $(1+px)^{1/2}= \sum_{k \ge 0} {1/2 \choose k} p^k x^k$ converges for $x\in \Bbb{Z}_p$ thus $(1+p\Bbb{Z}_p)=(1+p\Bbb{Z}_p)^2$.

$\mu_{p-1}^2= \mu_{(p-1)/2}$

$(p^\Bbb{Z})^2= p^{2\Bbb{Z}}$

Whence $$\Bbb{Q}_p^{\times 2} = p^{2\Bbb{Z}}\times \mu_{(p-1)/2}\times (1+p\Bbb{Z}_p)$$ is of index $4$ in $\Bbb{Q}_p^\times$.

• For $p=2$ it is $(1+8 x)^{1/2}$ that converges and $ \Bbb{Z}_2^\times= (1+2\Bbb{Z}_2)$, $(1+2\Bbb{Z}_2)^2 = 1+8\Bbb{Z}_2$ whence $$\Bbb{Q}_2^{\times 2} = 2^{2\Bbb{Z}}\times (1+8\Bbb{Z}_2)$$ is of index $8$ in $\Bbb{Q}_2^\times=2^{\Bbb{Z}}\times (1+2\Bbb{Z}_2)$.