Today during class we proved that there were exactly three quadratic field extensions of the $p$-adic number field $\mathbb{Q}_p$. To prove this it was stated that it was enough to look at the group \begin{align} \mathbb{Q}_{p}^{*}/(\mathbb{Q}_{p}^{*})^2, \end{align} but I dont understand what this group really means? Does $(\mathbb{Q}_{p}^{*})^2$ stand for $\mathbb{Q}_{p}^{*}\times \mathbb{Q}_{p}^{*}$, and if so, how is this embedded in $\mathbb{Q}_{p}^{*}$.
Thank you in advance!
If $G$ is a group, then sometimes $G^n$ does not denote the cartesian product $G^{\times n} = G \times \dotsc \times G$, but rather $\{g^n : g \in G\}$. For example, if $K$ is a field, then $(K^*)^2$ is the set of all squares of non-zero elements of $K$. The group $K/(K^*)^2$ is connected to field extensions of degree $\leq 2$ of $K$ as follows: If $u \in K$, then $K(\sqrt{u})$ is a field extension of degree $\leq 2$. If $v \in K^*$, then $\sqrt{uv^2}= \pm \sqrt{u}$ and hence the generated extension fields agree. In characteristic $\neq 2$, every field extension of degree $\leq 2$ is generated by a square root $\sqrt{u}$. Hence, we get a surjective map from $K/(K^*)^2$ to the set of field extensions of degree $\leq 2$ of $K$ up to isomorphism.