# quadratic field extensions of $\mathbb{Q}_p$

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 $\mathbb{Q}_{p}^{*}$.

• $(\mathbb{Q}_{p}^{*})^2=\{x^2\mid x\in\mathbb{Q}_{p}^{*}\}$. – user26857 Oct 23 '14 at 18:26
• It is not an uncommon notation in field theory when applied to the multiplicative group. It would be better for the teacher to explain it when first used. May be they had used a generic ${K^*}^n$ earlier (or in an earlier course)? Now that you have seen it, you will later easily see it from the context whether the exponent means this or a cartesian power! – Jyrki Lahtonen Oct 23 '14 at 18:31
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.