$k^*/k^{*2}$ is a vector space?

Question

I am reading about Hilbert symbols in Serre's A Course in Arithmetic. Let $k$ denote either the reals, $R,$ or the field $Q_p$ of $p$-adic numbers. I can see how the Hilbert symbol is a map from $k^*/k^{*2} \times k^*/k^{*2}$ to $\{\pm 1\}.$ However, Serre says that $k^*/k^{*2}$ is a vector space over $F_2?$ First of all, I don't think it has an additive abelian group structure on it. Elements are of the form $ak^{*2}, a \in k^*.$ And clearly $ak^{*2} - ak^{*2} = 0k^{*2} \notin k^{*2}.$

Answer 1

Apparently vector spaces don't require 'addition' in the conventional sense, but rather in regards to group property, must be an abelian group. Clearly, $k^*$ is an abelian group by multiplication. Hence, the quotient group $k^*/k^{*2}$ is an abelian group by multiplication as well. The scalar property of $F_2$ determines the exponent. That is, $0 \cdot a = a^0, 1 \cdot a = a^1$ and modding out by 2 works out because we are modding the quotient by $k^{*2}.$ It should be clear from this that the quotient is an $F_2$ vector space.