Let $K=F(\alpha)$ be an abelian extension of $F$ and let $\sigma$ be a map (could be any map) from $K^\times$ (the multiplicative group of $K$) to itself. Define an $F$-vector space $V$ to be $F$-span of the set of pairs $$\{(k\sigma(k),\sigma(k)):k\in K^\times\}.$$ Is it always true that $\dim_F(V)\geq \dim_F(K)$?
Some examples:
If $Im(\sigma)$ is finite, say $Im(\sigma)=\{\sigma(k_i):1\leq i\leq n\}$ for some integer $n$, then $$K=\langle K_1\rangle\cup\cdots\cup \langle K_n\rangle$$ where $\langle K_i \rangle$ is the $F$-span of $\{k\in K:\sigma(k)=\sigma(k_i)\}$, and so $K$ as an $F$-space must equals some $K_{i_0}$ for $1\leq i_0\leq n$. Therefore $K=K_{i_0}\sigma(k_{i_0})$ and so $\dim_F(V)\geq \dim_F(K)$.
Let $\tau:K^\times\rightarrow K^\times$ be a map such that $\tau(k)=k\sigma(k)$. Similarly, if $Im(\tau)$ is finite, then $\dim_F(V)\geq \dim_F(K)$. For instance, if $\sigma(k)=k^{-1}$, then $Im(\tau)=\{1\}$.
Here is what I tried:
Let $L$ be an intermediate field intermediate field. Since $K=F(\alpha)$ is an abelian simple extension of $F$, $L$ is an abelian simple extension of $F$ and meanwhile $K$ is also an abelian simple extension of $L$. So I would like to use induction on $\dim_F(K)$. However, I need to choose the intermediate field $L$ carefully.
If $F\subset F(\sigma(k):k\in K)\subset K$. I take $L=F(\sigma(k):k\in K)$. And I consider the restriction maps $\sigma_L$ and $\tau_L$. Observe that $\tau(k)=k\sigma(k)\in L$ if only if $k\in L$, and hence $\tau_L$ is a map from $L^\times $ to itself. For the top part, I tried to consider $L$-span of $\{(k\sigma(k),\sigma(k)):k\in K\}$. By induction, the $L$-dimension of the $L$-span of $\{(k\sigma(k),\sigma(k)):k\in K\}$ is larger than or equal to $\dim_L(K)$. So maybe I could do something by induction. Unfortunately, I couldn't go any further. Also, I don't know whether there are some tools, for example tools in number theory, to handle this problem.