I am trying to use Exercise 11.1.G in "The Rising Sea: Foundations of Algebraic Geometry" (by R.Vakil) to solve the following problem.
Problem Let $F$ be a field, $K$ be a finite extension of $F$ and let $\sigma$ be an endomorphism of the set $K^{\times}=K-\{0\}$. If $V$ is an $F$-space spanned by the set of pairs $\{(x\sigma(x), \sigma(x)):x\in K^{\times}\}$, then $\dim_F(V)\geq \dim_F(K)$.
Exercise 11.1.G from Ravi Vakil's FOAG says: If $X$ is an affine scheme over $k$, a field and $K|k$ is an algebraic field extension, then $X$ is of pure dimension $n$ iff $X_K:=X \times{ }_k K$ is of pure dimension $n$.
Here is my attempt to solve the problem by using Exercise 11.1.G:
Let $F[V]$ be the polynomial ring of $V$, $F[K]$ be the polynomial ring of $K$, $X=Spec(F[V])$ and $Y=Spec(F[K])$. Then $\dim_F(V)=\dim_F (X)$ and $\dim_F(K)=\dim_F (Y)$. Let $k$ be an algebraic closure of $K$. Then by Exercise 11.1.G $\dim_k (X_k)=\dim_F (X)$ and $\dim_k (Y_k)=\dim_F (Y)$. As $X_k=Spec(F[V]\otimes k)$ and $Y_k=Spec(F[K]\otimes k)$, $X_k=Spec(k[V])$ and $Y_k=Spec(k[K])$.
I guess $X_k$ and $Y_k$ are affine $k$-space. Then I might use dominant rational map from $X_k$ to $Y_k$ to define a $k$-algebra homomorphism from $k(K)$ to $k(V)$. Since this $k$-algebra homomorphism is an embedding, we conclude that $\dim_k (X_k)\geq \dim_k (Y_k)$.
However, since my lack of solid fundamentals in Algebraic Geometry, I am not sure what I am trying to do is correct. Even if it is true, I don't know how should I write down the proof properly. So, I need some detailed explanations. Also, I expect an elementary proof without using algebraic geometry. I really appreciate your help!
Is this argument correct? (I am wondering how to solve it with Vakil’s exercise too)
Given $V$ the subspace spanned by $$ \left\{(x \sigma(x), \sigma(x)): x \in K^\times\right\} $$ in $K \times K$. One can turn $V$ into a field: define $$(x \sigma(x), \sigma(x)) \cdot(y \sigma(y), \sigma(y))=(x y \sigma(x y), \sigma(x y))$$ Then $V$ is a $F$-algebra. Since it's also a finite dimension vector space on $F, V$ is a field extension of $F$. Then $K \rightarrow V$ defined by $x \mapsto (x\sigma(x),\sigma(x))$ is also a field extension. Hence $\operatorname{dim}_F V \geqslant \operatorname{dim}_F K$.