Let $K$ be a field of characteristic different from 2, $F$ an algebraic extension of $K$ and $L$ a subset of $F$ with the following properties: $L$ contains $K$, $L$ is a K-vector space, $\forall v \in L, $ and all positive integers $n$, $v^n \in L$. Prove that L is a subfield of F.
Since $K$ is a field, and $K$ is contained in L, we have $1 \in K, 0 \in K \rightarrow 1\in L. 0 \in L$
From the answer below, I know that since L is a vector space, it is an abelian group under addition. Hence all that remains is to show L is an abelian group under multiplication.
How can I utilize
F an algebraic extension of K. (Every element of F, satisfies some nonzero polynomial $\in K[x]$
characteristic of K play a role?
to show L is closed under multiplication?
Closure of$~L$ under multiplication follows in characteristic${}\neq2$ from additive closure and closure under squaring: $xy=\frac12((x+y)^2-x^2-y^2)$. Closure of$~L$ under inverses of nonzero elements follows from the fact that $K/F$ is algebraic: every $x\in L$ satisfies a polynomial equation, that we can assume with nonzero constant term by dividing out a power of$~x$, say $a_nx^n +\cdots+a_1x+a_0 =0$ with $a_0\neq0$. Isolating $a_0$ and factoring $x$ out of the rest gives $x(a_nx^{n-1}+\cdots+a_1)=-a_0$, so we get a polynomial $-a_0^{-1}(a_nx^{n-1}+\cdots+a_1)$ in$~x$ that is the inverse of$~x$, and it lies in$~L$ by the preceding point.