Let $F$ be a field, and $K$ a field extension of $F$. Prove that $[K:F] = 1$ iff $K=F$.

Question

Thus far this is what I have.

$[K:F] = 1 \Rightarrow K$ is a vector space over $F$ of dimension $1$ with $\{1\}$ as a basis. But am stuck as to how to continue..

Answer 1

$[K:F]=1 \Rightarrow$ all $x\in K$ has a minimal polynomial of degree $1$, say $ax+b=0$ with $a \ne 0$ so $x=\frac{-b}{a}\in F$.

Reciprocally if $K=F$ it is clear that $[K:F]=1$ by definition of algebraic extension of a field.

Answer 2

As Phillip noted, if $F=K$, then every element $a \in F$ can be written as $a \cdot 1$ with $a \in K$ and $1 \in F$. Thus $\{1\}$ is a basis for $F/K$ and, therefore, $[K:F]=1$.
For the other direction, suppose that $[F:K]=1$ and that $F \supset K$. Let $a \in F$ but $a \notin K$. Because $K \subseteq K(a) \subseteq F$ we have $1=[F:K]=[F:K(a)][K(a):K] \Rightarrow [K(a):K]=1$. Thus, $a$ is the root of a first degree monic polynomial with constant term from $K$, a contradiction since $a \notin K$. Thus $F=K$.