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