Let $K/k$ extension of field and $F$ the set of subfeilds. Suppose that for every $M,N \in F$ we have $M \subset N$ or $N \subset M$. Show that: K is simple extension i.e $K=k(x)$ for some $x \in K$.
My idea: is to considere $K$ is Algebraic extension of $k(S)$ (transendance basis). And we have $card(S) \leq 1$ because for every $x,y$ transcendance we have $k(x) \subset k(y)$ or $k(y) \subset k(x)$. So, if you start like me, i want to prouve K/k(x) is finite extension. Can you help me pleas ?
Let $k=\Bbb F_p$, the finite field of $p$ elements, and let $K_n=\Bbb F_{p^{2^n}}$, the finite field of order $p^{2^n}$. The $K_n$ form a chain: $k=K_0\subseteq K_1 \subseteq K_2\subseteq\cdots$. Let $K=\bigcup_{n=1}^\infty K_n$. The proper subfields of $K$ are the $K_n$. These are totally ordered under inclusion. But $K/k$ is not a simple extension. A simple algebraic extension is finite, but $K/k$ is infinite algebraic.