Show that extension is simple Let $E/k$ be a finite extension and for any sub field $F_1$ and $F_2$ of $E$ containing $k$ either $F_1 \subset F_2$ or $F_2 \subset F_1$ then $E=k(a)$ for some $a$


Solution: Assume, $E=k(x_i, 1 \leq i \leq n)$ and $F_i=k(x_i)$. By given condition of comparability we can compare $F_i$ and therefore there will be a “largest” $F_i$ which will be equal to $E$

Is this solution correct? Any other solution?
Let $E= k(x_1,\ldots,x_n)$ be a finite extension of $k$. Since the simple extensions $F_i=k(x_i)$ are pairwise comparable, there is largest element. W.l.o.g.\ let $F_n=k(x_n)$ be the largest one. Then $k(x_i)\subseteq k(x_n)$ for all $1\leq i\leq n$ and so $x_1,\ldots,x_{n-1},x_n\in F_n$. Hence, $E=F_n$ since both are the smallest extensions of $k$ containing $x_1,\ldots,x_n$.
Let $E= k(x_1,\ldots,x_n)$ be a finite extension of $k$. Since the simple extensions $F_i=k(x_i)$ are pairwise comparable, there is largest element. W.l.o.g.\ let $F_n=k(x_n)$ be the largest one. Then $k(x_i)\subseteq k(x_n)$ for all $1\leq i\leq n$ and so $x_1,\ldots,x_{n-1},x_n\in F_n$. Hence, $E=F_n$ since both are the smallest extensions of $k$ containing $x_1,\ldots,x_n$.