= Solution
A finite <field extension> $L/K$ is a <separable field extension> if every element of $L$ has a <minimal polynomial> over $K$ with distinct roots in an <algebraic closure>. The <primitive element theorem> says that a finite <separable field extension> $L/K$ is $K(\alpha)$ for some $\alpha\in L$.
If $K$ is finite, then so is $L$. The multiplicative group of a <finite field> is cyclic, so a generator of $L^\times$ generates $L$ as a field. To justify the group fact, a finite subgroup of a field's multiplicative group is cyclic: in a <finite abelian group> its exponent is attained as an element order, while the <polynomial> root bound forces its size to be at most that exponent.
If $K$ is infinite, first consider $L=K(a,b)$. The <minimal polynomials> of $a$ and $b$ have distinct roots $a_i,b_j$ in an <algebraic closure>. Choose $c\in K$, $c\ne0$, avoiding the finitely many values for which $a_i+cb_j=a+cb$ for a pair other than $(a,b)$. Set $\alpha=a+cb$. Over $K(\alpha)$, the <polynomials> $f(T)$ and $g((\alpha-T)/c)$ have precisely one common root, $a$. Their monic <greatest common divisor> is therefore $T-a$, so $a\in K(\alpha)$ and $b=(\alpha-a)/c\in K(\alpha)$. Thus $K(a,b)=K(\alpha)$. Induct over finitely many generators of the finite <separable field extension>. This proves \b[every finite separable extension is simple].
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