Finite algebraic extension and simple extension. Suppose that $L/K$ is finite algebraic extension and $\alpha$ is algebraic and separable over K.
If $L/K$ is simple algebraic extension, $L(\alpha)/K$ is simple.
Does the converse holds true?
That is, if $L(\alpha)/K$ is simple, then $L/K$ is simple.
 A: You have a tower of extensions $K \subseteq L \subseteq L(\alpha)$ with $\alpha$ separable and algebraic over $K$ from which it follows that $L(\alpha)/K$ is a finite separable extension. It now follows that $L/K$ is finite and separable and so by the primitive element theorem is simple.
Added for OP: Theorem. Let $L/K$ be a finite extension, so that $L = K(\alpha_1,\ldots,\alpha_n)$ for some $\alpha_1,\ldots, \alpha_n \in L$. Then $L$ is a separable extension of $K$ iff the minimal polynomial of each $\alpha_i$ is separable over $K$.
A: There's a characterization of finite simple extensions. It says if $K/L$ is a finite simple extension then $K=L(\theta)$ if and only if there are finitely many fields $F$ s.t. $L \subset F \subset K$. So if $L(\alpha)/K$ is a finite simple extension then so is $L/K$. 
If $L(\alpha)/K$ is an infinite simple extension then $L(\alpha) \cong K(x)$. I'm going to refer you to this small handout for the proof that every nontrivial subfield of $K(x)$ is also a simple transcendental extension of $K$. So the converse holds.
