# 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.

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this is direct consequence of primitive element theorem – Ram Dec 25 '12 at 12:26
why the first assertion true??...I mean why L/K simple algebraic extension implies L(alpha)/k is also simple?? – sayak Feb 26 '14 at 22:41

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 $K$.

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Why $L(\alpha)/K$ is a separable extension?Even if $L/K$ is not separable, is $L(\alpha)/K$ separable? – user53216 Dec 25 '12 at 12:30
@user53216 See the edit. – user38268 Dec 25 '12 at 12:38

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.

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Thank you!! : D – user53216 Dec 26 '12 at 15:31