field extension is simple I am working on the following problem:
Let $L/K$ be a field extension of degree $6$. Show that $L/K$ has a primitve element.
My idea is to use the theorem of primitive elements. So, if I prove that $L=K(a,b_1,...,b_n)$ for $a$ algebraic over $K$ and $b_1,...,b_n$ separable it follows that $L/K$ is simple. Unfortunately, I do not know how to start my proof. Any hint is greatly appreciated!
 A: Let $L/K$ be a field extension of degree $6$.

Claim:$\;L$ is a simple extension of $K$.

Proof:

Suppose otherwise.

By the theorem of the primitive element, it follows that

*

*The extension $L/K$ is not separable.$\\[4pt]$

*There are infinitely many intermediate fields between $K$ and $L$.

But there can't be two intermediate fields of degree $2$ over $K$ between $K$ and $L$, else the join of those two subfields of $L$ would have degree $4$ over $K$, which is impossible since $4{\,\not\mid\,}6$.

It follows that there are infinitely many intermediate fields of degree $3$ over $K$ between $K$ and $L$.

Let $K(a),K(b)$ be distinct subfields of $L$ both having degree $3$ over $K$.

Necessarily $K(a,b)=L$.

Since $L/K$ is not separable, at least one of $a,b$ is not separable over $K$.

Without loss of generality, assume $a$ is not separable over $K$

It follows that the minimal monic polynomial for $a$ over $K$ is $f=(x-a)^3$.

Let $g$ be the minimal monic polynomial for $b$ over $K$. and let
$$
g=(x-b_1)(x-b_2)(x-b_3)
$$
be a complete factorization of $g$ in $\overline{L}[x]$, where $\overline{L}$ is an algebraic closure of $L$.

From
$$
6=[L:K]=[K(a,b):K(a)][K(a):K]=3[K(a,b):K(a)]
$$
we get $[K(a,b):K(a)]=2$, hence if $h$ is the minimal monic polynomial for $b$ over $K(a)$. we can assume, without loss of generality, that
$$
g=(x-b_1)(x-b_2)
$$
Then by Vieta's formulas we get

*

*$b_1+b_2+b_3\in K$.$\\[4pt]$

*$b_1+b_2\in K(a)$.

hence $b_3\in K(a)$.

It's immediate that $K(a)$ is a splitting field over $K$ of $f$, so $K(a)$ is a normal extension of $K$, hence since $g\in K[x]$ has a root $b_3\in K(a)$, it follows that $b_1,b_2,b_3\in K(a)$.

But then we get $b\in K(a)$, contradiction, since $K(a),K(b)$ are distinct subfields of $L$ both having degree $3$ over $K$, which implies $K(a)\cap K(b)=K$.

This completes the proof.
