# 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!

• If $K$ is a number field, then this immediately follows from the primitive element theorem. If $K$ isn't a number field, then in complete generality this is false and it's necessary to give further details. I suspect that this is about verifying a case of the primitive element theorem, probably following the work in your textbook/class/whatever your problem source is. Commented Dec 20, 2022 at 16:41

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.