Is every field extension of degree $2018$ primitve? The question is whether every field extension $L/K$ of degree $2018$ is primitive; i.e. $K\subset L$ field extension and $[L:K]=2018$ implies $L=K(\alpha)$ for some $\alpha\in L$, true or false. The prime factorisation is $2\cdot1009=2018$. I know at least two theorem that could help me here: the theorem of the primitive element and another lemma about intermediate fields.

If $K\subset L$ is a finite separable extension (hence also algebraic), then there exists an $\alpha\in L$ such that $L=K(\alpha)$.

Second theorem:

Let $K\subset L$ be a finite field extension. Then equivalent are:
  [$L$ is primitive]$\iff$[there are only finitely many intermediate fields $E$ with $K\subset E\subset L$].

I don't now which one of these two will be easiest to show and I wouldn't know how to do either one of them. Any suggestions? Thanks in advance.

Any extension of degree $pq$, with $p$,$q$ distinct primes is primitive.
Suppose $[K:F]=pq$, it suffices to assume it is inseparable. Let $S$ denote the separable closure of $K/F$, then $K/S$ is purely inseparable.
Without loss of generality, assume $\text{char }K = p$, then $[K:S]=p, [S:F]=q$. Choose any $\alpha\in S-F, \beta\in K-S$, then $S=F(\alpha), K=S(\beta)$, so $K=F(\alpha,\beta)$, with $\alpha$ separable over $F$. The proof is concluded by using

If $F(\alpha,\beta)/F$ is algebraic with $\alpha$ separable over $F$, then $F(\alpha,\beta)/F$ is primitive.

I  remember seeing the assertion on this site, but cannot find it at the moment. I can add a proof if someone needs it.
