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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.

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    $\begingroup$ It's now 2019, so I think your question is a year late. :) $\endgroup$ – KCd Mar 26 at 12:14
  • $\begingroup$ @KCd hahaha, you're right! However, this was a bonus question on the midterm of Galois Theory that I took a year ago which I came accross in my archive; so, it was up to date back then. In addition, since $2019=3\cdot673$ in prime factorisation, by the answer below it should hold that also every field extension of degree $2019$ is simple. Voilá, we've solved this year's bonus question too lol $\endgroup$ – Algebear Mar 26 at 12:21
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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.

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    $\begingroup$ Is this the link? $\endgroup$ – Thomas Shelby Mar 26 at 7:38
  • $\begingroup$ I think I'm very confused here. First of all, the separable closure of $K$ is defined as the set of all algebraic elements over $K$ which are separable over $K$, right? This includes all elements of $K$ plus an additional set of algebraic and separable elements over $K$. Then why is it the case that $[K:S]<\infty$? I would think that $S$ is a bigger field than $K$, so we would write $S/K$ and talk about $[S:K]$. Furthermore, I don't understand why this $S$ cannot be an infinite degree field extension over $K$, if what I said above in this comment is true. Or should I read $S$ over $F$ sep cl. $\endgroup$ – Algebear Mar 26 at 10:54
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    $\begingroup$ @ThomasShelby yes, thank you for finding the thread. $\endgroup$ – pisco Mar 26 at 11:35
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    $\begingroup$ @Algebear I edited my answer, $S$ is the separable closure of $K$ over $F$, the set of elements of $K$ that is separable over $F$. $\endgroup$ – pisco Mar 26 at 11:35
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    $\begingroup$ @Algebear (1) Because $K/S$ is purely inseparable, so its degree is a power of the characteristic. (2) $[S:F] = [K:F]/[K:S]$. (3) Because the degrees $[S:F], [K:S]$ are primes, so any element not in the base field generated whole extension. Your concerns will all be eradicated if you have a more consolidate understanding of basic field theory. $\endgroup$ – pisco Mar 28 at 10:05
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Using pisco's notation and considering the original question we can assume that $p=2$. Then $\beta^2\in K(\alpha)$, which implies $K(\alpha)\subseteq K(\alpha+\beta)$ because of

$(\alpha+\beta)^2=\alpha^2+\beta^2\in K(\alpha)$

and

$K(\alpha^2)=K(\alpha)$.

But $K(\alpha)\neq K(\alpha+\beta)$, hence the assertion.

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