# 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.

• It's now 2019, so I think your question is a year late. :) – KCd Mar 26 at 12:14
• @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 – Algebear Mar 26 at 12:21

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.

• Is this the link? – Thomas Shelby Mar 26 at 7:38
• 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. – Algebear Mar 26 at 10:54
• @ThomasShelby yes, thank you for finding the thread. – pisco Mar 26 at 11:35
• @Algebear I edited my answer, $S$ is the separable closure of $K$ over $F$, the set of elements of $K$ that is separable over $F$. – pisco Mar 26 at 11:35
• @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. – pisco Mar 28 at 10:05

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.