# Finite field extension over a field

Let $$L=\mathbb F_3[X]/(X^3+X^2+2)$$ be a field (which has order $$3^3=27$$). Is there a finite field extension $$M\supseteq L$$ of degree $$n\geq 2$$?

Is the following argument correct?

There exists such an extension. Let $$f=X^5-1=(X-1)(X^4+X^3+X^2+X+1)\in L[X]$$. Then, $$f$$ only has root $$1$$ (because $$5$$ does not divide $$|L^\times|=26$$); thus, the second factor must have an irreducible, monic factor $$g\in L[X]$$ of degree $$\geq 2$$. Now, take $$M:=L/(g)$$. Then, $$[M:L]=\deg(g)\geq 2$$.

If it is correct, I was wondering if there a more fundamental or "straight forward" way to prove this. The argument above requries some "guess work" of finding such an irreducible polynomial of degree $$\geq 2$$; am I missing an easy point here?

• A more structural approach is possible. There exists an algebraic closure of $L$. And algebraically closed fields are infinite. So the degree of the closure as an extension of $L$ is infinite. Apr 1, 2021 at 16:56
• Thanks, but I forgot to mention that I am only looking for finite extensions here. :) Apr 1, 2021 at 16:57
• @Vercassivelaunos I think the argument in the question is slightly better, since the construction of algebraic closures is not trivial. Apr 1, 2021 at 16:58
• Other than changing $M=L/(g)$ to $M=L[X]/(g)$, I agree that the proof is correct (and a nice proof if we want to use as little machinery as possible). I guess technically the reasoning should be "because $\gcd(5,26)=1$" instead of "because $5$ does not divide $26$". Apr 1, 2021 at 17:12
• This is a good argument. A nitpicker may want you to add that $\gcd(f,f')=1$ and thus the roots of $f(x)$ are all simple. This extra check would fail if, instead of $x^5-1$, you would look at factors of $x^3-1=(x-1)(x^2+x+1)$. The condition $\gcd(3,26)=1$ holds in this case as well. But this time we won't get an irreducible quadratic factor because $$x^3-1=(x-1)^3\in \Bbb{F}_3[x].$$ Apr 2, 2021 at 4:13

Here's an argument with the same idea without any explicit calculations:

As you have realised, it suffices to show the existence of an irreducible polynomial $$g \in L[x]$$ such that $$\deg(g) \ge 2$$. In fact, we show that given any $$n \in \Bbb N$$, there exists an irreducible polynomial with degree $$\ge n$$.

Claim. The set $$S = \{g \in L[x] : g \text{ is irreducible}\}$$ is infinite.
Proof. (Essentially Euclid's proof of the infinitude of primes.) Suppose $$g_1, \ldots, g_s \in L[x]$$. Consider the polynomial $$f = g_1\cdots g_s + 1.$$
Then, $$f$$ has an irreducible factor which is not one of $$g_1, \ldots, g_s$$. $$\Box$$

Now, since the set $$L_n[x] = \{g \in L[x] : \deg(g) \le n\}$$ is finite (because $$L$$ is finite), it follows that $$S \setminus L_n$$ is non-empty for all $$n.$$

Thus, given any $$n$$, pick an element $$g \in S \setminus L_n$$ to get a field extension $$M = L[x]/(g)$$ of $$L$$ of degree $$\ge n$$.

Note that all I required in the above proof was that $$L$$ is finite. Thus, we have proven the following theorem.

Let $$L$$ be a finite field and $$n \in \Bbb N$$. Then there exists a field extension $$L \subset M$$ with $$[M : L] \ge n$$.

In fact, more is true, you can actually always find a field extension of degree exactly $$n$$.