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?

  • 2
    $\begingroup$ 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. $\endgroup$ Apr 1, 2021 at 16:56
  • $\begingroup$ Thanks, but I forgot to mention that I am only looking for finite extensions here. :) $\endgroup$
    – log_math
    Apr 1, 2021 at 16:57
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    $\begingroup$ @Vercassivelaunos I think the argument in the question is slightly better, since the construction of algebraic closures is not trivial. $\endgroup$
    – Kenta S
    Apr 1, 2021 at 16:58
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    $\begingroup$ 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 $26$". $\endgroup$ Apr 1, 2021 at 17:12
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    $\begingroup$ 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 $$x^3-1=(x-1)^3\in \Bbb{F}_3[x].$$ $\endgroup$ Apr 2, 2021 at 4:13

1 Answer 1


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 $n$.


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