10
$\begingroup$

Suppose $F_{p^k}$ is a finite field. If $F_{p^{nk}}$ is some extension field, then the primitive element theorem tells us that $F_{p^{nk}}=F_{p^k}(\alpha)$ for some $\alpha$, whose minimal polynomial is thus an irreducible polynomial of degree $n$ over $F_{p^k}$.

Is there an alternative to showing that irreducible polynomials of arbitrary degree $n$ exist over $F_{p^k}$, without resorting to the primitive element theorem?

$\endgroup$
5
  • 5
    $\begingroup$ Yes. See this for the case $k=1$. The same argument works when $k>1$. I like to think that the more general version has been covered here. The Möbius inversion formula for the number of irreducible polynomials works out irrespective of whether $k=1$ or not, and the linked question explains, how to deduce that the number is always positive. $\endgroup$ Commented Jan 27, 2013 at 15:35
  • $\begingroup$ Alternatively you can use the argument of the linked question to prove the existence of an irreducible polynomial of degree $nk$ over the prime field. The factors of that polynomial (over the extension field) are then necessarily of degree $n$. $\endgroup$ Commented Jan 27, 2013 at 15:39
  • $\begingroup$ Does the fact that the multiplicative group is cyclic count as a use of the primitive element theorem? $\endgroup$
    – Lior B-S
    Commented Jan 27, 2013 at 18:00
  • 2
    $\begingroup$ For $n=2$ and $n=3$ there exists a truly elementary proof, which is unfortunately hard to generalize... $\endgroup$
    – N. S.
    Commented Jan 27, 2013 at 18:34
  • 3
    $\begingroup$ @HailieMathison I think the following is one way to answer your question. Let $n>0$ be given. Since $k$ divides $kn$, we know that $F_{p^k}$ embeds in $F_{p^{kn}}$. Knowing that $F_{p^{kn}}^*$ is cyclic, let $\alpha\in F_{p^{kn}}^*$ generate $F_{p^{kn}}^*$. Then we must have $F_{p^k}(\alpha)=F_{p^{kn}}$ and thus the minimal polynomial of $\alpha$ over $F_{p^k}$ has degree $[F_{p^{kn}}:F_{p^k}]=n$. $\endgroup$ Commented Mar 14, 2016 at 13:54

1 Answer 1

16
$\begingroup$

A very simple counting estimation will show that such polynomials have to exist. Let $q=p^k$ and $F=\Bbb F_q$, then it is known that $X^{q^n}-X$ is the product of all irreducible monic polynomials over$~F$ of some degree$~d$ dividing $n$. The product$~P$ of all irreducible monic polynomials over$~F$ of degree strictly dividing $n$ then certainly divides the product over all strict divisors$~d$ of$~n$ of $X^{q^d}-X$ (all irreducible factors of$~P$ are present in the latter product at least once), so that one can estimate $$ \deg(P)\leq\sum_{d\mid n, d\neq n}\deg(X^{q^d}-X)\leq\sum_{i<n}q^i=\frac{q^n-1}{q-1}<q^n=\deg(X^{q^n}-X), $$ so that $P\neq X^{q^n}-X$, and $X^{q^n}-X$ has some irreducible factors of degree$~n$.

I should add that by starting with all $q^n$ monic polynomials of degree $n$ and using the inclusion-exclusion principle to account recursively for the reducible ones among them, one can find the exact number of irreducible polynomials over $F$ of degree $n$ to be $$ \frac1n\sum_{d\mid n}\mu(n/d)q^d, $$ which is a positive number by essentially the above argument (since all values of the Möbius function $\mu$ lie in $\{-1,0,1\}$ and $\mu(1)=1$). A quick search on this site did turn up this formula here and here, but I did not stumble upon an elementary and general proof not using anything about finite fields, although I gave one here for the particular case $n=2$. I might well have overlooked such a proof though.

$\endgroup$
4
  • $\begingroup$ I don't know much about finite fields but would like to understand his answer: The claim in the second line seems to be a general fact, can anyone give a reference or a proof for that? And I am pretty confused about the estimate: How does the $q$ appears? We consider the degree of the product of all irreducible polynomials over F of degree strictly dividing n. Don't they look like $X^d+...$ where $d|n$? And the degree is then just the sum over the $d$'s? I probably missed something, hope someone can help $\endgroup$
    – Algebra
    Commented Jun 28, 2018 at 7:47
  • 1
    $\begingroup$ The claim about the factorization of $X^{q^n}-X$ is staple finite field theory; basically, every element of $\Bbb F_{q^n}$ is a root of that polynomial, and for every such element$~x$ its minimal polynomial over $\Bbb F_q$ has degree$~d$ where $d$ is minimal such that $x$ is contained in the subfield $\Bbb F_{q^d}$, and this only happens when $d\mid n$ (and that minimal polynomial contributes its $d$ roots to those of $X^{q^n}-X$). $\endgroup$ Commented Jul 2, 2018 at 9:28
  • 1
    $\begingroup$ Also, in the estimate $q^d$ is the degree of $X^{q^d}-X$, the sum of the degree of all irreducible polynomials over $F$ of degree dividing$~d$; this is an upper bound for the sum of the degree of all such polynomials of degree exactly$~d$. $\endgroup$ Commented Jul 2, 2018 at 9:37
  • 1
    $\begingroup$ Indeed the argument could have been given without mentioning irreducible polynomials directly: the finite field of order $q^n$ contains at most one copy of the field of order $q^i$, and the inequality shows that even summing that over all $i<n$ gives too few elements to account for everyone. It follows that no finite field can be the union of its proper subfields, and then the minimal polynomial of an element not contained in any proper subfield provides an irreducible polynomial of degree$~n$. $\endgroup$ Commented Jul 2, 2018 at 9:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .