# Roots of irreducible polynomial over finite field extension.

Let $$K$$ be finite field and $$L$$ be an extension of $$K$$ of degree $$n$$. Fix a monic irreducible polynomial $$f(x)\in K[X]$$ of degree d dividing n. Show that there is element $$\alpha \in L$$ which has minimal polynomial $$f$$ over $$K$$.

I know that $$K$$ is isomorphic to field $$\mathbb{F}_{p^m}$$ for some $$m$$. If $$m=1$$ then $$K=\mathbb{F}_{p}$$ and we get result from the fact that $$L$$ is given by roots of polynomial $$X^{p^n}-X$$, which is product of all irreducible polynomials over $$\mathbb{F}_{p}$$ of degree $$d$$ dividing $$n$$, and hence has to contain roots of any irreducible polynomial of such degree.

I have trouble with general case when $$m\neq1$$.

If I take a root $$\alpha$$ of polynomial $$f$$ then I get extension $$K(\alpha)$$ of degree $$d$$ over $$K$$, which is isomorphic to $$\mathbb{F}_{p^{md}}$$. Field $$\mathbb{F}_{p^{md}}$$ is given by roots of polynomial $$X^{p^{md}}-X$$ which is product of all irreducible polynomials over $$\mathbb{F}_{p}$$ of degree dividing $$md$$. Hence minimal polynomial of $$\alpha$$ over $$\mathbb{F}_{p}$$ has to be of degree dividing $$md$$, and hence also dividing $$mn$$. Because of that, similarly as in case $$m=1$$, $$L$$ has to contain $$\alpha$$. Is my reasoning correct?

Is there is another, quicker approach?

• As a note on correct syntax (i.e. the rules of formal language), lower-case notation is not very desirable for the indeterminate of a polynomial ring; instead, upper-case letters are (to be) preferred: X, Y, Z etc. Furthermore, given a polynomial $f$ in indeterminate $X$ over ring $A$, the object $f(x)$ would in principle mean the substitution of $x$ in the place of the indeterminate $X$ inside $f$ (in more concise language, this is the image of $f$ through the unique ring morphism of ''substitution'' taking $X$ to $x$ from $A[X]$ to whatever ring $B$ happens to contain $x$). – ΑΘΩ Jan 7 '20 at 5:01
• (continuation of the above) hence, $f(X)$ is simply equal to $f$, the underlying substitution morphism being none other than the identity of the polynomial ring $A[X]$. Therefore, it is quite unnecessary to write $f(X)$; in order to specify the notation for the indeterminate, that should be declared from the onset, before even introducing the polynomial itself. – ΑΘΩ Jan 7 '20 at 5:04
• Thanks, I changed notations. Does my solution look correct? – OSBM Jan 7 '20 at 5:07
• It is correct in principle of thought, but a bit sloppy with the details. Without wanting to come off as an arrogant person, I would dare say that the solution below neatly and concisely expresses what you wanted to say. – ΑΘΩ Jan 7 '20 at 5:10
• @Lubin But of course, sir. May I toast a glass of red wine to the spirit of your invitation (and you are indeed not mistaken about me being a man). – ΑΘΩ Oct 31 '20 at 2:46

## 2 Answers

It is most convenient to consider an algebraic closure $$F$$ of $$L$$; then $$F$$ will automatically be an algebraic closure of $$K$$ containing $$L$$ as a subextension. The structure of the algebraic extension of a finite field is remarkably simple: for any $$n \in \mathbb{N}^*$$ there exists a unique subextension $$E_n$$ of degree $$n$$ over $$K$$, given explicitly as the set of all roots of the polynomial (separable over $$K$$) $$X^{q^n}-X$$, where $$q=|K|$$; furthermore, one has

$$F=\bigcup_{n \in \mathbb{N}^*} E_n$$

and

$$E_m \subseteq E_n \Leftrightarrow m|n$$

Considering an arbitrary root $$x \in F$$ of your given polynomial $$f$$, it is clearly the case that $$[K(x):K]=d$$ whence $$K(x)=E_d \subseteq E_n=L$$; therefore, all the roots of $$f$$ lie in the subextension $$L$$.

Here is a direct proof, for which you need only to know that a finite subgroup of the multiplicative group $$L^*$$ of a field is necessarily cyclic $$(*)$$. Here your $$L$$ is an extension of degree $$n$$ (missing in the first sentence of your post) of a finite field $$K=\mathbf F_q$$, where $$q$$ is a power of the characteristic $$p$$ of $$L$$. In a fixed algebraic closure $$\bar K$$, property $$(*)$$ implies that $$L=K(\mu_s)$$, where $$\mu_s$$ denotes the group of ($$q^n -1$$)-th roots of unity. Now let $$f(X)\in K[X]$$ be irreducible of degree $$d$$. Because of $$(*)$$, the splitting field of $$f$$ in $$\bar K$$ is of the form $$N=K(\mu_r)$$, with $$r=q^d -1$$. If $$d$$ divides $$n$$, then $$q^d -1$$ divides $$q^n -1$$, hence $$N\subset L$$ and we are done.

• How do you know that $q^d-1$ divides $q^n-1$? – OSBM Jan 8 '20 at 19:49
• If $d$ divides $n$, write $n=dm$, so that $q^n-1=(q^d)^m-1$. But $X^m-1=(X-1)(X^{m-1}+...+X+1)$. – nguyen quang do Jan 8 '20 at 22:20