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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?

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  • $\begingroup$ 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 $x$). $\endgroup$ – ΑΘΩ Jan 7 '20 at 5:01
  • $\begingroup$ (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. $\endgroup$ – ΑΘΩ Jan 7 '20 at 5:04
  • $\begingroup$ Thanks, I changed notations. Does my solution look correct? $\endgroup$ – OSBM Jan 7 '20 at 5:07
  • $\begingroup$ 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. $\endgroup$ – ΑΘΩ Jan 7 '20 at 5:10
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    $\begingroup$ @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). $\endgroup$ – ΑΘΩ Oct 31 '20 at 2:46
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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 $L$.

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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.

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

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