Dtyjlui

The problem goes like the following:

Let $F$ be a finite field, and $L/F$ an extension of degree $n$.

(a) Show that any irreducible polynomial in $F[x]$ of degree $n$ is the minimal polynomial of exactly $n$ elements of $L$.

(b) Suppose $|F|=p$. Determine, in terms of $p$, the number of irreducible polynomials in $F[x]$ of degree 3, and also the number of irreducible polynomials in $F[x]$ of degree 9, respectively.

My approach for (a) is: If $f(x)$ is an irreducible polynomial of degree $n$ in $F[x]$ and if $alphain L$ is a root of the polynomial, then we have $F(alpha)cong dfrac{F[x]}{(f(x))}$ and $[F(alpha):F]=n$. Since $[L:F]=n$, $L=F(alpha)$. I wanted to say that $L$ is Galois over $F$ so that it would contain all the roots of $f(x)$ and thus $f$ is irreducible and has exactly $n$ roots in $L$. But I don't know how to justify that $L/F$ is indeed a Galois extension. Also, it seems to me that if $|F|=p$, then $F=mathbb{F}_p$ and $L=mathbb{F}_{p^n}$ but I don't know if these would be helpful.

For part (b), using Mobius inversion formula, the number of irreducible polynomials in $F[x]$ of degree 3 is given by
$$frac{1}{3}sum_{d|3} mu(d)p^{3/d} = frac{1}{3}(p^3-p).$$
However, I don't think this would be the purpose of this problem. Now I've noted that $|L|=p^n$ as it can be viewed as a vector space over $F$ of dimension $n$. Using (a), since the irreducible polynomial in $F[x]$ of degree 3 is the minimal polynomial of 3 elements of $L$, there would be $dfrac{p^3}{3}$ such irreducible polynomials as $|L|=p^3$, which does not match with the result given by the Mobius inversion formula. Can someone help me out here? Many thanks!

Edit: I still need some help on proving part (a), though. If anyone has anything, please leave a comment or a solution. That would be really helpful!

Well, since no solution so far has been posted, I'll give what I figured out.

First, Part (a):

Proof: Suppose $|F| = q$ (note that here $q$ is not necessarily a prime so in turn not necessarily a prime subfield of a finite field, it could be a power of prime, since any finite field is isomorphic to $mathbb{F}_{p^r}$ for some prime $p$ and integre $rge 1$), and suppose $L/F$ is an extension of degree $n$, i.e. $[L:F]=n$. Viewing $L$ as a vector space over $F$ of dimension $n$, we see that $|L| = q^n$. This says that $L cong mathbb{F}_{q^n}$. We claim that $L/F$ is a Galois extension. Note that $L^{times}$ (is cyclic being the mutiplicative group of $L$, which is finite) has order $q^n-1$. So for any $thetain L$,
$$theta^{q^n-1}=1implies theta^{q^n}=thetaimplies text{$theta$ is a root of the separable polynomial (by derivative test) $x^{q^n}-x$ over $F$.} $$
Therefore, $L$ is a subfield of the splitting field of $x^{q^n}-x$ hence is the splitting field as $|L|=q^n$. So such $L$ (an extension of $F$) exists and is Galois (also we have, by definition, $|$Gal$(L/F)|=[L:F]=n$).

Now suppose that $f(x)in F[x]$ is an irreducible polynomial of degree $n$ having $alphain Lsetminus F$ as a root. Then all roots of $f$ are contained in $L$ since the extension is Galois. These $n$ roots are distinct elements in $L$ since $f$ is irreducible over a perfect field (namely, the finite field $F$) thus separable. It then follows that $f$ is the irreducible polynomial for the $n$ roots in $L$ and so $f$ is the minimal polynomial of $n$ elements of $L$. (a) is proved.

Part (b) is solved (see the comments given by Lubin above).