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I'm studing for an exam and I am stuck on the following practice problem.

Consider the the ring $R=\mathbb{F}_{p}[x]$. How many irreducible polynomials of degree 4 exist in $R$?

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Neat. There are other, more laborious, ways to count. – Geoff Robinson Aug 31 '12 at 18:45
A related answer. This answer gives the general formula for irreducible (monic) polynomials of a given degree. A vaguely related question. And another one (with high quality answers). And one more. – Jyrki Lahtonen Aug 31 '12 at 18:46
And finally a related question where one answer gives links to proofs of the general formula. – Jyrki Lahtonen Aug 31 '12 at 18:57
up vote 8 down vote accepted

Irreducible quartic polynomials (assumed monic) in $R$ are exactly the minimal polynomials of those elements of the (up to isomorphism) unique finite field $L=\mathbb{F}_{p^4}$ of $p^4$ elements that don't belong to a proper subfield.

From the basic facts about containment of one finite field in another we immediately see that $K=\mathbb{F}_{p^2}$ is (isomorphic to) the unique maximal subfield of $L$. Therefore there are exactly $p^4-p^2$ elements in $L$ with quartic minimal polynomials. As the said minimal polynomials are separable (owing to the fact that the base field is perfect), each and every one of them is the minimal polynomial of exactly four conjugate elements. Therefore there are $$ N_p=\frac{p^4-p^2}4 $$ irreducible monic quartic polynomials with coefficient in $\mathbb{F}_p$.

Note also that uniqueness of $L$ means that any irreducible quartic is the minimal polynomial of some element of $L$. This is because any such quartic has a root in some quartic extension of $\mathbb{F}_p$, but $L$ is the only one.

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I failed in my search for a close enough duplicate, so I expanded my comment to an answer. Deleted that comment attached to OP. – Jyrki Lahtonen Aug 31 '12 at 20:54
A general formula (due to Gauss ? involving Möbius inversion) for the number of irreducible monic polynomials with coefficients in a fixed finite field is known. The legend (see the links in my earlier comments) says that a proof is in e.g. Dummit and Foote (and probably many other books on basic algebra). – Jyrki Lahtonen Aug 31 '12 at 20:57
I think you also need to mention that two different irreduciable polynomials don't share roots for your counting argument – Belgi Aug 31 '12 at 21:06
May be? If two irreducibles shared roots they would have a non-trivial gcd and couldn't be irreducible! I thought that would be common enough knowledge, but may be not? – Jyrki Lahtonen Aug 31 '12 at 21:57

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