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1.Let us consider the set $F[x]_{n,k}$ of all polynomials over $F$ of degree $n$ having preciesly $k$ irreducible factors. The question is: Fix some $n$. Let $p$ be a big prime number.How the cardinality of the sets $\mathbb F_p[x]_{n,k}, 1\leq k\leq n$ looks like? Is it true that $|\mathbb F_p[x]_{n,1}|>0.99\cdot \sum\limits_{k=1}^n |\mathbb F_p[x]_{n,k}|$ when $p$ is very big?

Can we compute the limit of $\dfrac{|\mathbb F_p[x]_{n,j}|}{ \sum\limits_{k=1}^n |\mathbb F_p[x]_{n,k}|}$ for fixed $j,n$ when $p\to +\infty$?

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    Did you ever see $\sum_{d| n} \mu(d) p^{n/d}$ ? It is the number of elements of $\Bbb{F}_{p^n}$ whose $\Bbb{F}_p$-minimal polynomial has degree $n$. From its obvious asymptotic $p^n+O(p^{n/2} \tau(n))$ you can find the asymptotic of your sets. – reuns Apr 08 '21 at 22:06

1 Answers1

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No.

In fact, there is a formula (in terms of mobius inversion and the mobius function $\mu$) for the number of monic irreducibles of degree $n$ in $\mathbb{F}_p$:

$$ \frac{1}{n} \sum_{d \mid n} \mu(d) p^{n/d} $$

You can see here for a more detailed discussion of this formula.

In the case $n$ is prime, this has a simpler closed form (since the divisors $d \mid n$ are easy to understand):

$$ \frac{p^n - p}{n} $$

(again, see here for a more detailed discussion).

Since there are $p^n$ many monic polynomials of degree $n$, this tells us that the fraction of irreducible polynomials (if we continue assuming $n$ is prime, for simplicity) is:

$$\frac{1}{n} \left ( 1 - \frac{1}{p^{n-1}} \right )$$

Which, for large $p$, is roughly $1/n$.

With a little bit more work, you can use this same formula in order to compute the asymptotic size of your sets of interest.

I hope this helps ^_^

HallaSurvivor
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