Prove that there exists an irreducible polynomial $f\in\mathbb{F}_p[x]$ of degree $2014$

Question

Let $p$ be a prime and let $F=\mathbb{F}_p$. Prove that there exists an irreducible polynomial $f\in F[x]$ with $\deg(f)=2014$.

Attempt:

Assume in contradiction that every polynomial $f\in F[x]$ of degree $2014$ is reducible. Let $f\in F[x]$. (My problem is to find a proper $f$ such that I will get a contradiction). There exist $g,h\in F[x]$ s.t. $$ F[x]\ni f(x)=g(x)\cdot h(x)$$ If we look at $f$ as an element in $\mathbb{Z}[x]$ then we get that there exists some $r\in\mathbb{Z}[x]$ s.t. $$\mathbb{Z}[x] \ni f(x)=g(x)\cdot h(x) +p\cdot r(x)$$

Can't you just do $x^{2014}-2$ and appeal to Eisenstein or something? — mathworker21, Jul 23 '19 at 17:37
@reuns The degree is $p^N$, isn't it? And I'm sorry, I didn't really get your point. — J. Doe, Jul 23 '19 at 17:42
How much are you allowed to use? Can you use the existence of finite fields of any cardinality (which is the power of a prime number)? — Aphelli, Jul 23 '19 at 17:46
It would appear that the prime factorization is $2014=2\cdot 19\cdot 53$. You honestly might be able to do this by literally counting the reducible polynomials and the total number of polynomials. — , Jul 23 '19 at 17:46
@Mindlack Yes I'm allowed to use the existence of $\mathbb{F}_{p^n}$. — J. Doe, Jul 23 '19 at 17:48
Then consider the minimal polynomial of a generator of the cyclic group $\mathbb{F}^*_{p^{2014}}.$ — Aphelli, Jul 23 '19 at 17:49

score 1 · Accepted Answer · answered Jul 23 '19 at 17:51

Hint Consider the extension $$\mathbb{F} \subseteq \mathbb{F}_{p^{2014}}$$

Since the multiplicative group of $\mathbb{F}_{p^n}$ is cyclic, there exists some generator $b$. Then $\mathbb{F}_{p^n}=\mathbb{F}[b]$.

What can you say about the minimal polynomial of $b$?

Note The existence of an irreducible polynomial of degree $n$ ove $\mathbb{F}$ is equivalent to the existence of $\mathbb{F}_{p^n}$.

reuns · Answer 2 · 2019-07-23T18:19:17.093

Let $\Bbb{F}_{p^n}$ be the splitting field of $x^{p^n}-x$, from $(c+b)^p = c^p+b^p$ we obtain the roots of that polynomial is a field, as $\gcd(x^{p^n}-x,(x^{p^n}-x)') = \gcd(x^{p^n}-x,-1)=1$ the polynomial is separable so it has $p^n$ distinct roots and $\Bbb{F}_{p^n}$ is the unique field with $p^n$ elements
For $d \le p^n-1$, the polynomial $x^d-1$ has at most $d$ roots in $\overline{\Bbb{F}_p}$ thus its roots contain all the elements of $\Bbb{F}_{p^n}^*$ iff $d = p^n-1$ which means $\Bbb{F}_{p^n}^*$ is cyclic.
Let $a \in \Bbb{F}_{p^n}^*$ of order $p^n-1$, then $\Bbb{F}_{p^n} = \Bbb{F}_p(a) \cong \Bbb{F}_p[x]/(f(x))$ where $f$ is the minimal polynomial of $a$,
$\Bbb{F}_p[x]/(f(x))$ has $p^{\deg(f)}$ elements thus $\deg(f) = n$.
Since $\Bbb{F}_p$ is the fixed field of the Frobenius we have $g(x)=\prod_{m=1}^n (x-a^{p^m}) \in \Bbb{F}_p[x]$ and hence $f = g$

Prove that there exists an irreducible polynomial $f\in\mathbb{F}_p[x]$ of degree $2014$

2 Answers2