There are infinitely many irreducible polynomials in $\Bbb{F}_p[X]$ I can intuitively understand that there are infinitely many irreducible polynomials in $\Bbb{F}_p[X]$, where $p$ is a prime number, but I'm having a hard time actually proving it. What proof strategy should I take for this?
 A: Assume that there are finitely many (monic) irreducible polynomials in $\mathbf F_{p}[X]$,$p_{1}(x), p_{2}(x), \dots, p_{n}(x)$. Consider $f(x)=p_{1}(x)p_{2}(x) \cdots p_{n}(x)+1$. Now $f(x)$ is not divisible by any irreducible, and hence is irreducible and is not in the list of irreducibles. Contradiction.
A: Mimic Euclid: $\, f_{n+1}\! = 1+f_1\cdots f_n\,$ is an infinite sequence of coprimes $\,f_i.\,$ Choosing $\,g_i\,$ to be an irreducible factor of $\,f_i\,$ yields an infinite sequence of coprime (so nonassociate) irreducibles.

Alternatively, recall that the sequence of polynomials $\rm\:f_n = (x^n\!-\!1)/(x\!-\!1)\:$ is a strong divisibility sequence, i.e. $\rm\:(f_m,f_n) = f_{(m,n)}$ in $\rm\mathbb Z[x].\:$ Hence the subsequence with prime indices yields an infinite sequence of pairwise coprime polynomials. Further the linked proof shows the gcd has linear (Bezout) form $\rm\:(f_m,f_n) = f_{(m,n)}\! = g\, f_m + h\, f_n,\,$ $\rm\, g,h\in\mathbb Z[x],\:$ so said coprimality  persists in every ring $\,R[x].\,$ Thus, for each prime $\rm\:q,\:$ choosing a factor of $\rm\:f_q\:$ irreducible in $R[x]$ yields infinitely many irreducible polynomials, pariwise nonassociate (being pairwise coprime). 
