infinite number of irreducible polynomials in $\mathbb{Z}/2{\mathbb Z}[X]$ For $A= \mathbb{Z}/2{\mathbb Z}[X]$ ring of polynomials with coefficient in the field $\mathbb{Z}/2{\mathbb Z},$ I need to show that there are infinite number of irreducible polynomials in $A.$
How do I show that? I didn't come to any conclusion. I though of series of polynomials but since it it modulo $2$ they were not suitable.
Any direction?
(And: does any one have a link to a web where I can choose Latex symbols and see how they are written? I had one, but I lost it, and can't find it in Google)
 A: Besides Euclid's classical method, here's another approach. 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 mod $2;\,$ indeed $\rm\:(p,q)=1\:$ for primes $\rm\:p\ne q,$ so
$$\rm\,mod\ 2\!:\ \ d\:|\:f_p,f_q\ \Rightarrow\ d\:|\:g\,f_p\!+\!h\,f_q = f_{(p,q)}\! = f_1 = 1.\,$$
Thus, for each prime $\rm\:p,\:$ choosing a prime factor of $\rm\:f_p\:$ yields infinitely many prime polynomials mod $2,\,$ none associate (being pairwise coprime). Note that this argument works quite generally.
A: A variant of Euclid's proof should work fine.  Assume there are only finitely many irreducible polynomials $p_1,...,p_n$ in $A$, and consider the irreducible factors of $\prod_{i=1}^n p_i + 1$.
