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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)

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    @JyrkiLahtonen: Thanks again :)2012-07-05

2 Answers 2

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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 $\prod_{i=1}^n p_i + 1$.

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    Doesn't $\mathbb{Z}$/2$\mathbb{Z}$ only have finite elements, since $X$ would equal $X^2$, would equal $X^3$, ...?2018-10-16
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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.