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)

  • 9
    $\begingroup$ Hint: mimic a very, very old proof that there are infinitely many primes in some other ring. $\endgroup$
    – user29743
    Jul 3, 2012 at 13:17
  • 3
    $\begingroup$ Others have described a simple existential proof (+1). Another non-constructive one is to prove that finite fields of cardinality $2^n$ exist for all $n$. If you are interested in an infinite family of explicit polynomials, I refer you to an earlier answer of mine (a solution of an exercise from Lidl & Niederreiter). $\endgroup$ Jul 3, 2012 at 13:20
  • $\begingroup$ Google for "latex math symbols", perhaps adding the name of the particular symbol you want help on. See also this part of the meta Math FAQ. $\endgroup$
    – hardmath
    Jul 3, 2012 at 13:28
  • $\begingroup$ Someone here gave me once alink to a web where I can click the symbols/matrices and it shows as math, "cdg" something. I really liked it. $\endgroup$
    – Jozef
    Jul 3, 2012 at 13:33
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    $\begingroup$ Then there's detexify. You don't even need to search a table -- just draw what you want. $\endgroup$
    – tomasz
    Jul 3, 2012 at 13:41

2 Answers 2


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

  • $\begingroup$ Besides showing that none of $p_1,...p_n$ divide the polynomial you build above, Do I need to mention or show something else? $\endgroup$
    – Jozef
    Jul 5, 2012 at 14:29
  • $\begingroup$ @Jozef That's really all you need - $\prod p_i + 1$ has an irreducible factor which can't be one of the $p_i$s, so you have a contradiction. $\endgroup$
    – Cocopuffs
    Jul 5, 2012 at 14:55
  • $\begingroup$ Cocopuffs what would be different if I was asked to prove the same claim for $\mathbb Z / 3 \mathbb Z[x]$? $\endgroup$
    – Jozef
    Jul 5, 2012 at 14:58
  • $\begingroup$ @Jozef Nothing. $\endgroup$
    – Cocopuffs
    Jul 5, 2012 at 15:47
  • $\begingroup$ Doesn't $\mathbb{Z}$/2$\mathbb{Z}$ only have finite elements, since $X$ would equal $X^2$, would equal $X^3$, ...? $\endgroup$
    – Tristan
    Oct 16, 2018 at 15:04

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.


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