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Show that there are infinitely many irreducible polynomials in $\Bbb Z_2[x]$.

I have a strategy about how to do this but I need some help finding a contradiction. I want to assume there are finitely many irreducible polynomials. $p_1,...p_n$. Then I want to consider the irreducible factors of $(p_1p_2...p_n)+1$. I don't really know where to go from here. Any help would be appreciated.

TfwBear
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    Same as for the proof of infinitely many primes. None of the $p_i$ can be a factor of the polynomial you have constructed, so it must have some factor (hence, some irreducible factor) outside your list. – Gerry Myerson Dec 07 '15 at 06:24
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    @Gerry: you need to say a bit more than this; namely, you need to rule out the possibility that this element is a constant (which you can do by looking at its degree). This is one of the things that can go wrong if you try to run this proof in an arbitrary ring. – Qiaochu Yuan Dec 07 '15 at 08:38
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    @Nid, by "$\mathbb{Z}_2$" do you mean the 2-adic integers, or the field with 2 elements? – Noah Schweber Dec 07 '15 at 21:48

2 Answers2

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$x+n$ is irreducible for all $n\in \mathbb Z$. They are also different irreducibles.

If question is about finite field with two element: each of root of your polynomial will be in some extension of $\mathbb F_2$. Think bigger extensions.

vudu vucu
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  • Rofl that is also true, but your answer also suffers from the same assumption I made, stemming from the fact that for some reason a lot of introductory algebra textbooks like to use $\mathbb{Z}_2$ to denote $\mathbb{F}_2$ the field with 2 elements. – oxeimon Dec 07 '15 at 23:51
  • It is okay. They are right. It unfortunately has two meanings. – vudu vucu Dec 08 '15 at 00:01
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EDIT 2: Thanks everyone for downvoting a legitimate answer and an honest attempt to help. Great community here.

You can use Eisenstein's criterion to note that $x^n - 2$ is irreducible in $\mathbb{Z}_2[x]$ for every integer $n\ge 2$.

Also if you don't require your irreducible polynomials to be monic, then for any one irreducible polynomial $f$, $uf$ is also irreducible for any unit $u\in\mathbb{Z}_2[x]$. If you know that every element of $\mathbb{Z}_2$ can be expressed as a power series $\sum_{k\ge 0}a_k 2^k$ where $a_k\in\{0,1\}$, then by Hensel's lemma every power series with $a_0 = 1$ corresponds to a unit in $\mathbb{Z}_2$. Thus $\mathbb{Z}_2$ has infinitely many units, so the family $\{uf : u\in\mathbb{Z}_2^\times\}$ is already an infinite family of irreducible polynomials (though not monic).

EDIT: If by $\mathbb{Z}_2$ you mean the field $\mathbb{F}_2$ with two elements, then your strategy is probably easiest. If $p_1,\ldots,p_n$ denote all the primes of $\mathbb{F}_2[x]$, then none of the $p_i$'s can divide $(p_1p_2\cdots p_n)+1$, since if say $p_1$ divides $(p_1p_2\cdots p_n)+1$, then $(p_1p_2\cdots p_n)+1 = p_1q$ for some $q\in\mathbb{F}_2[x]$, so $$1 = p_1q - (p_1p_2\cdots p_n)$$ so $$1 = p_1(q - (p_2p_3\cdots p_n))$$ which is to say that $p_1$ is a unit (ie, $p_1$ is invertible), which of course cannot be the case if $p_1$ is prime/irreducible (these are the same in $\mathbb{F}_2[x]$ because $\mathbb{F}_2[x]$ is a UFD)

oxeimon
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