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A) $x^4+x+1\in \Bbb Z_2[x]$ is irreducible

Proof:

$x^4+x+1\in \Bbb Z_2[x]$ is primitive. The mod 2 reduction of $x^4+x+1\in \mathbb{Z}[x]$ is $f(x)=x^4+x+1\in \Bbb Z_2[x]$. Since $f(a) = 1 \neq 0$ for all $a\in\Bbb Z_2$ it follows that $f(x)$ has no linear factors. Suppose that $f(x)$ is reducible. Then it must be the product of quadratic factors. There are $3$ quadratic reducible polynomials in $\Bbb Z_2[x]$. The irreducible one is $x^2+x+1$ since this polynomial has no roots in $\Bbb Z_2$. Therefore $f(x)=(x^2+x+1)^2=x^4+x+1$ which is not the case. Thus it is irreducible.

B) $\displaystyle\sum_{i=1}^{42}x^{i-1} \in \Bbb Z_2[x]$ is irreducible

I was able to prove problem A which I needed your guidance which on whether or not it was true.

But how to prove problem B?

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  • $\begingroup$ Yes your problem A is correct (albeit a pain to edit). $\endgroup$ – anon Apr 8 '14 at 1:30
  • $\begingroup$ Thank you anon. But will you give an idea to how to solve B. I looked over wikipedia link(below). But confused to how to prove that. $\endgroup$ – user1413 Apr 8 '14 at 1:35
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    $\begingroup$ Given $f(x)=x^{42}+\cdots+x+1=\frac{x^{43}-1}{x-1}$ I would show $f(x+1)=\frac{(x+1)^{43}-1}{x}$ is irreducible (using binomial theorem and then Eisenstein with the prime $2$) hence $f(x)$ is irreducible. This is the same proof on Wikipedia. If you're confused about it (which is understandable), point out specific parts where you are confused so we can help you. $\endgroup$ – anon Apr 8 '14 at 1:38
  • $\begingroup$ Oh ok I got it. I just got an example of this dpmms.cam.ac.uk/~par31/notes/eis.pdf $\endgroup$ – user1413 Apr 8 '14 at 1:40
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Cyclotomic polynomials are irreducible. It is elaborated here:

http://en.wikipedia.org/wiki/Eisenstein's_criterion

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  • $\begingroup$ Firstly, is my problem A right? Secondly for problem B I looked over to generalization which was in the link you posted and I came up with this question but still confused how to solve. $\endgroup$ – user1413 Apr 8 '14 at 0:48
  • $\begingroup$ Check out the markup for your link now. you could change the part in the square braces to be any other text. $\endgroup$ – alex.jordan Apr 8 '14 at 0:58
  • $\begingroup$ I visited wikipedia site before uploading this post. I would like to know if i have problem like problem B then how to solve the question. $\endgroup$ – user1413 Apr 8 '14 at 1:17
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Even I am interested in this question. Btw what do you mean by markup?

I think first one is right. But about Second I think it should be

Let

Q=$\sum_{i=1}^{42}$x$^{i-1}$

be an element of D[x], the polynomial ring with coefficients in D. Suppose there exists a prime ideal p of D such that x$^{i-1}$ ∈ p for each i ≠ 42, Then Q cannot be written as a product of two non-constant polynomials in D[x]. If in addition Q is primitive (i.e., it has no non-trivial constant divisors), then it is irreducible in D[x]. If D is a unique factorization domain with field of fractions F, then by Gauss's lemma Q is irreducible in F[x], whether or not it is primitive (since constant factors are invertible in F[x]); in this case a possible choice of prime ideal is the principal ideal generated by any irreducible element of D.

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