Abstract Algebra: Polynomials are irreducible

Question

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?

Answer 1

Cyclotomic polynomials are irreducible. It is elaborated here:

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

Answer 2

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.