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For this problem, I was given three polynomials to determine which were irreducible:

  1. $x^4 + x + 1\in\mathbb Z/(2)[x]$
  2. $1 + x + x^2 + ... + x^{41}\in\mathbb Z/(2)[x]$
  3. $x^{42} + 42x + 4x^2 + 42\in\mathbb Q[x]$

I know that using Gauss' Lemma and Eisenstein's criterion are necessary for this problem, but I don't know how and where to apply them - I never worked with $\mathbb Z/(2)[x]$ polynomials before. I don't need direct answers for this question, I just want to know how I would go about finding them.

user126731
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    Hints: If $x^4+x+1$ is reducible, does it have a factor of degree $1$? If not, does it have a factor of degree $2$? Why don't you need to check for a factor of degree $3$? In the second case, how many nonzero coefficients does the polynomial have? – Dilip Sarwate Apr 06 '14 at 16:41
  • is $\mathbb{Z}/(2)[x]$ actually $\mathbb{Z}_2[x]$? – Ellya Apr 06 '14 at 17:00
  • If it would be reducible, then it could have a factor of $(x + ...)$ and a factor of $(x^3 + ...)$. For degree two, couldn't it have factors of $(x^2 + ...)$? Unless the mod 2 concept fits in somehow. – user126731 Apr 06 '14 at 17:11
  • And ellya, it is written as $\mathbb Z/(2)[x]$. – user126731 Apr 06 '14 at 17:12

1 Answers1

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You only use Gauss/Eisenstein for problem 3. Gauss tells you that if you have a primitive polynomial, it is irreducible over $\mathbb{Q}$ if and only if it is irreducible over $\mathbb{Z}.$ Then Eisenstein tells you that if you can find a prime $p$ that doesn't divide the leading term but divides all the rest, and $p^2$ doesn't divide the constant term then your polynomial is irreducible over $\mathbb{Z}.$ Can you find such a prime for question 3?

You can't really use (the generalization of) Eisenstein for Q1 and Q2, since $\mathbb{Z}/(2)$ is a field. In a field an element is divisible by $x$ implies that it's divisible by $x^2$ as well, which makes satisfying Eisenstein futile.

For Q1, you know that if it was reducible, it would factor either as a cubic times a linear term, or as the product of two quadratics. The linear term could only be $x$ or $x+1$ (there are no other linear polynomials over $\mathbb{Z}/(2)$). These are factors if and only if $0$ or $1$ correspondingly are roots of the polynomial (this is the root theorem), which you can check is not true. So if $x^4+x+1$ factors over $\mathbb{Z}/(2),$ it must be as the product of two irreducible quadratics (why must they be irreducible?). It's easy to list out all the quadratic polynomials, and then it's easy to check which ones are irreducible (the ones with no roots, by similar steps to what we just did). You'll see there is only one irreducible quadratic, and you can verify that $x^4+x+1$ is not the square of that.

For Q2, sum the geometric series to get $p(x)=\dfrac{x^{42}-1}{x-1}.$ Prove that for any polynomial, $p(x)$ is irreducible if and only if $p(x+1)$ is irreducible and apply that here to reduce our problem to checking the irreducibility of $\dfrac{ (x+1)^{42}-1 }{x}.$ Use the binomial theorem to simplify this, and try to decide if what you have if irreducible.

Ragib Zaman
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  • I understood Q1, but how would Q2 work? Could the geometric series be summed to that p(x)? – user126731 Apr 06 '14 at 18:28
  • @user126731 Sure. It's a polynomial identity that $(x-1)(1+x+\cdots + x^{41} ) = x^{42}-1,$ expand out the brackets and everything works out fine even working over $\mathbb{Z}/(2).$ – Ragib Zaman Apr 06 '14 at 18:51
  • @user126731 I just realised a much quicker way of reaching the same conclusion. Hint: Substitute $x=1.$ – Ragib Zaman Apr 06 '14 at 18:54
  • Okay. Also, how can I prove that P(x) would be irreducible if P(x+1) is irreducible? – user126731 Apr 06 '14 at 19:04
  • @user126731 If P(x) is reducible, write P(x) = f(x) g(x), and then P(x+1) = f(x+1) g(x+1) so P(x+1) is reducible. – Ragib Zaman Apr 06 '14 at 19:06
  • If $f(x) = 1 + x + x^2 + x^3 + \cdots + x^{41} \in \mathbbZ/(2)[x]$, then $f(1) = 1 + 1 + 1 + \cdots + 1 = 0$ and so $(x-1) | f(x)$. – Dilip Sarwate Apr 07 '14 at 02:27