Let $F= \Bbb Z_2[x]/ \langle f \rangle$ with $f=x^3+x+1 \in \Bbb Z_2[x]$. Now consider f as an element of $F[x]$ and

a) show that there exists $\alpha \in F$ with $f(a)=0$
b) find $g \in F[x]$ with $f=(x-\alpha)g$
c) show that also $\alpha^2$ and $\alpha^4$ are roots of $f$ over $F$ and write $f$ as a product of irreducible elements of $F[x]$

So I tried finding the root of $f$ over $F$ and it seems like $x^2+x$ and $x^2$ are both roots but I'm stuck on part b and c, when I divide $f$ by $x-\alpha$ I always get a remainder.

Am I approaching this the right way or maybe I'm missing something. Also I'm having some trouble wrapping my head around the $F[x]$, what should be the modulus of it?

Thanks in advance!

  • $\begingroup$ I don't quite follow: as an element of $\;\Bbb F[x]\;$ , we have that $\;f=0\;$ ... $\endgroup$ – DonAntonio Sep 23 '13 at 12:39
  • $\begingroup$ @DonAntonio That's what I thought when I had a look at it, part of the reason why I have no idea what to do :S $\endgroup$ – whyisuckatmath Sep 23 '13 at 12:44
  • 2
    $\begingroup$ @DonAntonio, I believe the purpose of the question is to show that $F$ is the splitting field of $f$. $\endgroup$ – Jonathan Y. Sep 23 '13 at 12:44
  • $\begingroup$ What do you mean by the "modulus" of $F[x]$? $\endgroup$ – Gerry Myerson Sep 23 '13 at 12:44
  • $\begingroup$ As for $g$, work out the degree of $g$; then work out its leading coefficient; then the next coefficient; continue until you get to its constant term, for which you will want to make use of $\alpha^3+\alpha+1=0$. $\endgroup$ – Gerry Myerson Sep 23 '13 at 12:46

It is better to use two indeterminates, so let us say that $F = \mathbb{Z}_{2}[y]/\langle f(y) \rangle$.

a) One of the roots of $f(x)$ is $y$ (or more precisely $y + \langle f(y) \rangle$).

b) Yes, divide as you suggest by $x - y = x + y$ to get $$ x^{3} + x + 1 = (x + y) (x^{2} + y x + (1 + y^{2})). $$

c) This is a general fact, as over a commutative ring of characteristic two, the map $u \mapsto u^{2}$ will be a ring homomorphism. So if $\alpha$ is a root of $f(x)$ you have $0 = f(\alpha) = \alpha^{3} + \alpha + 1$, and thus $$ 0 = (\alpha^{3} + \alpha + 1)^{2} = \alpha^{6} + \alpha^{2} + 1 =(\alpha^{2})^{3} + \alpha^{2} + 1 = f(\alpha^{2}). $$

| cite | improve this answer | |
  • $\begingroup$ Thank you! It made so much more sense when I used a different notation for the x inside F. $\endgroup$ – whyisuckatmath Sep 23 '13 at 12:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.