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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!

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I don't quite follow: as an element of $\;\Bbb F[x]\;$ , we have that $\;f=0\;$ ... –  DonAntonio Sep 23 '13 at 12:39
    
@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 –  whyisuckatmath Sep 23 '13 at 12:44
2  
@DonAntonio, I believe the purpose of the question is to show that $F$ is the splitting field of $f$. –  Jonathan Y. Sep 23 '13 at 12:44
    
What do you mean by the "modulus" of $F[x]$? –  Gerry Myerson Sep 23 '13 at 12:44
    
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 $\alpha^3+\alpha+1=0$. –  Gerry Myerson Sep 23 '13 at 12:46

1 Answer 1

up vote 3 down vote accepted

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 $$ 0 = (\alpha^{3} + \alpha + 1)^{2} = \alpha^{6} + \alpha^{2} + 1 =(\alpha^{2})^{3} + \alpha^{2} + 1 = f(\alpha^{2}). $$

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Thank you! It made so much more sense when I used a different notation for the x inside F. –  whyisuckatmath Sep 23 '13 at 12:58

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