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The task was:

Polynomials over the body $Z_2$ are viewed.

Determine for $p(z) = z^4 + z^3 + z$ and $q(z) = z^2 + 1$

  1. $f(z) = p(z) + q(z)$
  2. $g(z) = p(z) * q(z)$
  3. $h \equiv p\ mod\ q$ with minimal degree(h)

And factorize the polynomials if possible

What i did was:

  1. $f(z) = z^4 + z^3 + z^2 + z + 1$
  2. $g(z) = z^6 +z^5 +z^4 + z^3 + z^3 + z $
  3. $h \equiv (z^4 + z^3 + z):(z^2 + 1) = z^2 + z +$$ {-z^2}\over{z^2 + 1}$

$h \equiv -z^2 $

My question is:

How should i factorize the polynomials? I dont think i can factorize it more! And is 3 correct? Thanks

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Modulo 2, $q(z) = (z+1)^{2}$ –  neelp Dec 29 '13 at 14:28
    
About question 2: What is $z^3+z^3 \mod 2$? –  TonyK Jan 6 at 9:00
    
$(z^4 + z^3 + z) \equiv (z^2 + 1) *(z^2 + z +1)+1\pmod{2}$, so $h \equiv 1 \pmod{2}$ –  miracle173 Jan 6 at 23:47

1 Answer 1

up vote 1 down vote accepted

Factorizing in $\mathbb Z_2$ is fun. Let's try $f$. It has a linear factor iff it has a root. It doesn't. So if it factors, it factors into two quadratic, say: $$ (ax^2+bx+c)(dx^2+ex+F) $$

Now $ad=1$, so $a=d=1$. $ae+bd=b+e=1$, so $b\neq e$. Similarly, $c=F=1$. Next, $aF+be+cd=be=1$, so $b=e=1$. This is a contradiction. So $f$ can't be factored.

There's probably a less stupid way to do this, but those ways are boring.

share|improve this answer
    
Thanks! Im not sure if i get you 100%! Could you please give me a example with my numbers? Thanks –  user2724695 Dec 29 '13 at 14:36
2  
This was with your numbers... –  Gaffney Dec 29 '13 at 14:37

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