2
$\begingroup$

So plugging in $1$ gives $f(1) = 0$ which means $1$ is a root and $f$ has a factor $(x-1)$ which is $\equiv (x+4)$ in $\mathbb Z_{5}[x]$ ?

I then divide $f(x)$ by $(x+4)$ using polynomial long division and I get

$x^4-4x^3+16x^2-64x+255$ with remainder $-1020$

And that's looking kinda strange.. not sure what I'm doing at this point

I notice $f(0)$ also $= 0$ so maybe if I divide $f(x)$ by $x$ instead I would get $(x^4-1)$ but then I'm still stuck and don't know what else to do

$\endgroup$
  • $\begingroup$ $-4\equiv 1$, $16\equiv 1$, $-64\equiv 1$, $255\equiv 0$. But most impotantly, there is no remainder: $-1020\equiv 0$. You can divide $x^5-x$ by $x-1$ and then by $x$ to obtain $x^3+x^2+x+1$. Or in full: $x^5-x=(x+1)\cdot x\cdot (x^3+x^2+x+1)$. Now try to split $x^3+x^2+x+1$. $\endgroup$ – Hagen von Eitzen Nov 10 '12 at 7:51
5
$\begingroup$

You started out on a good track, but the coefficients of your division result are in $\mathbb Z$ instead of $\mathbb Z_5$.

The key to doing this more efficiently is Fermat's little theorem, which states that all residues $\bmod\,5$ are roots of this polynomial, so it factors as $x(x-1)(x-2)(x-3)(x-4)$.

$\endgroup$
5
$\begingroup$

$f(x) = x (x-1)(x^3+x^2 + x+1)=x(x-1)(x+1)(x^2+1)$

$(x+a)(x+b)=x^2 + (a+b)x + ab = x^2 +1$ in ${\bf Z}_5[x]$

So $a+b =0(5)$ and $ab=1(5)$ Hence $a=2$ and $b=3$

$ f(x) =x(x-1)(x+1)(x+2)(x+3)$

$\endgroup$
4
$\begingroup$

Well, $$x^5-x=x(x^4-1)=x\bigl((x^2)^2-1)\bigr)=x(x^2-1)(x^2+1)=x(x-1)(x+1)(x^2+1),$$ just by standard factoring methods. All you need to do is determine how (if at all) $x^2+1$ can be factored into linear terms.


As a more general (and more difficult) result, we can show that for any prime $p$, we have $$x^p-x=x\prod_{k=1}^{p-1}(x-k)$$ in $\Bbb Z_p[x]$.

$\endgroup$
2
$\begingroup$

$$x^5-x=x(x^4-1)=x(x^2+1)(x+1)(x-1)=x(x^2-4)(x+1)(x-1)=x(x-2)(x+2)(x+1)(x-1)$$ where at a middle step a 1 was replaced by its congruent -4.

$\endgroup$
2
$\begingroup$

$f(x)=x^5-x=x(x^4-1)=x(x+1)(x-1)(x^2+1)$ over any ring. Now over $\mathbb{Z}_5$, what can we say about the quadratic factor $x^2+1$? Note that it actually has roots $\pm2$, Thus $f(x)=\prod_{r=0}^4(x-r)$. This also follows from Fermat's Little Theorem since $5$ is prime.

$\endgroup$

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.