Factorise $f(x)=x^5-x$ into a product of irreducibles in $\mathbb Z_{5}[x]$

Question

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

Answer 1

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 $x(x-1)(x-2)(x-3)(x-4)$.

Answer 2

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.

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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 $\Bbb Z_p[x]$.

Answer 3

$f(x)=x^5-x=x(x^4-1)=x(x+1)(x-1)(x^2+1)$ 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.

Answer 4

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

Answer 5

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

$(x+a)(x+b)=x^2 + (a+b)x + ab = x^2 +1$ 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)$