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Let $f = 15x^4+22x^3-x=0$ a polynomial in $\mathbb Z_p[x]$, find the first prime $p$ value that will make $f$ result in being grade 3 and monic. Then factorize $f$ in $\mathbb Z_3[x]$ as product of irreducible factors.

Find the $p$ value

In order for $f$ to be monic and grade 3, I need to find a $p$ value so that:

$$\begin{aligned} 15 \equiv_p 0 \end{aligned}$$ $$\begin{aligned} 22 \equiv_p 1 \end{aligned}$$

it's easy to see that $p = 3$, so I will be working in $\mathbb Z_3[x]$ and $f = x^3+2x$.

Factorizing $f$

I need to find all the solutions to $x^3+2x=0$ in order to get it factorized as product of irreducible factors.

$$\begin{aligned} x^3+2x = x(x^2+2) = 0 \Leftrightarrow x = 0 \vee (x^2+2) = 0 \end{aligned}$$

Let's examine $\Delta = b^2-4ac$ of $(x^2+2)$

$$\begin{aligned} \Delta = 0 - 8 = -8 \equiv_3 = 1 \Rightarrow \end{aligned}$$ $$\begin{aligned} \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{0 \pm \sqrt{1}}{2} \Rightarrow x_1 = +\frac{1}{2} \equiv_3 2 \text{, } x_2=-\frac{1}{2} \equiv_3 1 \end{aligned}$$

So $f = x^3+2x= x(x+2)(x+1)$.

As my algebra exam day is fast approaching I really am in need of knowing if I've got this right or not. Sorry if I incuded trivial calculations in it, but I have to be extra sure I have understood everything. I have a question, though. Is it wrong to assume $x^3 = 1$ as we are dealing with $\mathbb Z_3$ elements? Doesn't the modulo apply to the exponents as well?

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up vote 0 down vote accepted

Your calculation seems right, though you could have avoided using the quadratic formula by noting that $x^2+2 = x^2-1 = (x+1)(x-1)$ in $\mathbb Z/3\mathbb Z$.

To answer your question on reducing exponents: In $\mathbb Z/3\mathbb Z$ every nonzero element $x$ satisfies $x^2 = 1$ so you can reduce exponents modulo $2$ (not modulo $3$). In general, if $m$ is an integer and $x$ is coprime to $m$ then $x^i \equiv x^j \mod m$ if $i \equiv j \mod {\varphi(m)}$. Here, $\varphi$ is Euler's totient function. If $m=p$ is a prime then $\varphi(p) = p-1$, so $x^{p-1} \equiv 1 \mod p$ if $p\nmid x$ (this is Fermat's little theorem).

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what does $\varphi(m)$ mean? – haunted85 Jul 14 '12 at 9:52
$\varphi$ is Euler's totient function. I added that to my answer. – marlu Jul 14 '12 at 9:57

Your answer seems correct.

For your second question you can reduce the exponents using Fermats Little Theorem which says : $a^p\equiv a\mod p$, not $a^p\equiv1\mod p$ like you said.${} {} {}$

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