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I am trying to show that the polynomial $x^3 - 3$ splits into linear factors over $ \mathbb{Z}_7[x]/ \langle x^3 - 3 \rangle $, but am having trouble doing so, as I'm not very familiar with rings.

First, the question as given to me actually asks me to show that $x^3 - 3$ splits into linear factors over $ \mathbb{Z}_7[y]/ \langle y^3 - 3 \rangle $, but I'm assuming this is a typo and that $y$ is meant to be replaced by $x$. Assuming $x^3 - 3$ splits into linear factors, I'm thinking this means we can write $$ x^3 - 3 + I = (x - a)(x - b)(x - c) + I $$ where $ I = \langle x^3 - 3 \rangle$. This happens if and only if $ x^3 - (a+b+c)x^2 + (ab+ac+bc)x -abc$ is an element of $I$, which leads me to the congruences $$ a+b+c \equiv 0, ab + ac + bc \equiv 0, abc \equiv 3 $$ where the congruences are all modulo 7. Now attempting to solve these congruences leads me to $a^3 \equiv 3 \mod 7$, which does not seem to have any solutions for $ a \in \mathbb{Z}_7$.

Any help is appreciated!

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You are over-using $x$. In $\mathbb{Z}_7[x]/ \langle x^3 - 3 \rangle$, $x^3-3=0$. I think you mean to split $y^3-3$ over this field. –  Thomas Andrews Feb 6 '13 at 18:00

2 Answers 2

up vote 1 down vote accepted

Let $\mathbb F=\mathbb{Z}_7[x]/ \langle x^3 - 3 \rangle$. Then in $\mathbb F[y]$, the polynomial $y^3-3$ factors as $(y-x)(y-2x)(y-4x)$.

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Hint 1: $y$ is a root of the desired polynomial.

Hint 2: $2^3\equiv 1 \pmod 7$

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