Proving that $E=\mathbb{Z}[x]/(x^3-2x+1)$ is not an integral domain and that $x$ is a unit.
Let $r=(x^3-2x+1)$
So to show that $\mathbb{Z}[x]/(x^3-2x+1)$ is not an integral domain I must find a $p(x)\neq0$ and $q(x)\neq0$ such that $p(x)q(x)=0$ mod r
I also want to show that $x$ mod r is a unit in $E$.
To show that it is not an integral domain, I tried to factor $x^3-2x+1$, but am pretty sure it is irreducible. That means that I need to find a $p(x)$ and $q(x)$ such that $p(x)q(x)$=$r*f(x)$ for some $f(x) \in \mathbb{Z}[x]$. I am having trouble with this.
To show that $x$ mod r is a unit in $E$, I need to find some $g(x) \in \mathbb{Z}[x]$ such that $x*g(x)=0$ mod r, or in other words, such that $x*g(x)= r*h(x)$ for some $h(x) \in \mathbb{Z}[x]$
Is this all correct? I could really use a hand with this, i'm really scrubbing it up at the moment!!
(1) The rational roots theorem tells you to test two possible roots for $f(x)$. One is indeed a root, which you can use to get a factorization of $f(x)$.
(2) No, the equation $xg(x)=0$ would say that $x$ is a zero divisor (if $g\ne0$). To say $x$ is a unit, you need to write down the equation $xg(x)=1$. Hint: factor $x$ out of the nonconstant terms of $f(x)$.