Describe the quotient ring $F[x]/(x^3-x-1)$, where $F[x]=\mathbb{Z}/3\mathbb{Z}$. How many elements are in the ring? Is it an integral domain? a field? Which primitive roots of unity does it contain?
The quotient ring $F[x]/(x^3-x-1) \cong \{a+b\alpha+c\alpha^2: a,b,c\in\mathbb{Z}/3\mathbb{Z}, \alpha^3-\alpha-1=0 \}.$
$(i)$ I believe that $\left|\mathbb{Z}/3\mathbb{Z}\right|^{\deg (x^3-x-1)}=3\times 3\times 3=27$ elements in $R$.
$(ii)$ Multiplication is defined as \begin{align*} (a+b\alpha+c\alpha^2)(a'+b'\alpha+c'\alpha^2)&=aa'+(ab'+ba')\alpha+(ac'+bb'+ca')\alpha^2+(bc'+cb')\alpha^3+cc'\alpha^4\\ &=aa'+(ab'+ba')\alpha+(ac'+bb'+ca')\alpha^2+(bc'+cb')(\alpha+1)+cc'\alpha(\alpha+1)\\ &=aa'+cb'+bc'+(ab'+ba'+bc'+cc'+cb')\alpha+(ac'+bb'+ca'+cc')\alpha^2. \end{align*} By properties of addition of functions, we clearly have associativity, commutativity, and distributivity.
The ring clearly contains the additive identity 0 and the multiplicative identity 1.
For any $f=a+b\alpha+c\alpha^2\in R$, we clearly have the additive inverse of $f$ in $R$, where $$f^{-1}=(3-a)\mod 3+\alpha(3-b)\mod 3+\alpha^2(3-c)\mod 3.$$
For any $f$, we have a multiplicative inverse of $f$ in $R$. I confirmed this for all $27-1=26$ nonzero elements. I was wondering if there is a better approach to this.
Thus, we have a field.
Since $R$ is a field, it is also an integral domain. I found it difficult to show that the multiplication of any nonzero elements is nonzero.
I also am not sure what it means to be a primitive root of unity in this field and would like some clarification on these points.