Cardinality of a ring obtained by quotienting $\Bbb Z[x]$ Let $R$ be the ring $\Bbb Z[x]/((x^2+x+1)(x^3 +x+1))$ and I be the ideal generated by $2$ in $R$. What is the cardinality of the ring $R/I$?


*

*27

*32

*64

*infinite  
Now I was thinking $R$ could be written as $(\mathbb Z[x]/(x^2 +x+1)/(x^3 + x+1) = \Bbb Z[i]/(x^3+x+1)$. Now if I quotient  it  with  the $2$ of this ring, $2$ is not irreducible anymore. $2=(1+i)(1-i)$. So I don't think I can write it as $\Bbb Z_2[i]/(x^3+x+1)$.  So what is going wrong? 
P.S. Thanks  for all  your  help . It is done now.I have  another  doubt  here. The  ideal  we  quotiented  the  ring $Z_2[x]  by  is  not  generated  by  an  irreducible  polynomial .  So,can  the  generator being  reducible or  irreducible make  any  difference  or not? 
 A: $$R/2R\simeq\mathbf Z/2\mathbf Z[x]/\bigl((x^2+x+1)(x^3+x+1)\bigr)$$
Now in $\mathbf Z/2\mathbf Z[x]$, $x^2+x+1$ and $x^3+x+1$ are coprime, hence by the Chinese Remainder Theorem, $R/2R$ is the product of two fields:
$$R/2R\simeq \mathbf Z/2\mathbf Z[x]/\bigl(x^2+x+1)\times\mathbf Z/2\mathbf Z[x]/(x^3+x+1)\simeq \mathbf F_4\times \mathbf F_8. $$
The factors have $4$ and $8$  elements, respectively.
A: Hint: The order in which you quotient does not matter. So first consider $\mathbb{Z}[x]/(I)$, where $I$ is defined above. What does this give you?

 $$ \mathbb{Z}[x]/(I)=\mathbb{F}_2[x]$$ where $\mathbb{F}_2$ is the finite field of two elements.

Now quotient off by the ideal $((x^2+x+1)(x^3+x+1))$. Let $S$ be the ring obtained. Note that $(x^2+x+1)(x^3+x+1)$ has degree $5$, so every polynomial in $((x^2+x+1)(x^3+x+1))$ has degree $\geq 5$. Thus in $S$, we have some relation that allows us to write 
$$
\overline{x}^5=p(\overline{x}),
$$
where $p(\overline{x})$ is a polynomial of degree $\leq 4$ and $\overline{x}$ is the residue of $x$. Therefore every polynomial in $S$ has degree at most $4$, at $S$ has scalars from $\mathbb{F}_2$. Can you finish? 
