Number of elements in the quotient ring $\mathbb{Z}_6 [x]/\langle 2x +4\rangle$ I am confused about this quotient ring.
I know that $\mathbb{Z}_7 [x]/\langle x^2 + 1\rangle =  \{ f(x) + \langle x^2 + 1\rangle \mid f(x) \in \mathbb{Z}_7[x] \}$. Here $x^2 + 1$ is a zero element in $\mathbb{Z}_7 [x]/\langle x^2 + 1\rangle$, so if we replace $x$ by $i$ we get $\mathbb{Z}_7 [x]/\langle x^2 + 1\rangle =\mathbb{Z}_7 (i) $.
I know the distinct element of $ \mathbb{Z}_6 [x]/\langle 2x +4\rangle$ are $\langle 2x +4\rangle,\;x + \langle 2x +4\rangle, \;x^2 + \langle 2x +4\rangle, \ldots$ but I do not  know how to prove this.
Please tell what could we do similar here in case of $ \mathbb{Z}_6 [x]/\langle 2x +4\rangle$? Since $1$ is the zero element of the polynomial $2x +4 $. Why we can not replace here $x$ by $1$?
Any help would be appreciated. Thank you.
 A: Here is an isomorphism using the Chinese Remainder Theorem.
\begin{align*}
\frac{(\mathbb{Z}/6\mathbb{Z})[x]}{(2x+4)} &\cong \frac{\mathbb{Z}[x]}{(6, 2x+4)} = \frac{\mathbb{Z}[x]}{(2)(3, x+2)} \cong \frac{\mathbb{Z}[x]}{(2)} \times \frac{\mathbb{Z}[x]}{(3, x+2)}\\
&\cong (\mathbb{Z}/2\mathbb{Z})[x] \times \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x+2)}  \cong (\mathbb{Z}/2\mathbb{Z})[x] \times \mathbb{Z}/3\mathbb{Z}
\end{align*}
The first isomorphism is the Third Isomorphism Theorem and the second equality follows since the ideals $(6, 2x+4) = (2)(3, x+2)$ are equal in $\mathbb{Z}[x]$.  Since $1 = -2 + 3 \in (2) + (3, x+2)$, then $(2)$ and $(3, x+2)$ are comaximal, so we can apply the Chinese Remainder Theorem.  Finally, the isomorphism $\frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x+2)} \cong \mathbb{Z}/3\mathbb{Z}$ is induced by the evaluation map at $-2$.
A: In principle, problems of this type ar given either over fields, or at least for ideals generated by polynomials with invertible leading coefficient, and they are generally solved by using Euclid's algorithm for polynomial division. Alas, the fact that $2$ is not invertible in $\Bbb Z _6$ makes this approach infeasible.
Note though that if a polynomial $f$ has all the coefficients $0$, $2$ or $4$, then you would write it as $2g$ and you could perform the Euclidean division not of $f$ by $2x + 4$, but rather of $g$ by $x + 2$. This is the idea that we shall follow. Split any polynomial $f \in \Bbb Z _6 [x]$ into a part $f _\text{even}$ with even coefficients and a rest $f_2$with coefficients either $0$ or $1$ (this splitting is unique). Next, there exist $q \in \Bbb Z _6 [x]$ and $r_f \in \Bbb Z _6$ such that $f _\text{even} = (2x+4) q + r_f$. Note that $r_f = f _\text{even} (1) \in \{0,2,4\} \subset \Bbb Z _6$. Since this set is a subring isomorphic to $\Bbb Z _3$, identify $r_f$ with $\widehat {r_f} \in \Bbb Z _3$, its class modulo $3$.
Consider now the morphism $\Bbb Z _6 [x] \to \Bbb Z _2 [x] \times \Bbb Z _3$, given by $f \mapsto (f_2, \widehat {f _\text{even} (1)})$. Its kernel is made of all those polynomials that have no "odd" part and when divided by $2x+4$ produce no remainder, therefore are multiples of $2x+4$. Thus, the kernel is exactly the ideal $(2x+4)$ and by the fundamental isomorphism theorem this gives $$\frac {\Bbb Z _6 [x]} {(2x+4)} \simeq \Bbb Z _2 [x] \times \Bbb Z _3 .$$
In particular, this ring has infinitely many elements.
