Ring $\frac{\mathbb{Z}}{ \mathbb{6Z}}[x] / \langle 2x+4 \rangle\cong\ ?$ TRUE/FALSE:  Ring $\frac{\mathbb{Z}}{ \mathbb{6Z}}[x] / \langle 2x+4 \rangle$ has infinitely many elements?
To find this ring we have $\frac{\mathbb{Z}}{ \mathbb{6Z}}[x]$ and put $x=-2=4$ in the elements of $\frac{\mathbb{Z}}{ \mathbb{6Z}}[x]$. From here I do not see infinitely many elements. So it must be false , right?
What is this ring isomorphic to?
 A: You don’t put “$x = -2 = 4$”. I know what you’re getting at, but this a dangerous heuristic. It’s still useful and good, of course. But apply it with care. Let’s unwrap it.
It think you imagine $\frac ℤ {6ℤ}[x] / ⟨2x + 4⟩$ to be “the ring $\tfrac ℤ {6ℤ}[x]$ where $2x + 4$ is artificially made zero”. So in this ring, you have “$2x + 4 = 0$”, hence “$2x = -4$”. I suspect that you now conclude “$x = - 2$” and then “since $6 = 0$ in $\tfrac ℤ {6ℤ}$, also $-2 = 4$”. (The latter is fine.) But you can not conclude “$x = -2$” because you cannot cancel by $2$ in $\tfrac ℤ {6ℤ}[x] / ⟨2x + 4⟩$ because $2$ is a zero divisor in $\tfrac ℤ {6ℤ}$.
Yes $2x = -4$ for $x = -2$, but also for $x = -2 + 3x$ and $x = -2 + 3x^2$ and so on … (And here you can already see that you might get infinite elements …)
So, the formal answer is what Hagen already hinted at: The elements $x, x^2, x^3, …$ are all pairwise different in $\tfrac ℤ {6ℤ}[x] / ⟨2x + 4⟩$. To see this, note that $x^m = x^n$ in $\tfrac ℤ {6ℤ}[x] / ⟨2x + 4⟩$ if and only if $x^m - x^n ∈ ⟨2x + 4⟩$ in $\tfrac ℤ {6ℤ}[x]$. Now you can compare coefficients with multiples of $2x + 4$ …
