Is $\Bbb Z[x]/(2x-3)\cong\Bbb Z$? Consider the following problem.
I know that $\mathbb{Q}[x]/ (2x-3) $ is isomorphic to $\mathbb{Q}$. Use the evaluation map $x \to \frac{3}{2}$.
But why $\Bbb Z[x]/(2x-3)\cong\Bbb Z[1/2]$? Can I use the same evaluation map here and get $\Bbb Z[x]/(2x-3)\cong\Bbb Z$?
Please help me?

Well, how do you use the evaluation map in the rational case? You consider the map
$$
e_{3/2}: \mathbb{Q}[x] \to \mathbb{Q} 
$$
given by $e_{3/2}(f) = f(3/2)$. Now, you notice that $e_{3/2}$ is surjective (why?), that $\ker e_{3/2} = (2x - 3)$ and you use the first isomorphism theorem to conclude that
$$
\frac{\mathbb{Q}[x]}{(2x - 3)} \simeq \mathbb{Q}.
$$
Now, consider the same map with domain $\mathbb{Z}[x]$. What can you say about its image ? Is it well defined as a function with co-domain $\mathbb{Z}$? Can you apply the same argument?
