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?

Your ring is $\mathbb{Z}[x]/(2x -3) = \mathbb{Z}[x-1]/(2(x-1)-1)$.  Now in general $A[x]/( a x -1) \simeq A[1/a]$.
