Isomorphism between $\Bbb Z[X,Y ]/(2Y−1)$ and $\Bbb Z[\frac{1}{2}][X]$ [closed]

This is a little tricky to prove, so I'll give you a few hints and ideas. Let $A=\Bbb Z[X]$.

Define the map $$\phi:A[Y]\rightarrow A[1/2]\\\phi(f(Y))=f(1/2)$$

Then you need so show that $\ker(\phi)=(2Y-1)$. By the first isomorphism theorem you're then done.

The inclusion $(2Y-1)\subseteq\ker(\phi)$ is OK, for the other direction suppose that $f(Y)\in\ker(\phi)$. Then find $n$ such that $2^nf(Y)=g(Z)$, where $Z=2Y$. You then have that $g(1)=0$. Now use the factor theorem.

Example Suppose that $f(Y)=4Y^4-4Y^3+21Y^2-20Y+5$, we check that $f(1/2)=0$ is true. Now compute $2^4f(Y)$, which yields $$2^4f(Y)=4(2Y)^4-4\cdot 2(2Y)^3+21\cdot 2^2 (2Y)^2-20\cdot 2^3(2Y)+5\cdot 2^4$$ Letting $Z=2Y$, we arrive at $$g(Z)=4Z^4-8Z^3+84Z^2-160Z+80$$ since $Y=1/2$ is a root of $f$, we know that $Z=1$ is a root of $g$. Therefore $g(Z)=(Z-1)h(Z)$, and so $$2^4f(Y)=(2Y-1)h(2Y)$$ you then only have to conclude that $\ker(\phi)\subseteq (2Y-1)$.