Consider the ideal $I=(1+2x)\cdot \Bbb Z[x]$ in the polynomial ring $\Bbb Z[x]$. I am trying to show that $\Bbb Z[x]/I$ is isomorphic to $R=\{\frac{a}{2^r}:a\in \Bbb Z, r\in \Bbb N_0\}$.
My approach:
Define a ring homomorphism $f:\Bbb Z[x]\to R$ by $X^r\mapsto (-1)^r \frac{1}{2^{r+1}}$ for $r\in \Bbb N$ and $X^0\mapsto 1$. Then this descends to a homorphism on $\Bbb Z[x]/I$ since $f(1+2x)= 1- 2\cdot \frac{1}{2}=0$.
An inverse to the descent of $f$ is given by $g:R\to \Bbb Z[x]/I$ with $g(\frac{1}{2^r})=(-1)^{r-1}x^{r-1}$
Is this a valid proof?
The idea is good. What you have to show is
Surjectivity should be easy enough. Since the map $f$ satisfies $f(X)=-\frac{1}{2}$, it is clear that $1+2x\in\ker f$, so $\ker f\supseteq I$. In order to show the converse, suppose $p(x)\in\ker f$. Then write $$ p(x)=(1+2x)q(x)+r $$ in $\mathbb{Q}[x]$, with $r$ a constant. Then $p(-1/2)=0$ implies $r=0$. Now you just have to prove that $q$ has integer coefficients.