I am looking at describing the ring obtained from $\mathbb{Z}/12\mathbb{Z}$ by adjoining the inverse of 2. I know the final answer is supposed to be that it is isomorphic to $\mathbb{Z}$ but here is what I think so far:
We look at ring $(\mathbb{Z}/12\mathbb{Z})[x]/(2x-1)$=$\mathbb{Z}[x]/(12,2x-1)$. We see that 3 is in $(12,2x-1)$ since 3 = $(2x-1)^2 - 12x^2 + 12x$ when we plug in the inverse of 2 for x.
However, I don't see how $(12,2x-1)$ contains $3x-(2x-1) = x+1$ and then how the ideal $(3,x+1)$ contains $12$ and $2x-1=3x-(x+1)$ so then $\mathbb{Z}[x]/(12,2x-1) = \mathbb{Z}[x]/(3,x+1) = \mathbb{Z}/3\mathbb{Z}$
but from this step above I see that $\mathbb{Z}[x]/(x+1)$ is isomorphic to $\mathbb{Z}$.
