Let $R=\mathbb{Z}[\sqrt{-3}]$ be a ring with the maximal ideal $(2,1+\sqrt{-3})$. I am trying to prove that $$R_M\cong \mathbb{Z}_{(2)}[\pi|\pi^2=2\pi-4]$$ where $R_M$ is the localization $R_M=(R-M)^{-1}R$
I found out that $\pi = 1+\sqrt{-3}$ so the now I need to prove: $$R_M\cong \mathbb{Z}_{(2)}[1+\sqrt{-3}]$$ Can someone give me a direction?
Note that $\mathbb{Z}_{(2)}[1+\sqrt{-3}]=\mathbb{Z}_{(2)}[\sqrt{-3}]$ is a localization of $R$, namely the localization with respect to the set $\mathbb{Z}-(2)$. Since $\mathbb{Z}-(2)\subseteq R-M$ (proof: the quotient $R/M$ is a field of characteristic $2$ so every odd integer is nonzero in it), it suffices to show that inverting each element of $\mathbb{Z}-(2)$ will invert every element of $R-M$, so that these two localizations have equivalent universal properties. Now if $a+b\sqrt{-3}\in R-M$, note that $a$ and $b$ must have different parity (if they were both even then $a+b\sqrt{-3}$ would be a multiple of $2$, and if they were both odd then $a+b\sqrt{-3}$ would be a multiple of $2$ plus $1+\sqrt{-3}$). It follows that $(a+b\sqrt{-3})(a-b\sqrt{-3})=a^2+3b^2$ is an odd integer. That is, $a^2+3b^2$ is inverted in the localization with respect to $\mathbb{Z}-(2)$, and thus so is $a+b\sqrt{-3}$ since any factor of a unit is a unit.