Localization of $\mathbb{Z}_6$ with respect to the powers of $2$

Question

I want to localize $\mathbb Z_6$ with respect to the powers of $2$. Now if $\frac{2}{2} = \frac{a}{b}$, by $c(2a-2b)= 0$ and letting $c=3$ we conclude if $a \neq 0$ then $\frac{a}{b} =1$. Thus $(\mathbb Z_6)_2 \cong \mathbb Z_2$. But my book says $(\mathbb Z_6[X])_2 \cong \mathbb Z_3[X]$.

Answer 1

The localization certainly cannot be $\mathbb{Z}_2$, because $[1]/[1]\ne[2]/[1]$ (where $[x]$ denotes the residue class of $x$ modulo $6$). Indeed, if they were equal, there would exist $n$ such that $$ [2]^n([2]-[1])=0 $$ but $[2]$ is not nilpotent in $\mathbb{Z}_6$.

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On the other hand, $[3]/[1]=[6]/[2]=[0]/[1]$. Now note that the canonical map $$ \mathbb{Z}_6\to \mathbb{Z}_6/[3]\mathbb{Z}_6\cong\mathbb{Z}_3 $$ satisfies the universal property of the localization.