I have a question which I suspect has the same underlying problem as my question on ring adjuncts yesterday.
Artin asks us to consider adjoining $1/2$ to $R=\mathbb{Z}/12\mathbb{Z}$. So I construct $R[1/2]\cong R[x]/(2x-1)$. For any $f\in R[x]$ we consider $f(x)=(2x-1)g(x)+r(x)$ whose residue in $R[x]/(2x-1)$ is just $r(x)$. $r$ must be a constant polynomial, since its degree must be less than $2x-1$, so it seems like $R[x]/(2x-1)$ is simply isomorphic to $\left\{r(x)=c|c\in R\right\}$ which in turn is isomorphic to $R$.
It seems to me that I've constructed an isomorphism $R\cong R[x]/(2x-1)$, which should indicate that $R\cong R[1/2]$. If I consider $R[1/2]=\left\{\sum r_i (1/2)^i|r_i\in R\right\}$
this isomorphism is patently false, so I think I'm doing something wrong.
Unlike the question I asked yesterday, $(2x-1)$ is certainly in $R[x]$
Best Answer
HINT $\ 12 = 0\ \Rightarrow\ 0 = 12\:(1/2)^2 = 3\:.\ $ Now you can apply the division algorithm since $\:2\:x-1\: =\: -(x+1)\:$ has unit leading coef. Conclude $R[1/2]\cong (\mathbb Z/3)[x]/(x+1) \cong \mathbb Z/3\:.\:$
Alternatively compute the kernel of the evaluation map from $\mathbb Z[x]$ as $(12,2x-1) = (3,x+1)\:$ hence $R[1/2] \cong \mathbb Z[x]/(3,x+1)\cong \mathbb Z/3\:.$
REMARK $\ $ In the same way one can construct fields of fractions and localizations presented as quotient of (multivariate) polynomial rings, i.e. to force $s$ to be a unit we pass to $R[x]/(s\:x-1)$ and repeat this for all elements to be forced to units. As above, when $R$ isn't a domain, this may force some elements of $R$ to become $0$, i.e. the image of $R$ in the constructed ring generally is not an embedding. See my post here for references.