Total quotient ring of $\mathbb Z_{2^n}$ I want to characterize the total classical quotient rings of the   (commutative) rings $R=\mathbb Z_4$, $R=\mathbb Z_8$ or any $R=\mathbb Z_{2^n}$. 
In fact, if we get $S$ to be the regular elements of $R$, by the classical quotient ring we mean $RS^{-1}$ (which is the ring of fractions of $R$).
Thanks for any good answer! 
 A: As @user26857 has pointed out in the comments, the total ring of fractions of a finite ring $R$ (commutative and with $1$) is $R$ itself, since the set of non-zero-divisors is precisely the set of units.
We have an even more remarkable result for a finite ring $R$: Let $S \subset R$ be any multiplicative subset. Then $|S^{-1}R| \leq |R|$ and equality holds if and only if $S$ contains no zero-divisors.
For the proof, we show that $R \to S^{-1}R$ is surjective, hence the inequality. The assertion about the equality is then a direct consequence of the well known fact that $R \to S^{-1}R$ injective if and only if $S$ contains no zero-divisors.
To show surjectivity it suffices to give pre-images of $\frac{1}{s}$ for any $s \in S$, i.e. find some $r \in R$ with $\frac{r}{1}=\frac{1}{s}$.
The ring is finite, hence the descending chain of ideals
$$(s) \supset (s^2) \supset (s^3) \supset \dotsc $$
eventually stabilizes, i.e. $(s^n)=(s^{n+1})$ for some $n \in \mathbb N$. Hence there is some $r \in R$ with $s^n=rs^{n+1}$, i.e. $s^n(rs-1)=0$. By the definition of the localization, this yields the desired $\frac{r}{1}=\frac{1}{s}$.

The classical picture of localizations, that we add elements - namely fractions - cannot be maintained for finite rings.

If you happen to be a bit advanced in commutative algebra, you will of course notice that one can immediately generalize this (with exactly the same proof) to any artinian ring, i.e. $R \to S^{-1}R$ is surjective and it is an isomorphism if and only if $S$ contains no zero-divisors.
