Determining isomorphism of ring of fractions/quotients In a homework problem, I was asked to show that if $R=\mathbb{Z}_6$ and $S=\lbrace 2,4 \rbrace$, then $S^{-1}R\cong\mathbb{Z}_3$.
I was able to determine which fractions are equivalent and used that fact to develop the following function $f:\mathbb{Z}_3\to S^{-1}R$, which I believe is an isomorphism:
\begin{align*}
f(0) &= 0/2 = 0/4 \\
f(1) &= 2/2 = 1/4 \\
f(2) &= 1/2 = 2/4
\end{align*}
However, this manual rule of assignment is kind of awkward to work with - I'm not even sure what is required to show that $f$ is a ring isomorphism.
1) If this "manual" rule of assignment is really the best way to go, what do I need to do to show that it is an isomorphism?
2) If this is not the best way to go about showing that $S^{-1}R\cong\mathbb{Z}_3$, what is? I'd prefer a hint to an outright answer.
3) I know that when $S'$ is the set of all nonzero elements of $R'$ which are not zero divisors, then there is a universal property for $S'^{-1}R'$ which dictates the existence of a homomorphism into commutative unital rings (under certain conditions). Is there a general way to create homomorphism, or even better, determine a "familiar" ring to which a ring of quotients is isomorphic, if the $S'^{-1}R'$ is not the complete ring of quotients?
 A: In this answer I show that the map $R \to S^{-1}R, r \mapsto \frac{r}{1}$ is always surjective for a finite ring $R$, hence you can compute $S^{-1}R$ by computing the kernel. In your case, you can just go through all six elements and check whether they are mapped to zero. You are correct that the answer is $\mathbb Z_3$.
A: Hint: Use the Chinese Remainder Theorem.  What happens if you localize $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ at $S$?
Full solution:

By the Chinese Remainder Theorem, we have $\mathbb{Z}/6 \mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$.
Since $2 = 0$ in $\mathbb{Z}/2\mathbb{Z}$, localizing at $S$ produces the zero ring, and since $2$ is already a unit in $\mathbb{Z}/3\mathbb{Z}$, then localizing at $S$ leaves it unchanged. Thus
$$S^{-1}(\mathbb{Z}/6 \mathbb{Z}) \cong S^{-1}(\mathbb{Z}/2\mathbb{Z}) \times S^{-1}(\mathbb{Z}/3\mathbb{Z}) \cong 0 \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \, .$$

$$
%(a,b) \mapsto -2b + 3a \qquad (0,1) \mapsto -2 = 4 \qquad (0,2) \mapsto -4 = 2
$$
