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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?

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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 $\mathbb Z_3$.

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  • $\begingroup$ Unfortunately, I'm not sure that map helps in this case, as $R=\mathbb{Z}_6$, so the mapping you mention is not a map between the right rings and thus fails to be injective (As $R$ is not isomorphic to $S^{-1}R$) $\endgroup$ – MightyTyGuy Dec 7 '16 at 8:04
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    $\begingroup$ You have not understood, what I was saying: You have a surjective map $R \to S^{-1}R$, hence you have $S^{-1}R \cong R/I$, where $I$ is the kernel. So you just have to compute $I$, which is a gimme in this case. $\endgroup$ – MooS Dec 7 '16 at 8:26
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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 $$S^{-1}(\mathbb{Z}/6 \mathbb{Z}) \cong S^{-1}(\mathbb{Z}/2\mathbb{Z}) \times S^{-1}(\mathbb{Z}/3\mathbb{Z}) \cong 0 \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \, .$$

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

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