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I am studying commutative algebra and I have a problem in calculating the fraction rings. For example, it would be a great help someone help me to calculate the ring of fraction if $ S= \{1,3,5\} $ is a multiplicative subset of $R=\mathbb{Z}_6$.

I know the definition of the ring of fractions, but I can not calculate it. Thanks for any help.

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  • $\begingroup$ Write all of the fractions that use one element of $S$ in the denominator. Note that $0 \in $\mathbb{Z}}_{6}$. $\endgroup$ – Jay Feb 18 at 16:06
  • $\begingroup$ Thanks for your help $\endgroup$ – f.j1995 Feb 18 at 16:22
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Well, it's a ring of fractions, so the elements are just that: fractions. The possible fractions are $$ \frac01, \frac11, \frac21, \frac31, \frac41,\frac51,\frac61,\\ \frac03, \frac13, \frac23, \frac33, \frac43,\frac53,\frac63,\\ \frac05, \frac15, \frac25, \frac35, \frac45,\frac55,\frac65 $$ Just like the fractions you're used to (rational numbers), some of these are the same. We are allowed to expand fractions using any element of $S$ and it won't change the values of the fractions. For instance, just like you would expect, we have $\frac01 = \frac03 = \frac05$. But what you might not expect is that $\frac23 = \frac{2\cdot 3}{3\cdot 3} = \frac03$. In other words, any fraction with an even numerator is equal to $\frac03$.

In fact, we can expand any fraction so that it has $3$ in the denominator, and what we end up with then are just $$ \frac03 = \frac23 = \frac43\\ \frac13 = \frac33 = \frac53 $$ So in the end we have only $2$ distinct elements in the final fraction ring. It's not difficult to see that it is isomorphic to $\Bbb Z_2$.

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  • $\begingroup$ Thank you, it clearly shows everything for me. $\endgroup$ – f.j1995 Feb 18 at 16:22
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    $\begingroup$ @f.j1995 This whole "$(a, b)$ is equivalent to $(c, d)$ if there is an $s\in S$ such that $s(ad-cb) = 0$" thing in the definition (which is usually how it's phrased) is just a fancy (and rigorous) way of saying "fractions are considered to be equal if you can expand them both to be the same fraction". If $S$ has zero-divisors, then it may be necessary to expand both of them before two equal fractions actually have the same numerator and denominator. But if $S$ has no zero-divisors you can always expand one to be the same as the other. $\endgroup$ – Arthur Feb 18 at 16:26
  • $\begingroup$ If it is possible for you, tell me what you mean by " to expand both of them before two equal fractions actually have the same numertor and denumerator"? $\endgroup$ – f.j1995 Feb 18 at 16:40
  • $\begingroup$ Thanks for your great help $\endgroup$ – f.j1995 Feb 18 at 16:41
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    $\begingroup$ @f.j1995 Note that you also need to prove that the ring is not the trivial one-element ring, i.e. said two elements really are distinct. $\endgroup$ – Bill Dubuque Feb 18 at 16:44

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