# Ring of fractions of $\mathbb Z_6$

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.

• Write all of the fractions that use one element of $S$ in the denominator. Note that $0 \in$\mathbb{Z}}_{6}$. – Jay Feb 18 at 16:06 • Thanks for your help – f.j1995 Feb 18 at 16:22 ## 1 Answer 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 $$\Bbb Z_2$$. • Thank you, it clearly shows everything for me. – f.j1995 Feb 18 at 16:22 • @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. – Arthur Feb 18 at 16:26
• 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"? – f.j1995 Feb 18 at 16:40
• Thanks for your great help – f.j1995 Feb 18 at 16:41
• @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. – Bill Dubuque Feb 18 at 16:44