I am trying to understand ring of fractions and I have this question:
From ring $\Bbb Z/9\Bbb Z$ determine the different rings of fractions that can be obtained according to the choice of $D$.
Dummit 7.5: Theorem 15. Let $R$ be a commutative ring. Let $D$ be any nonempty subset of $R$ that does not contain $0$, does not contain any zero divisors and is closed under multiplication. Then there is a commutative ring $Q$ with $1$ such that $Q$ contains $R$ as a subring and every element of $D$ is a unit in $Q$. (The ring $Q$ is called the ring of fractions of $D$ with respect to $R$ and is denoted $D^{-1}R$.)
So, the first thing I did was to find the subset $D$. I consider there are four subsets with this characteristics: $D_1 =$ {1}, $D_2 =$ {1, 8}, $D_3 =$ {1, 4, 7}, $D_4 =$ {1, 2, 4, 5, 7, 8}.
Then, I wrote the elements of $D_1^{-1}R$, nine elements, so I claim is isomorphic to $\Bbb Z/9\Bbb Z$. The same way, $D_2^{-1}R$, nine different elements again and this ring would be isomorphic to $\Bbb Z/9\Bbb Z$.
I am not quite sure of my procedure, feels like I'm doing something wrong. I would really appreciate your help to guide me on the right way, or any advice to find those different ring of fractions.
Best Answer
To form the ring of fractions of a commutative ring, you use its set of regular elements $S$.
We always have the image of $R$ in $RS^{-1}$, the elements of the form $r/1$.
For each unit $u\in R\subseteq S$, consider an element like $r/u\in RS^{-1}$. Using definitions it's easy to verify that $r/u= ru^{-1}/1$, and since $ru^{-1}\in R$ we see that every fraction of the form $r/u$ is nothing new, it's just something in (the image of) $R$.
So for a commutative ring whose regular elements are exactly the units, localization doesn't produce anything new.
This includes not only rings like $\mathbb Z/n\mathbb Z$ with $n>1$, but also every commutative Artinian ring, and also some broader classes are included.