# Complete Ring of Fractions of $\mathbb{Z}_{n}$ for any $n$

Recall first the following two definitions:

• Definition 1: Let $R$ be a commutative ring and $D$ a nonempty multiplicative sub-semigroup of $R$. The ring of fractions over $R$ with the set of denominators $D$ is a commutative ring with identity and is called the ring of fractions over $R$ with the set of denominators $D$. Notation: $D^{-1}R$
• Definition 2: If $D$ is the set of all non-zero elements of $R$ that are not zero divisors, then $D^{-1}R$ is referred to as the complete ring of fractions over $R$.

For the set $\mathbb{Z}_{n} = \{0,1,\dots, n-1\}$, the set of all nonzero elements of $\mathbb{Z}_{n}$ that are not zero divisors is the set $D = \{ m \in \mathbb{Z}_{n} \vert \gcd(m,n)=1\}$. So, $\forall d \in D$, $m \in \mathbb{Z}_{n}$, it seems to me that, according to the definition given above, the complete ring of fractions of $\mathbb{Z}_{n}$ for any $n$ is $$D^{-1}\mathbb{Z}_{n}=\left \{ \frac{m}{d}\, \;\middle|\; \gcd(d,n)=1\right \}$$

However, it just doesn't seem like it should be this easy. Am I wrong or not? Thanks for your time and patience!

Note: This is NOT a duplicate of that other question, because I am specifically asking whether my particular method works, I don't even know what a class module is, and that question was closed for being off-topic because it had no context.

• Hint $\ m/d = md^{-1}\,$ already lies on $\,\Bbb Z_n\$ Have you studied rings of fractions or localizations in general, e.g. do you know their universal properties? (update: see T. Gunn's edit) – Bill Dubuque Jul 7 '17 at 21:57
• @BillDubuque There is a section in my notes called "The Universal Property of Rings of Fractions" which consists of several theorems regarding commutative diagrams. The word "localization" is never used, but it looks as though one of the results is saying that $\phi: R \to D^{-1}R$, which it describes as "the canonical homomorphism of $R$ to the ring of fractions $D^{-1}R$" is invertible, which would then make it an isomorphism. – ALannister Jul 7 '17 at 22:18
• – MatheinBoulomenos Jul 8 '17 at 1:11
• @ALannister "class module $n$" is a bad translation of "equivalence classes modulo $n$" – Trevor Gunn Jul 8 '17 at 3:54
• @MatheiBoulomenos have you actually seen that post? It was closed as off-topic and the person who posted it gave absolutely zero context or work of his/her own. I am specifically asking whether the method by which I chose to approach this problem is correct. – ALannister Jul 8 '17 at 21:28

In $\mathbf{Z}_n$ and in general, in any finite ring $R$, an element $x \in R$ is either a zero divisor or a unit. So the complete ring of fractions of $R$ is $R$ when $|R| < \infty$. To see this, note that if $x$ is not a zero divisor then the map $y \mapsto xy$ is injective and hence surjective since $R$ is finite. So there is some $y$ such that $xy = 1$.
Edit: The universal property of localization says that if $S \subseteq R$ is a multiplicative set and $f : R \to R'$ is a homomorphism such that $f(s)$ is invertible for all $s \in S$ then $f$ factors through $S^{-1}R$: $$R \to S^{-1}R \xrightarrow{\tilde f} R'.$$ If $S \subseteq R^\times$ is a multiplicative group of units then apply the universal property to the identity map. Since $\operatorname{id}(s) = s$ is invertible in $R$, the map $\operatorname{id} : R \to R$ factors through $S^{-1}R$ as $$R \to S^{-1}R \to R.$$ It follows that $R \cong S^{-1}R$. The homomorphisms are $\frac{x}{s} \mapsto s^{-1}x$ and $x \mapsto \frac{x}{1}$.
• I'm not sure I understand what you're trying to say...or at least, it doesn't seem to go with Definition 2 I posted in my original question. How can the complete ring of fractions of $R$ be $R$ when the elements of $R$ themselves aren't even necessarily fractions? – ALannister Jul 7 '17 at 21:47
• I thought the ring of fractions consisted of all possible combinations of numerators (which come from the ring itself) and denominators, which I am here calling $D$. – ALannister Jul 7 '17 at 21:49
• @ALannister The complete (total) ring of fractions is determined up to isomorphism. The embedding $R\to D^{-1}R$ turns out to be an isomorphism, when $R$ is finite. – egreg Jul 7 '17 at 21:50
• If $x$ is already invertible in $R$ then $R[\frac{1}{x}] \cong R$ because $\frac{1}{x} = x^{-1}$. Similarly, if $D$ is the set of units in $R$ then $D^{-1}R \cong R$ via the map $\frac{x}{y} \mapsto xy^{-1}$. – Trevor Gunn Jul 7 '17 at 21:50
• @T.Gunn I might ask this as a separate question, but what if we were talking about $\mathbb{Z}_{p}$, which is not the p-adic integers, but the set $\mathbb{Z}_{n}$ in the case where $n$ is some prime $p$? Or if $n = p^{2}$? – ALannister Jul 7 '17 at 22:20