# Examples of rings of fractions

I wanted to come up with a few examples of rings of fractions $S^{-1}R$. Can you tell me if these are correct:

1.Let $R = \mathbb Z$, $S = (2 \mathbb Z \setminus \{0\}) \cup \{1\}$. Then every $[x] = \frac{r}{s} \in S^{-1}R$ consists of the elements: $[x] = \{ 2x, \frac12 x\}$. The ring homomorphism $f: R \to S^{-1}R$ is injective since if $f(r) = [\frac{r}{1}] = [\frac{r^\prime}{1}] = f(r^\prime)$ we have $2k r = 2k r^\prime$ for $2k \in \mathbb Z$ and since $\mathbb Z$ is an integral domain, $r=r^{\prime}$.

2.Now let $R=\mathbb Z / 12 \mathbb Z$ and $S = \{1,2,4,6,8,10\} = (2 R \setminus \{0\}) \cup \{1\}$. Then $f: R \to S^{-1}R$ is not injective since $f(6) = f(3)$.

Are there more interesting examples where $f: R \to S^{-1}R$ is not injective?

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Note: $R$ is a commutative unital ring and $S$ is a multiplicative subset of $R$. –  Rudy the Reindeer Jun 12 '12 at 11:51
I believe that you have non-injectivity when the ring in question has zero divisors. –  user38268 Jun 12 '12 at 11:55
@BenjaminLim Thank you, that's right. In fact, we have $$f(r) = \frac{r}{1} = 0 \iff \exists s \in S: sr = 0$$ –  Rudy the Reindeer Jun 12 '12 at 19:09
@BenjaminLim: Not necessarily; e.g., if $S$ consists only of elements that are not zero divisors. Since $\frac{as}{s} = \frac{0}{s}$ holds if and only if there exists $s'\in S$ such that $s'(sas - 0s)=0$, if neither $s$ nor $s'$ are zero divisors, then $sas$ cannot be zero, and therefore $s'(sas)$ cannot be zero. So even if $R$ has zero divisors, you still may get injectivity. E.g., taking $S$ to be the set of all nonzero divisors gives you the total ring of fractions of $R$, and every commutative ring embeds into its total ring of fractions. –  Arturo Magidin Jun 12 '12 at 19:09
@ArturoMagidin Thank you for this comment! –  Rudy the Reindeer Jun 13 '12 at 6:49

It's difficult to find a place to begin, because there are so many examples. For $R=\mathbb{Z}$,

1. there is of course the choice $S=R\setminus\{0\}$, producing $\mathbb{Q}$.

2. You could also just choose a prime ideal, say $P=2\mathbb{Z}$, and use $S=R\setminus P$. This yields the subring of $\mathbb{Q}$ consisting of all fractions of odd denominator.

3. You could fix a nonzero element, say $a\in R$ and use $S=\{a^i\mid i\in\mathbb{Z}^+\}$. This produces the subring of $\mathbb{Q}$ whose denominators are a power of $a$.

4. To break from $\mathbb{Z}$ for a second, I'd like to apply the last point to $R=F[x]$ and choose $a=x$. The ring of fractions based on the powers of $x$ then yield $F[x,x^{-1}]$, the Laurent polynomials.

In your second example you put $0$ in your multiplicative set... are you aware that will make $S^{-1}R=\{0\}$?

If the map is not injective, and you exclude $0$ from $S$, then there must be a nonzero zero divisor in $S$. If $a\neq b$ in $R$, but $(a,1)=(b,1)$ in $S^{-1}R$, then that says there exists $s\in S$ such that $as=bs$. This means $(a-b)s=0$.

Using that idea, let $as=0$ for some nonzero $a\in R$, nonzero $s\in S$. Let $b$ be anything nonzero in $R$. Note that $0=as=(a+b-b)s=(a+b)s-bs$. Then $(a+b,1)=(b,1)$.

A simple way to produce a nontrivial ring of fractions such that the canonical map is not an injection would be to find a nonzero zero divisor $x$ in a ring, such that $x$ is not nilpotent. Then localize at the nonnegative powers of $x$.

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For the first of your examples, I don't think that $S^{-1}R$ consists of just those two elements. For example you have $\frac{13}{4}$ in the localisation that is not of the form $2x$ or $\frac{x}{2}$. On second thoughts I don't get what you mean by $[x]$ is just those two elements. Do you mean to talk about the equivalence class of a fraction? In that case there are infinitely many elements in an equivalence class. For example

$$\frac{13}{4} = \frac{26}{8} = \frac{39}{12} = \ldots$$

As can be seen $\frac{13}{4}$ is not even of the form you claimed that an element is in an equivalence class. Other important examples of localisation:

1. Let $P$ be a prime ideal in a ring $R$. Then $S = R - P$ is a multiplicative set, and $S^{-1}R$ is a local ring. We call this "localisation at the prime ideal $P$". For example if $P$ is all the multiples of $5$ in $\Bbb{Z}$, then

$$R_P =\{q \in \Bbb{Q} | q = \frac{a}{b} \hspace{1mm} \text{with} \hspace{1mm} a,b\in \Bbb{Z}\}$$

1. Let $f$ be a non-zero non-nilpotent element. Then consider $S = \{1,f,f^2,\ldots \}$. When localising by this multiplicative subset, we sometimes write $R_f$. This example crops up a lot in algebraic geometry I believe.

2. More advanced example: Let $R$ be a $\Bbb{Z}$ - graded ring. Let $f$ be a non-zero homogeneous element of degree one. Notice that $R_f$ is naturally a graded ring. Then $R_f \cong (R_f)_0[x,x^{-1}]$ where $x$ is an indeterminate and $(R_f)_0$ we denote by the degree zero component of $R_f$. More generally, let $P$ be a homogeneous prime ideal in $R$ and $U$ the multiplicative subset of all homogeneous elements not in $P$. Then it is surprising that for a more general $U$ we than before we still have

$$U^{-1}R \cong (U^{-1}R)_0[x,x^{-1}]$$

where $(U^{-1}R)_0$ is the degree zero component of $U^{-1}R$.

1. Interesting example. Let $R$ be a PID and $K$ its field of fractions. Let $A$ be a subring of $K$ containing $R$. Then one can show that $A$ is actually isomorphic to $S^{-1}R$ where $S$ is the multiplicative set of all elements in $R$ that are units in $A$. You can use this to prove that $A$ must always be a PID!

By the way in your last example you have included zero in the multiplicative subset $S$. We claim that $S^{-1}R$ is the zero ring. Indeed, any

$$\frac{a}{b} = \frac{0}{1}$$

because there exists $0 \in S$ such that $(a\cdot 1- b\cdot 0)\cdot a = 0$.

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I don't follow what you mean here: if by $\,[x]\,$ you mean the equivalence class of $\,x\,$ in the localization then I can't see how $\,2x\,,\,\frac{1}{2}x\,$ are equivalent in $\,(1)$...?

If you're talking of localization of $\,R\,$ wrt $\,S\,$, then an element of $\,S^{-1}R\,$ is of the form $\,\displaystyle{\frac{r}{s}\,\,,\,r\in R\,,\,s\in S}$, and of course $\,S\,$ cannot contain zero and has to be multiplicatively closed...

So in $\,(1)\,$ , the elements we have there are of the form $$\,\displaystyle{\frac{m}{r}\,,\,m\in\mathbb{Z}\,,\,r\in 2\mathbb{Z}\cup \{1\}}$$ and $$\frac{m}{r}=\frac{n}{s}\Longleftrightarrow ms=nr\,\,\text{in}\,\,R$$

In your second example $\,S\,$ is not valid because it contains zero

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