## Construct and prove the universal propery of rings of fractions

Construct and prove the universal property of rings of fractions.

Let $R$ be a commutative ring with $1 \neq 0$. A subset $D \subseteq R$ is said to consist of denominators if (1) it is multiplicatively closed, (2) it does not contain 0, and (3) it does not contain any zero divisors. Given a set $D$ of denominators in $R$, there exists a commutative ring with $1 \neq 0$, $Q$, and an injective ring homomorphism $\iota : R \rightarrow Q$ such that if $d \in D$, then $\iota(d)$ is a unit in $Q$. Moreover, $(Q, \iota)$ has the following “universal property”: if $S$ is a commutative ring with $1 \neq 0$ and $\varphi : R \rightarrow S$ is an injective ring homomorphism such that $\varphi(d)$ is a unit in $S$ for all $d \in D$, then there exists a unique injective ring homomorphism $\Phi : Q \rightarrow S$ such that $\Phi \circ \iota = \varphi$.

Proof: Let $R$ be a commutative ring with $1 \neq 0$ and let $D \subseteq R$ be a set of denominators. Define a relation $\sigma$ on $R \times D$ as follows: $(r_1,d_1) \,\sigma\, (r_2,d_2)$ precisely when $r_1d_2 = r_2d_1$.

We claim that $\sigma$ is an equivalence relation.

Proof of claim:

1. ($\sigma$ is reflexive) Let $(r,d) \in R \times D$. Since $rd = rd$, we have $(r,d)\,\sigma\,(r,d)$. Thus $\sigma$ is reflexive.
2. ($\sigma$ is symmetric) Suppose $(r_1,d_1)\,\sigma\,(r_2,d_2)$. Then $r_1d_2 = r_2d_1$, and thus $r_2d_1 = r_1d_2$. Thus $(r_2,d_2)\,\sigma\,(r_1,d_1)$. Thus $\sigma$ is symmetric.
3. ($\sigma$ is transitive) Suppose $(r_1,d_1)\,\sigma\,(r_2,d_2)$ and $(r_2,d_2)\,\sigma\,(r_3,d_3)$. Then we have $r_1d_2 = r_2d_1$ and $r_2d_3 = r_3d_2$. Then $r_1d_2d_3 = r_2d_3d_1$ $= r_3d_2d_1$. Since $d_2$ is not a zero divisor in $R$, we have $r_1d_3 = r_3d_1$, so that $(r_1,d_1)\,\sigma\,(r_3,d_3)$. Thus $\sigma$ is transitive.

Thus $\sigma$ is an equivalence relation. We let $D^{-1}R = (R \times D)/\sigma$, and we denote the equivalence class $[(r,d)]_\sigma$ by $\frac{r}{d}$.

We now claim that for all $r \in R$ and all $d,d^\prime \in D$, $\frac{rd^\prime}{dd^\prime} = \frac{r}{d}$. To see this, note that $rd^\prime d = rdd^\prime$, so that $(rd^\prime,dd^\prime)\,\sigma\,(r,d)$.

We now define operators on $D^{-1}R$ as follows: $\frac{r_1}{d_1} + \frac{r_2}{d_2} = \frac{r_1d_2 + r_2d_1}{d_1d_2}$ and $\frac{r_1}{d_1} \cdot \frac{r_2}{d_2} = \frac{r_1r_2}{d_1d_2}$. We claim that $(D^{-1},+,\cdot)$ is a ring.

Proof of claim:

1. ($+$ is well-defined) Suppose $\frac{r_1}{d_1}$, $\frac{r_2}{d_2}$, $\frac{r_1^\prime}{d_1^\prime}$, $\frac{r_2^\prime}{d_2^\prime} \in D^{-1}R$ with $\frac{r_1}{d_1} = \frac{r_1^\prime}{d_1^\prime}$ and $\frac{r_2}{d_2} = \frac{r_2^\prime}{d_2^\prime}$. Then we have $r_1d_1^\prime = r_1^\prime d_1$ and $r_2d_2^\prime = r_2^\prime d_2$. Thus $r_1d_1^\prime d_2 d_2^\prime = r_1^\prime d_1d_2d_2^\prime$ and $a_2d_2^\prime d_1d_1^\prime = r_2^\prime d_2 d_1 d_1^\prime$, and thus $r_1 d_2 d_1^\prime d_2^\prime + r_2 d_1 d_1^\prime d_2^\prime = r_1^\prime d_2^\prime d_1 d_2 + r_2^\prime d_1^\prime d_1 d_2$. Thus $(r_1 d_2 + r_2 d_1)d_1^\prime d_2^\prime = (r_1^\prime d_2^\prime + r_2^\prime d_1^\prime)d_1d_2$. Thus $\frac{r_1 d_2 + r_2 d_1}{d_1d_2} = \frac{r_1^\prime d_2^\prime + r_2^\prime d_2^\prime}{d_1^\prime d_2^\prime}$, so that $\frac{r_1}{d_1} + \frac{r_2}{d_2} = \frac{r_1^\prime}{d_1^\prime} + \frac{r_2^\prime}{d_2^\prime}$. Thus $+$ is well-defined.
2. ($+$ is associative) Let $\frac{r_1}{d_1}$, $\frac{r_2}{d_2}$, $\frac{r_3}{d_3} \in D^{-1}R$. Note that $(\frac{r_1}{d_1} + \frac{r_2}{d_2}) + \frac{r_3}{d_3} = \frac{r_1d_2 + r_2d_1}{d_1d_2} + \frac{r_3}{d_3}$ $= \frac{(r_1d_2 + r_2d_1)d_3 + r_3d_1d_2}{(d_1d_2)d_3}$ $= \frac{r_1d_2d_3 + r_2d_1d_3 + r_3d_1d_2}{d_1d_2d_3}$ $= \frac{r_1d_2d_3 + (r_2d_3 + r_3d_2)d_1}{d_1(d_2d_3)}$ $= \frac{r_1}{d_1} + \frac{r_2d_3 + r_3d_2}{d_2d_3}$ $= \frac{r_1}{d_1} + (\frac{r_2}{d_2} + \frac{r_3}{d_3})$. Thus $+$ is associative.
3. (An additive identity exists) Let $d_0 \in D$. (We can do this because $D$ is nonempty.) Let $\frac{r}{d} \in D^{-1}R$. Then $\frac{r}{d} + \frac{0}{d_0} = \frac{rd_0 + 0 \cdot d}{dd_0}$ $= \frac{rd_0}{dd_0} = \frac{r}{d}$. Similarly, $\frac{0}{d_0} + \frac{r}{d} = \frac{r}{d}$. Thus $\frac{0}{d}$ is an additive identity in $D^{-1}R$.
4. (Every element has an additive inverse) Let $\frac{r}{d} \in D^{-1}R$. We have $\frac{r}{d} + \frac{-r}{d} = \frac{rd - rd}{d^2}$ $= \frac{0}{d^2} = 0$. Similarly, $\frac{-r}{d} + \frac{r}{d} = 0$. Thus every element of $D^{-1}R$ has an additive inverse.
5. ($+$ is commutative) For all $\frac{r_1}{d_1}, \frac{r_2}{d_2} \in D^{-1}R$, we have $\frac{r_1}{d_1} + \frac{r_2}{d_2} = \frac{r_1d_2 + r_2d_1}{d_1d_2}$ $= \frac{r_2d_1 + r_1d_2}{d_2d_1}$ $= \frac{r_2}{d_2} + \frac{r_1}{d_1}$. Thus $+$ is commutative. In particular, $(D^{-1}R,+)$ is an abelian group.
6. ($\cdot$ is well-defined) Suppose $\frac{r_1}{d_1} = \frac{r_1^\prime}{d_1^\prime}$ and $\frac{r_2}{d_2} = \frac{r_2^\prime}{d_2^\prime}$. Then $r_1d_1^\prime = r_1^\prime d_1$ and $r_2d_2^\prime = r_2^\prime d_2$. Now $r_1d_1^\prime r_2 d_2^\prime = r_1^\prime d_1 r_2^\prime d_2$, and since $R$ is commutative, $r_1r_2 d_1^\prime d_2^\prime = r_1^\prime r_2^\prime d_1d_2$. Thus $\frac{r_1r_2}{d_1d_2} = \frac{r_1^\prime r_2^\prime}{d_1^\prime d_2^\prime}$. Hence $\frac{r_1}{d_1} \cdot \frac{r_2}{d_2} = \frac{r_1^\prime}{d_1^\prime} \cdot \frac{r_2^\prime}{d_2^\prime}$. Thus $\cdot$ is well-defined.
7. ($\cdot$ is associative) We have $\frac{r_1}{d_1} (\frac{r_2}{d_2} \cdot \frac{r_3}{d_3}) = \frac{r_1}{d_1} \cdot \frac{r_2r_3}{d_2d_3}$ $= \frac{r_1(r_2r_3)}{d_1(d_2d_3)}$ $= \frac{(r_1r_2)r_3}{(d_1d_2)d_3}$ $= \frac{r_1r_2}{d_1d_2} \cdot \frac{r_3}{d_3}$ $= (\frac{r_1}{d_1} \cdot \frac{r_2}{d_2}) \cdot \frac{r_3}{d_3}$. Thus $\cdot$ is associative.
8. ($\cdot$ distributes over $+$) Note that $\frac{r_1}{d_1}(\frac{r_2}{d_2} + \frac{r_3}{d_3}) = \frac{r_1}{d_1} \cdot \frac{r_2d_3 + r_3d_2}{d_2d_3}$ $= \frac{r_1r_2d_3 + r_1r_3d_2}{d_1d_2d_3}$ $= \frac{r_1r_2d_1d_3 + r_1r_3d_1d_2}{d_1d_2d_1d_3}$ $= \frac{r_1r_2}{d_1d_2} + \frac{r_1r_3}{d_1d_3}$ $= \frac{r_1}{d_1} \cdot \frac{r_2}{d_2} + \frac{r_1}{d_1} \cdot \frac{r_3}{d_3}$. Thus $\cdot$ distributes over $+$ on the left; similarly we have distributivity on the right.

Thus $D^{-1}R$ is a ring. We claim that moreover it is a commutative ring with $1 \neq 0$.

Proof of claim:

1. ($\cdot$ is commutative) We have $\frac{r_1}{d_1} \cdot \frac{r_2}{d_2} = \frac{r_1r_2}{d_1d_2}$ $= \frac{r_2r_1}{d_2d_1}$ $= \frac{r_2}{d_2} \cdot \frac{r_1}{d_1}$. Thus $\cdot$ is commutative.
2. (A multiplicative identity exists) Let $d_1 \in D$, and let $\frac{r}{d} \in D^{-1}R$. Now $\frac{r}{d} \cdot \frac{d_1}{d_1} = \frac{rd_1}{dd_1} = \frac{r}{d}$. Thus $D^{-1}R$ has a multiplicative identity; in particular, $1 = \frac{d}{d}$ for all $d \in D$.
3. ($1 \neq 0$) Suppose $1 = 0$. Then for any $d \in D$, we have $\frac{d}{d} = \frac{0}{d}$, so that $d^2 = 0$. This is a contradiction because $D$ is closed under multiplication and $0 \notin D$.

Thus $D^{-1}R$ is a commutative ring with $1 \neq 0$.

Now let $d \in D$ and define $\iota_d \rightarrow R \rightarrow D^{-1}R$ by $\iota_d(r) = \frac{rd}{d}$. Suppose $d_1, d_2 \in D$; note that $\iota_{d_1}(r) = \frac{rd_1}{d_1}$ $= \frac{rd_1d_2}{d_1d_2}$ $= \frac{rd_2}{d_2}$ $= \iota_{d_2}(r)$; thus $\iota_{d_1} = \iota_{d_2}$. Thus we can speak of the map $\iota : R \rightarrow D^{-1}R$, without mentioning $d$, because the choice of $d$ does not matter. We claim that $\iota$ is an injective ring homomorphism, and that if $d \in D$ then $\iota(d)$ is a unit in $D^{-1}R$.

Proof of claim:

1. ($\iota$ preserves $+$) Let $r_1,r_2 \in R$ and $d \in D$. Then $\iota(r_1 + r_2) = \frac{(r_1+r_2)d}{d}$ $= \frac{r_1d}{d} + \frac{r_2d}{d}$ $= \iota(r_1) + \iota(r_2)$.
2. ($\iota$ preserves $\cdot$) Let $r_1,r_2 \in R$ and $d \in D$. Then $\iota(r_1r_2)$ $= \frac{r_1r_2d^2}{d^2}$ $= \frac{r_1d}{d} \cdot \frac{r_2d}{d}$ $= \iota(r_1) \cdot \iota(r_2)$.
3. ($\iota$ is injective) Suppose $\iota(r_1) = \iota(r_2)$. Then $\frac{r_1d}{d} = \frac{r_2d}{d}$, so that $r_1d^2 = r_2d^2$, and since $d$ is not a zero divisor, $r_1 = r_2$. Thus $\iota$ is injective.
4. (If $d \in D$ then $\iota(d)$ is a unit) Let $d \in D$. Now $\iota(d) = \frac{d^2}{d}$, and we have $\frac{d^2}{d} \cdot \frac{d}{d^2} = \frac{d^3}{d^3} = 1$. Thus $\iota(d)$ is a unit.

We now attend to the universal property of $D^{-1}R$. Let $S$ be a commutative ring with $1 \neq 0$, and suppose we have an injective ring homomorphism $\varphi : R \rightarrow S$ such that $\varphi(d)$ is a unit for all $d \in D$. Now define a map $\Phi : D^{-1}R \rightarrow S$ by $\Phi(\frac{r}{d}) = \varphi(r)\varphi(d)^{-1}$. We claim that $\Phi$ is an injective ring homomorphism.

Proof of claim:

1. ($\Phi$ is well-defined) Suppose $\frac{r_1}{d_1} = \frac{r_2}{d_2}$. Then $r_1d_2 = r_2d_1$, so that $\varphi(r_1)\varphi(d_2) = \varphi(r_2)\varphi(d_1)$. Since $\varphi(d_1)$ and $\varphi(d_2)$ are units in $S$, we have $\varphi(r_1)\varphi(d_1)^{-1} = \varphi(r_2)\varphi(d_2)^{-1}$; thus $\Phi(\frac{r_1}{d_1}) = \Phi(\frac{r_2}{d_2})$. Hence $\Phi$ is well-defined.
2. ($\Phi$ preserves $+$) We have $\Phi(\frac{r_1}{d_1} + \frac{r_2}{d_2}) = \Phi(\frac{r_1d_2 + r_2d_1}{d_1d_2})$ $= \varphi(r_1d_2+r_2d_1)\varphi(d_1d_2)^{-1}$ $= \varphi(r_1)\varphi(d_1)^{-1} + \varphi(r_2)\varphi(d_2)^{-1}$ $= \Phi(\frac{r_1}{d_1}) + \Phi(\frac{r_2}{d_2})$.
3. ($\Phi$ preserves $\cdot$) We have $\Phi(\frac{r_1}{d_1} \cdot \frac{r_2}{d_2}) = \Phi(\frac{r_1r_2}{d_1d_2})$ $= \varphi(r_1r_2)\varphi(d_1d_2)^{-1}$ $= \varphi(r_1)\varphi(d_1)^{-1}\varphi(r_2)\varphi(d_2)^{-1}$ $= \Phi(\frac{r_1}{d_1}) \Phi(\frac{r_2}{d_2})$.
4. ($\Phi$ is injective) Suppose $\Phi(\frac{r_1}{d_1}) = \Phi(\frac{r_2}{d_2})$. Then $\varphi(r_1)\varphi(d_1)^{-1} = \varphi(r_2)\varphi(d_2)^{-1}$, so that $\varphi(r_1d_2) = \varphi(r_2d_1)$. Since $\varphi$ is injective, we have $r_1d_2 = r_2d_1$, and thus $\frac{r_1}{d_1} = \frac{r_2}{d_2}$. Thus $\Phi$ is injective.

Finally, note that for all $r \in R$, we have $(\Phi \circ \iota)(r) = \Phi(\frac{rd}{d}) = \varphi(r)\varphi(d)\varphi(d)^{-1}$ $= \varphi(r)$. Thus $\Phi \circ \iota = \varphi$.

• Tuan Nguyen  On December 29, 2010 at 8:48 pm

I love your lemmas which my professors actually don’t talk much in classes. Thank you very much for your proofs.

• RyanK  On April 26, 2011 at 6:12 pm

Is this proof similar to R being a commutative ring with a 1 not = 0 and D is a set of non-zero divisors, thus proving D is multiplicatively closed subset of R and that every element of D^-1R is either a unit or a zero divisor?

• nbloomf  On April 27, 2011 at 1:00 am

In a word, yes.

We can carry out the same process using any subset $D$ which is multiplicatively closed and does not contain zero divisors and obtain a larger ring in which the elements of $D$ become units. I think we even get an appropriate universal property (with respect to the subset $D$ of course).

The last idea – that every element is either a unit or a zero divisor – I’m not so sure about. That might be true if $D$ is the set of all non zero divisors, but if it isn’t, then I imagine $D^{-1}R$ will contain elements which are neither units nor zero divisors.

The next thing I’m going to say is way out of my element: I think this has something to do with localization.

• RyanK  On April 27, 2011 at 9:39 pm

Sweet thanks, any ideas on Let k = Z/2Z. Then the number of prime ideals in k[x]/(x+1)4k[x]?

• nbloomf  On April 27, 2011 at 10:45 pm

I’m a little confused by the notation.

• RyanK  On April 28, 2011 at 8:56 am

Sorry I meant, Let k = Z/2Z. Then the number of prime ideals in k[x]/(x+1)^4?

• nbloomf  On April 28, 2011 at 2:59 pm

I see.

Since $k = \mathbb{Z}/(2)$ is a field, $k[x]$ is a PID. Using the lattice isomorphism theorem (and the lemma below, which I had to prove for myself to be sure of), the prime ideals of $R = k[x]/((x+1)^4)$ correspond to the prime ideals of $k[x]$ containing $((x+1)^4)$. Recall that $(\alpha(x)) \supseteq ((x+1)^4)$ if and only if $\alpha$ divides $(x+1)^4$. Clearly this polynomial has 5 divisors, three of which are nontrivial, and (noting that $x+1$ is not a unit in $k[x]$) only one of which is not obviously reducible. Since $x+1$ is irreducible in the UFD $k[x]$, the ideal $(x+1)$ is prime.

So the prime ideals in $R$ are $R$ and $(\overline{x+1})$. (0 is not prime since $R$ is not an integral domain.)

More generally, you can show that if $R$ is a PID (hence UFD) and $\alpha \in R$, then the prime ideals in $R/(\alpha)$ are $(p + (\alpha))$ for the irreducible factors $p$ of $\alpha$.

Lemma: If $I \subseteq J \subseteq R$ are ideals, then $J/I \subseteq R/I$ is prime if and only if $J \subseteq R$ is prime. Suppose $J \subseteq R$ is prime and $(a+I)(b+I) ab+I \in J/I$. Then in particular $ab \in J$, so that either $a \in J$ or $b \in J$, hence either $a+I \in J/I$ or $b+I \in J/I$. Conversely, if $J/I \subseteq R/I$ is prime and $ab \in J$, then $ab+I = (a+I)(b+I) \in J/I$ so that either $a+I \in J/I$ or $b+I \in J/I$. Thus either $a \in J$ or $b \in J$.

• RyanK  On April 29, 2011 at 3:01 pm

So would it be two or just one… cause it seem like you said R and then (x+1) were prime ideals, right?

• nbloomf  On April 29, 2011 at 3:08 pm

That is correct.