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.

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Comments

  • 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.

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