Construct and prove the universal property of rings of fractions.

Let be a commutative ring with . A subset 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 of denominators in , there exists a commutative ring with , , and an injective ring homomorphism such that if , then is a unit in . Moreover, has the following “universal property”: if is a commutative ring with and is an injective ring homomorphism such that is a unit in for all , then there exists a unique injective ring homomorphism such that .

Proof: Let be a commutative ring with and let be a set of denominators. Define a relation on as follows: precisely when .

We claim that is an equivalence relation.

Proof of claim:

- ( is reflexive) Let . Since , we have . Thus is reflexive.
- ( is symmetric) Suppose . Then , and thus . Thus . Thus is symmetric.
- ( is transitive) Suppose and . Then we have and . Then . Since is not a zero divisor in , we have , so that . Thus is transitive.

Thus is an equivalence relation. We let , and we denote the equivalence class by .

We now claim that for all and all , . To see this, note that , so that .

We now define operators on as follows: and . We claim that is a ring.

Proof of claim:

- ( is well-defined) Suppose , , , with and . Then we have and . Thus and , and thus . Thus . Thus , so that . Thus is well-defined.
- ( is associative) Let , , . Note that . Thus is associative.
- (An additive identity exists) Let . (We can do this because is nonempty.) Let . Then . Similarly, . Thus is an additive identity in .
- (Every element has an additive inverse) Let . We have . Similarly, . Thus every element of has an additive inverse.
- ( is commutative) For all , we have . Thus is commutative. In particular, is an abelian group.
- ( is well-defined) Suppose and . Then and . Now , and since is commutative, . Thus . Hence . Thus is well-defined.
- ( is associative) We have . Thus is associative.
- ( distributes over ) Note that . Thus distributes over on the left; similarly we have distributivity on the right.

Thus is a ring. We claim that moreover it is a commutative ring with .

Proof of claim:

- ( is commutative) We have . Thus is commutative.
- (A multiplicative identity exists) Let , and let . Now . Thus has a multiplicative identity; in particular, for all .
- () Suppose . Then for any , we have , so that . This is a contradiction because is closed under multiplication and .

Thus is a commutative ring with .

Now let and define by . Suppose ; note that ; thus . Thus we can speak of the map , without mentioning , because the choice of does not matter. We claim that is an injective ring homomorphism, and that if then is a unit in .

Proof of claim:

- ( preserves ) Let and . Then .
- ( preserves ) Let and . Then .
- ( is injective) Suppose . Then , so that , and since is not a zero divisor, . Thus is injective.
- (If then is a unit) Let . Now , and we have . Thus is a unit.

We now attend to the universal property of . Let be a commutative ring with , and suppose we have an injective ring homomorphism such that is a unit for all . Now define a map by . We claim that is an injective ring homomorphism.

Proof of claim:

- ( is well-defined) Suppose . Then , so that . Since and are units in , we have ; thus . Hence is well-defined.
- ( preserves ) We have .
- ( preserves ) We have .
- ( is injective) Suppose . Then , so that . Since is injective, we have , and thus . Thus is injective.

Finally, note that for all , we have . Thus .

## Comments

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

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?

In a word, yes.

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

allnon zero divisors, but if it isn’t, then I imagine 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.

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

I’m a little confused by the notation.

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

I see.

Since is a field, 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 correspond to the prime ideals of containing . Recall that if and only if divides . Clearly this polynomial has 5 divisors, three of which are nontrivial, and (noting that is not a unit in ) only one of which is not obviously reducible. Since is irreducible in the UFD , the ideal is prime.

So the prime ideals in are and . (0 is not prime since is not an integral domain.)

More generally, you can show that if is a PID (hence UFD) and , then the prime ideals in are for the irreducible factors of .

Lemma: If are ideals, then is prime if and only if is prime. Suppose is prime and . Then in particular , so that either or , hence either or . Conversely, if is prime and , then so that either or . Thus either or .

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

That is correct.