## Fractions

One thing we haven’t given good examples of is fields. We can get some from factoring out a maximal ideal from a commutative ring with unit, but the most familiar example — rational numbers — comes from a different construction.

First we define a multiplicatively closed set. This is a subset of a commutative ring with unit which is, predictably enough, closed under the ring multiplication. We also require for technicality’s sake that contains the unit . A good place to get such multiplicatively closed sets is as complements of prime ideals — given two elements and in but not in the prime ideal , their product must also be outside . Another good way is to start with some collection of elements and take the submonoid they generate under multiplication.

In general not all the elements of will be invertible in . What we want to do is make a bigger ring that properly contains (a homomorphic image of) in which all elements of *do* have inverses. We’ll do this sort of like how we built the integers by adding negatives to the natural numbers.

Consider the set of all elements with and . We’ll think of this as the “fraction” . Now of course we have too many elements. For example, should be “the same” as for all . We introduce the following equivalence relation: if and only if there is a with . Notice that if contained no zero-divisors we could do away with the “there is a ” clause, but we might need it in general.

So as usual we pass to the set of equivalence classes and assert that the result is a ring. The definitions of addition and multiplication are exactly what we expect if we remember fractions from elementary school. Choose representatives and , and define and . From here it’s a straightforward-but-tedious verification that these operations are independent of the choices of representatives and that they satisfy the ring axioms.

We call the resulting ring by a number of names. Two of the most common are and . If is generated by some collection of elements we sometimes write . There are a few more, but I’ll leave them alone for now.

It comes with a homomorphism , sending to . If contains no zero-divisors then this is an isomorphism onto its image, since then would imply that . That is, a copy of sits inside . This homomorphism has a nice universal property: if is any homomorphism of commutative rings with units sending each element of to a unit, then factors uniquely as . That is, is the “most general” such homomorphism.

Now let’s say we start with an integral domain . This means that the ideal consisting of only the zero element is prime. Then its complement — all nonzero elements of — is a multiplicatively closed set . We construct the field of fractions by adding inverses to all the nonzero elements. Now every nonzero element has an inverse, so this really is a field. In fact, it’s the “most general” field containing .

And, finally, let’s apply this construction to the integers. They are an integral domain, so it applies. Now the field of fractions consists of all fractions with , with the above-defined sum and product. That is, it consists of the fractions we all know from elementary school. We call this field : the field of rational numbers.

…contains no zero-divisors then this is an isomorphismNo, just injective.

Comment by Graham | May 19, 2007 |

Damn, you’re right. I was thinking “isomorphic onto its image” and didn’t type that.

Comment by John Armstrong | May 19, 2007 |

[...] are (I still haven’t defined them here) use them, but otherwise you can get away for now with rational numbers. We consider the category of finite-dimensional vector spaces over and -linear maps between [...]

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