The Unapologetic Mathematician

Mathematics for the interested outsider


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 S of a commutative ring with unit R which is, predictably enough, closed under the ring multiplication. We also require for technicality’s sake that S contains the unit 1. A good place to get such multiplicatively closed sets is as complements of prime ideals — given two elements a and b in R but not in the prime ideal P, their product ab must also be outside P. 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 S will be invertible in R. What we want to do is make a bigger ring that properly contains (a homomorphic image of) R in which all elements of S 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 (r,s) with r\in R and s\in S. We’ll think of this as the “fraction” \frac{r}{s}. Now of course we have too many elements. For example, (s,s) should be “the same” as (1,1) for all s\in S. We introduce the following equivalence relation: (r_1,s_1)\sim(r_2,s_2) if and only if there is a t\in S with t(r_1s_2-r_2s_1)=0. Notice that if S contained no zero-divisors we could do away with the “there is a t” 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 (r_1,s_1) and (r_2,s_2), and define (r_1,s_1)+(r_2,s_2)=(r_1s_2+r_2s_1,s_1s_2) and (r_1,s_1)(r_2,s_2)=(r_1r_2,s_1s_2). 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 S^{-1}R and R_S. If S is generated by some collection of elements \{x_1,...,x_n\} we sometimes write R[x_1^{-1},...,x_n^{-1}]. There are a few more, but I’ll leave them alone for now.

It comes with a homomorphism \iota:R\rightarrow R_S, sending r to (r,1). If S contains no zero-divisors then this is an isomorphism onto its image, since then (r_1,1)\sim(r_2,1) would imply that r_1-r_2=0. That is, a copy of R sits inside R_S. This homomorphism has a nice universal property: if f:R\rightarrow R' is any homomorphism of commutative rings with units sending each element of S to a unit, then f factors uniquely as \bar{f}\circ\iota. That is, \iota:R\rightarrow R_S is the “most general” such homomorphism.

Now let’s say we start with an integral domain D. This means that the ideal \mathbf{0} consisting of only the zero element is prime. Then its complement — all nonzero elements of D — is a multiplicatively closed set D^{\times}. We construct the field of fractions D_{D^{\times}} 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 D.

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 \frac{m}{n} with m,n\in\mathbb{Z}, with the above-defined sum and product. That is, it consists of the fractions we all know from elementary school. We call this field \mathbb{Q}: the field of rational numbers.

May 19, 2007 - Posted by | Ring theory


  1. …contains no zero-divisors then this is an isomorphism

    No, just injective.

    Comment by Graham | May 19, 2007 | Reply

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

    Comment by John Armstrong | May 19, 2007 | Reply

  3. […] 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 […]

    Pingback by Ordered Linear Spaces I « The Unapologetic Mathematician | September 23, 2007 | Reply

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