Localization, An Introduction


Once upon a time an ancient bookkeeper wanted to talk about half an apple, or half a cow, or whatever; and fractions were born. This is a natural extension of the integers. If an object is divided into 7 pieces, put the 7 down below and write 1/7.

Now that we've reached the age of abstract algebra, this process can be generalized to various rings. Most of these rings are commutative, in fact most are integral domains. That's why I placed this topic below integral domains in the hierarchy. However, we will, on occasion, take the fractions of a ring with zero divisors; maybe even a noncommutative ring. But that's somewhat rare, so on we go.

Let R be a ring and let S be a nonempty subset of elements in the center of R, excluding 0, such that S is closed under multiplication. The new ring, S inverse of R, is based on the cross product of R and S. The ordered pairs are written a/b, rather than a,b or a.b or any of the other standard conventions. The / is merely a delimitor separating the two elements of an ordered pair, rather than an arithmetic operator, but the resemblence to fractions is unmistakable, and deliberate.

Consider the rationals, where 4/7 is the same as 8/14. The two fractions look different, but they are equal. Cross multiply to show a/b and c/d are equivalent iff ad = bc. In other words, 4×14 = 8×7. Let's generalize this to arbitrary rings and multiplicative sets.

Define an equivalence relation on the set R cross S as follows. Let a/b = c/d when u*(ad-bc) = 0 for some u in S.

If S contains 0 then all fractions are equivalent, which isn't very interesting. That's why S excludes 0. And if S contains a zero divisor it better not contain the "other" zero divisor, else S contains 0.

When S has no zero devisors we can forget about the factor u, and write ad-bc = 0, or ad = bc. This is the (simplified) formula when R is an integral domain.

Since S is nonempty, select any u in S, and a/b is equal to itself. That is, u(ab-ba) = 0. We are using the fact that b comes from S, and is in the center of R, hence b commutes with a.

Symmetry is also straightforward: u(ad-bc) = 0 iff u(cb-da) = 0.

Now let's look at transitivity. To get started, assume S has no zero divisors. Thus a/b = c/d is the same as saying ad = bc. Similarly, c/d = e/f means cf = ed. Remember that b d and f are all in the center of R, so we have some latitude here. Multiply the first equation through by f, giving adf = cfb. Remember that cf = ed, so by substitution, adf = edb. Again, d is in the center, so afd = ebd, and since d is not a zero divisor, af = eb, and a/b = e/f.

Next assume S contains zero divisors. We are given a/b = c/d = e/f. Apply the definition.

1. uad = ucb

2. vcf = ved

Multiply (1) by f on the right and v on the left.

3. uvadf = uvcbf

Use (2) to substitute for vcf in (3) giving:

4. uvadf = uvedb

5. uvd(af-eb) = 0

This completes the proof. The relation is an equivalence relation, and the equivalence classes are well defined. These classes are referred to as the fractions of R by S, or S inverse of R, or R/S. Returning to the fractions you know, 4/7 is equivalent to 8/14, hence both represent the same equivalence class, i.e. the same fraction.

Here is a small exercise, which you can skip if you like. Watch what happens when S is not commutative. Let R be the quaternions, with no zero divisors to worry about, and let S be ±(1,i,j,k). First, if we want i/j to be equivalent to itself, we have to change the equivalence formula to ad-cb, rather than ad-bc. Now ij = ij, and we're ok. But that only postpones the shipwreck. Using this formula, show that -i/1 = 1/i = k/j. However, setting -i/1 = k/j forces k = -k, which is a contradiction. We really need S to commute with R.

Ok, R/S forms a well defined set of equivalence classes, but that doesn't make it a ring. We're going to do that now.

Define a/b + c/d = (ad+bc)/bd, and a/b * c/d = ac/bd.

Begin by showing these operations are well defined. In other words, 2/3 + 4/7 will give the same "fraction" as 2/3 + 8/14. It doesn't matter which representative you choose, the result is the same. Let's replace c/d with an equivalent fraction e/f and watch what happens.

   ucf = ude
   ubcf = ubde
   uadf+ubcf = uadf+ubde
   uf(ad+bc) = ud(af+be)
   ubf(ad+bc) = ubd(af+be)
   (ad+bc)/bd = (af+be)/bf  ( definition of equivalence )

   ucf = ude
   uabcf = uabde
   ac/bd = ae/cf  ( definition of equivalence )

It doesn't matter which representatives we select; the sum and product produce the same equivalence class, the same element in the fraction ring.

Let w be any member of S, and show that 0/w is the additive identity. Similarly, w/w represents the multiplicative identity. The other ring properties, i.e. associative and distributive, are inherited from R. (You can prove them rigorously if you like.) The result is indeed a ring, and is called S inverse of R, or the fraction ring of R by S, or the complete fraction ring of R (if S contains everything other than the zero divisors), or the fraction field of R if the result is a field.

If S has no zero divisors, we can characterize 0 and 1 in the fraction ring. To find the class that is 0 in the new ring, let a/b = 0 and solve a/b + c/d = c/d, giving add+bcd = bcd, or a = 0. The fractions 0/b form the class that is 0.

If R has no zero divisors, set a/b = 1, so that a/b * c/d = c/d. This gives acd = bcd, hence a = b. The fractions b/b for all b in S form the class of 1 in the fraction ring.

When S has no zero divisors, R embeds in its fraction ring. Map R to R/1. Verify this is a ring homomorphism. If a/1 = b/1 then a = b, hence The map is a monomorphism, and R is a subring of its fractions.

If R is an integral domain, and S contains all nonzero elements of R, these fractions form a field. The inverse of a/b is b/a. Furthermore, R embeds in its fraction field via R/1. We sort of glossed over this step when proving gauss' lemma, but now we know that fraction fields are well defined.

What about the fraction field of the fraction field of R? In this case S contains only units, so step back and ask what happens when S contains only units. Equate a/b with ab-1/1, and each fraction is just another element of R. If two elements become 1 in R/S then x/1 = y/1, or x = y. (Remember S has no zero divisors.) So the map is 1-1 and onto, and the ring has not changed. We need to put some nonunits in the denominator, or we haven't made any progress.

Assume R has no zero divisors, and write ac/bd = 0/w. this means wac = 0, and since w is nonzero, a or c is 0. The fractions of a domain form another domain.

Note that the fraction ring is an R module. Let R act on the numerators.