Local Rings, An Introduction

Introduction

You have probably seen local rings defined this way: R is local if it has a unique maximal ideal. That's fine if R is commutative. Here is a more general definition.

The ring R is local if it has a unique maximal left ideal. Remember that the jacobson radical is the intersection of the maximal left ideals, and in this case J is our unique maximal left ideal.

Remember that J is automatically an ideal, so R/J is well defined. Let y be an element of R representing a nontrivial coset of J. In other words, y is nonzero in R/J. Let y and J generate the left ideal M. This is all of R, hence y has a left inverse in R/J. This holds for every y, hence R/J is a division ring.

Conversely, let J be the jacobson radical and assume R/J is a division ring. Suppose there is a maximal left ideal M beyond J, and let y lie in M-J. Since xy = 1 in R/J, y and J span all of R. This is a contradiction, hence J is already a maximal left ideal, and it is the only maximal left ideal.

By symmetry, R/J a division ring implies J is the only maximal right ideal. Thus a local ring R has a unique maximal left ideal, or a unique maximal right ideal, or a jacobson radical J such that R/J is a division ring. The definitions are equivalent.

Asserting a unique maximal ideal is no longer sufficient. Let R be the 2×2 matrices over the reals. This is a simple ring whose only proper ideal is 0, yet it has many maximal left ideals, one for each line in the plane passing through the origin. It is not a local ring.

Units

Let R be a local ring. If J is the maximal ideal, also serving as the jacobson radical, then the units of R map onto the units of R/J. Everything in R/J is a unit, hence everything in R-J is a unit. In other words, R minus its units is an ideal.

Conversely, assume R minus its left or right invertible elements is an ideal J. There can be no left ideals above J, except for R, hence J is the only maximal left ideal and R is local.

If R minus its units is an ideal J, then J has no left invertible elements, else it would contain all of R, including the unit 1. The left invertibles are all units, R minus its left invertibles is an ideal, and R is local.

If the nonunits form an ideal then they certainly form a closed subgroup under addition. Conversely, let the nonunits form a subgroup J. If x is in J and xy is a unit then x is also a unit, which is a contradiction. Therefore xy remains a nonunit, and lies in J. This makes J an ideal, and we are back to a local ring.

Nilpotent Nonunits

If all nonunits are nilpotent then R is local. Let x be a nonzero nonunit with xn = 0. The product (yx)*xn-1 = 0, and if yx is a unit then xn-1 has to equal 0. This means yx is a nonunit, and is nilpotent. The left ideal generated by x consists entirely of nilpotent elements. This is a nil left ideal, and all such ideals live in J. Every nonunit lies in J, and R is local.

Don't try the converse. The localization of Z about 2 is local, but the nonunits are not nilpotent. In fact the ring is an integral domain.

Dedekind Finite

A ring R is dedekind finite or von neumann finite if x*y = 1 implies y*x = 1. Thus a one sided inverse implies a unit.

Let R be local and suppose x is not right invertible. Embed x*R in J, the jacobson radical. Since J is an ideal, x cannot be left invertible either. Therefore R is dedekind finite.

Idempotents

If e is an idempotent in a local ring R, let f = 1-e, the "other" idempotent. Since e+f = 1 they cannot both lie in J, so let e be a unit. Now e*f = 0, hence f = 0 and e = 1. There are no nontrivial idempotents.

Don't try for the converse. A ring like Z is dedekind finite, with no nontrivial idempotents, yet it is not local.