Mathematicians are generally more familiar with the case of local commutative rings, so we’ll begin from there.
Definition. A commutative ring R is said to be local if it has a unique maximal ideal.
Note that every non-zero commutative ring has at least one maximal ideal; indeed, we can take the collection ∑ of all ideals of R not containing 1; Zorn’s lemma provides a maximal element I in ∑, which does not contain 1 and is maximal.
Lemma. If R is a commutative local ring with maximal ideal M, then its complement R-M is the set of units of R.
If x∈R is a unit, it cannot be contained in the proper ideal M, so it must lie in R–M. Conversely, if x∈R is not a unit, the ideal Rx is properly contained in R so it is contained in a maximal ideal. But since there’s only one maximal ideal we have Rx ⊆ M, so x∈M, and we have x∈R–M. ♦
Proposition. A commutative ring R is local if and only if whenever x and y are non-units, so is x+y. [ Equivalently, if x+y is a unit, then either x or y is a unit. ]
Suppose R is local with maximal ideal M. The above lemma says M is precisely the set of non-units of R. Thus this set is closed under addition.
Conversely, let M be the set of non-units of R and assume (x, y ∈ M ⇒ x+y ∈ M). We claim that M is an ideal. First it is closed under addition by condition. Next if x∈M, y∈R, we claim xy∈M; indeed if not, xy is a unit so both x and y are units (contradiction!). So the set of non-units of R is an ideal. On the other hand, every x∈R–M is a unit so cannot be contained in a maximal ideal. This shows that M is the unique maximal ideal of R. ♦
1. Let R be the set of rational numbers of the form a/b, where b is not divisible by 7. Then R is local; indeed, x∈R is a unit if and only if x=a/b, where a, b are not divisible by 7. So if x,y∈R are non-units, then x=a/b, y=c/d, where a, b are divisible by 7 and c, d are not. Hence, so is x+y = (ad+bc)/(bd). Thus, the unique maximal ideal M of R is given by a/b, where a is divisible by 7 and b is not.
2. Let R be the set of rational functions in x of the form f(x)/g(x) where f(x), g(x) are polynomials and g has non-zero constant term. An element of R is non-unit iff it is of the form f(x)/g(x) where f has zero constant term and g does not. By the same reasoning as in example 1, this forms the unique maximal ideal of R.
3. More generally let S be an integral domain and P be any prime ideal of S. Take K, the field of fractions of S. Now define:
The set of non-units of R is thus One sees that this is closed under addition (we need P to be prime so that the product of elements of S–P remains in S–P). So R is local with maximal ideal M.
Example 3 can even be generalized to a commutative ring with zero-divisors. This is the essence of localization, which is another story for another day.
Non-commutative Local Rings
We repeat the above definition for non-commutative R.
Definition. Let R be a ring, not necessarily commutative. We say that R is local if, whenever x, y ∈ R are non-unit, so is x+y.
In other words, R is local if whenever x+y is a unit, either x or y is a unit. Since x is a non-unit iff –x is, we see that the definition is unchanged when we replace x+y with x–y. Hence, we have the following:
- (non-unit) ± (non-unit) = (non-unit);
- (unit) ± (non-unit) = (unit). [ For if it were a non-unit, bringing the LHS non-unit term to the RHS contradicts the first property. ]
Lemma. Let R be a local ring. If x, y are non-units, so is xy.
[ Note: this is not as trivial as it seems! Indeed, without the local property, we can only conclude that x has a right inverse. ]
Assuming xy is a unit, we get:
- y non-unit, 1 unit ⇒ 1+y unit;
- xy unit, x non-unit ⇒ xy+x unit;
- xy+x = x(y+1) unit, y+1 unit ⇒ x unit (contradiction). ♦
Thus, in a local ring we have:
- (unit) × (unit) = (unit). [ Obvious. ]
- (non-unit) × (unit) = (non-unit). [ If this were a unit, divide by the inverse of the unit on the LHS to get a contradiction. ]
- (unit) × (non-unit) = (non-unit). [ Same as above. ]
- (non-unit) × (non-unit) = (non-unit). [ Above lemma. ]
Next we have:
Proposition. If R is a non-zero local ring, the set of non-units is precisely the Jacobian radical J(R), the set of “bad elements”.
Let N be the set of non-units of R. By definition, N is closed under addition/subtraction. By the above observation, we see that if x or y is a non-unit, so is xy. Thus N is a two-sided ideal of R.
Obviously no element of J(R) is a unit (for R must have a maximal left ideal). Thus J(R)⊆N. Conversely, if x lies in N, and not in some maximal left ideal M, we get Rx + M = R. So yx + z = 1 for some y in R, z in M. But yx and z are both non-units which sum up to a unit, contradicting what we know about local rings. ♦
Corollary. In an artinian local ring R, every element is either unit or nilpotent. The radical is precisely the set of nilpotent elements.
In an artinian ring, the Jacobian J is nilpotent. In particular, every element of J is nilpotent. But the above proposition says J is precisely the set of non-units! ♦
Finally, we have:
Lemma. Let R be a local ring, and I ⊂ R be a proper two-sided ideal. Then R/I is also local.
Suppose (x+I), (y+I) ∈ R/I are such that (x+y)+I is a unit, so there is z∈R such that z(x+y) – 1 and (x+y)z – 1 are elements of I. An element of I is not a unit. Since R is local, z(x+y) and (x+y)z are units (e.g. if z(x+y) and z(x+y)-1 are both non-units, then so is 1, which is absurd). Hence, x+y is a unit, since in general if ab and ba are both units, then a and b are both left- and right-invertible and units as well. Again since R is local, we know that x or y is a unit, and so x+I or y+I is a unit. ♦