## Local Rings

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 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.

Proof

If xR is a unit, it cannot be contained in the proper ideal M, so it must lie in RM. Conversely, if xR 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 xM, and we have xRM. ♦

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. ]

Proof

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 (xy ∈ M ⇒ x+y ∈ M). We claim that M is an ideal. First it is closed under addition by condition. Next if xMyR, we claim xyM; 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 xRM is a unit so cannot be contained in a maximal ideal. This shows that M is the unique maximal ideal of R. ♦

Examples

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, xR is a unit if and only if x=a/b, where ab are not divisible by 7. So if x,yR are non-units, then x=a/by=c/d, where a, b are divisible by 7 and cd 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: $R = \{\frac x y \in K : x \in S, y \in S-P\}.$

The set of non-units of R is thus $M := \{\frac x y\in K : x\in P, y\in S-P\}.$ One sees that this is closed under addition (we need P to be prime so that the product of elements of SP remains in SP). 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 xy. 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. ]

Proof

Assuming xy is a unit, we get:

• y non-unit, 1 unit ⇒ 1+y unit;
• xy unit, x non-unit ⇒ xy+x unit;
• xy+xx(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”.

Proof

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 RxMR. So yxz = 1 for some y in Rz 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.

Proof

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.

Proof

Suppose (x+I), (y+I) ∈ R/I are such that (x+y)+I is a unit, so there is zR 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. ♦

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