§ Local ring in terms of invertibility
Recall that a local ring is a ring with a unique maximal ideal .
This is supposedly equivalent to the definition:
A local ring is a ring such that and for all in ,
§ Stepping stone: If is a local ring then set of all units of is equal to
§ All elements of are units:
- Let be a local ring with unique maximal ideal .
- Let . [ for unit ].
- If is a unit, we are done.
- Otherwise, consider the ideal generated by , .
- must live in some maximal ideal. Since
- is the only maximal ideal, we have that .
- This is a contradiction, since cannot be both in and .
- Hence all elements are units.
§ All units are in :
- Let a unit.
- We cannot have since is a maximal ideal, .
- If then , hence .
§ Part 1: Local ring to to invertible:
- Let have a unique maximal ideal .
- We have already shown that all invertible elements are in .
- Hence if is invertible, it belongs to .
- We must have either or invertible.
- Suppose not: while .
- This is impossible because is an ideal and is thus closed under addition.
- So, we must have that if is invertible then either or is invertible.
§ Part 2: Invertible to Local ring.
- Let be a ring such that if is invertible then either or is invertible.
- Conversely, if neither nor are invertible then is not invertible.
- Hence the set of non-invertible elements form an ideal , as , sum of non-invertibles are not invertible (assumption), product of non-invertibles is not invertible (easy proof).
- This ideal is contained in some maximal ideal .
- This maximal ideal is such that every element in is invertible, since all the non-invertible elements were in from which was built.
- Formally, assume not: Some element is not invertible. Then . This contradicts assumption that .
- Hence is a unique maximal ideal and is a local ring.