§ Local ring in terms of invertibility

Recall that a local ring RR is a ring with a unique maximal ideal MM. This is supposedly equivalent to the definition:
A local ring is a ring RR such that 101 \neq 0 and for all x,yx, y in RR, x+y invertible    x invertibley invertiblex + y \text{ invertible} \implies x \text{ invertible} \lor y \text{ invertible}

§ Stepping stone: If (R,M)(R, M) is a local ring then set of all units of RR is equal to RMR - M

§ All elements of (RM)(R - M) are units:

  • Let RR be a local ring with unique maximal ideal MM.
  • Let uRMu \in R - M. [ uu for unit ].
  • If uu is a unit, we are done.
  • Otherwise, consider the ideal generated by uu, (u)(u).
  • (u)(u) must live in some maximal ideal. Since
  • MM is the only maximal ideal, we have that u(u)Mu \in (u) \subseteq M.
  • This is a contradiction, since uu cannot be both in MM and RMR - M.
  • Hence all elements uRMu \in R - M are units.

§ All units are in (RM)(R - M):

  • Let uu a unit.
  • We cannot have uMu \in M since MM is a maximal ideal, MRM \neq R.
  • If uMu \in M then u1u=1Mu^{-1} u = 1 \in M, hence M=RM = R.
  • Contradiction.

§ Part 1: Local ring to to invertible:

  • Let RR have a unique maximal ideal MM.
  • We have already shown that all invertible elements are in RMR - M.
  • Hence if x+yx + y is invertible, it belongs to RMR - M.
  • We must have either xx or yy invertible.
  • Suppose not: x,yMx, y \in M while x+y∉Mx + y \not \in M.
  • This is impossible because MM is an ideal and is thus closed under addition.
  • So, we must have that if x+yx + y is invertible then either xx or yy is invertible.

§ Part 2: Invertible to Local ring.

  • Let RR be a ring such that if x+yx + y is invertible then either xx or yy is invertible.
  • Conversely, if neither xx nor yy are invertible then x+yx + y is not invertible.
  • Hence the set of non-invertible elements form an ideal II, as 0I0 \in I, sum of non-invertibles are not invertible (assumption), product of non-invertibles is not invertible (easy proof).
  • This ideal II is contained in some maximal ideal MM.
  • This maximal ideal MM is such that every element in RMR - M is invertible, since all the non-invertible elements were in II from which MM was built.
  • Formally, assume not: Some element sRMs \in R - M is not invertible. Then sIMs \in I \subseteq M. This contradicts assumption that sRMs \in R - M.
  • Hence MM is a unique maximal ideal and RR is a local ring.

§ References