A local ring is a ring $R$ such that $1 \neq 0$ and for all $x, y$ in $R$, $x + y \text{ invertible} \implies x \text{ invertible} \lor y \text{ invertible}$

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

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

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

§ All units are in $(R - M)$:

• Let $u$ a unit.
• We cannot have $u \in M$ since $M$ is a maximal ideal, $M \neq R$.
• If $u \in M$ then $u^{-1} u = 1 \in M$, hence $M = R$.
• Contradiction.

§ Part 1: Local ring to to invertible:

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

§ Part 2: Invertible to Local ring.

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

§ References

• Using the internal language of toposes in algebraic geometry