Characterisation of Local Rings

Theorem
Let $R$ be a ring.

Let $J \lhd R$ be a maximal ideal.


 * $(1): \quad$ If the set $R \setminus J$ is precisely the group of units $R^\times$ of $R$, then $\left({R, J}\right)$ is a local ring.


 * $(2): \quad$ If $1 + x$ is a unit in $R$ for all $x \in J$ then $\left({R, J}\right)$ is local.

Proof
$(1): \quad$ Suppose that $J$ is the set of non-units of $R$.

Then by Ideal of Unit is Whole Ring, every ideal not equal to $R$ is contained in $J$.

Therefore $J$ is the unique maximal ideal of $R$, so $\left({R, J}\right)$ is local.

$(2): \quad$ Let $x \in R \setminus J$.

Then:
 * $J \subsetneq I \left({J \cup \left\{{x}\right\}}\right)\subseteq R$

where $I \left({S}\right)$ is the ideal generated by $S$.

Since $J$ is maximal, by definition, $x$ and $J$ generate all of $R$.

Therefore $t x + m = 1$ for some $m \in J$, $t \in R$.

Thus $t x = 1 - m \in 1 + J$ is a unit by hypothesis.

Therefore $x$ is a unit.

Now use part $(1)$.