Characterisation of Local Rings

Theorem

Let $R$ be a ring.

Let $\mathfrak m \lhd R$ be a maximal ideal.

$(1): \quad$ If the set $R \setminus \mathfrak m$ is precisely the Group of Units $R^\times$ of $R$, then $\left({R, \mathfrak m}\right)$ is a local ring.

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

Proof

$(1): \quad$ Suppose that $\mathfrak m$ 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 $\mathfrak m$.

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

$(2): \quad$ Let $x \in R \setminus \mathfrak m$. Then:

$\mathfrak m \subsetneq I \left({\mathfrak m \cup \left\{{x}\right\}}\right)\subseteq R$

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

Since $\mathfrak m$ is maximal, by definition, $x$ and $\mathfrak m$ generate all of $R$.

Therefore $tx + m = 1$ for some $m \in \mathfrak m$, $t \in R$.

Thus $tx = 1 - m \in 1 + \mathfrak m$ is a unit by hypothesis.

Therefore $x$ is a unit.

Now use part $(1)$.

$\blacksquare$