Equivalence of Definitions of Commutative Local Ring

Theorem
Let $A$ be a commutative ring with unity.

Definition 1 implies Definition 2
Let $\mathfrak m \subsetneq A$ be the unique maximal ideal.

First, $A$ is nontrivial, since $1 \notin \mathfrak m$.

Secondly, let $x, y \in A$ be non-units.

Let $\ideal x$ and $\ideal y$ be the principal ideals generated by $x$ and $y$, respectively.

In view of the unique maximality, $\ideal x \subseteq \mathfrak m$ and $\ideal y \subseteq \mathfrak m$.

In particular:
 * $x, y \in \mathfrak m$

which implies:
 * $x + y \in \mathfrak m$

Therefore $x + y$ is a non-unit.

Definition 2 implies Definition 3
First, by the assumption:
 * $x, y \in M \implies x + y \in M$

Secondly, let $a \in A$ and $x \in M$.

We shall show that $a x \in M$.

there exists a $u \in A$ such that:
 * $u \paren {a x} = 1$

Thus:
 * $\paren {u a} x = 1$

which means that $x$ is a unit.

This contradicts the fact that $x \in M$.

Therefore $a x \in M$.

Altogether $M$ is an ideal.

Finally, $1 \notin M$, since $A$ is nontrivial.

Therefore $M$ is a proper ideal.

Definition 3 implies Definition 1
Let $I$ be an arbitrary ideal such that $M \subsetneq I$.

Let $u \in I \setminus M$.

Then $u$ is a unit by the definition of $M$.

Therefore $M = A$.

Furthermore, $M$ is assumed to be a proper ideal.

Therefore $M$ is a maximal ideal.

On the other hand, let $N$ be an arbitrary maximal ideal.

As $N$ does not contain a unit:
 * $N \subseteq M$

which implies:
 * $N = M$

Therefore $M$ is the unique maximal ideal.