Proof.
Let R be a ring, and \mathfrak m a maximal ideal. If x \in R \setminus \mathfrak m, and x is not a unit then there is a maximal ideal \mathfrak m' containing x. Hence R has at least two maximal ideals. Conversely, if \mathfrak m' is another maximal ideal, then choose x \in \mathfrak m', x \not\in \mathfrak m. Clearly x is not a unit. This proves the equivalence of (1) and (3). The equivalence (1) and (2) is tautological. If R is local then (4) holds since x is either in \mathfrak m or not. If (4) holds, and \mathfrak m, \mathfrak m' are distinct maximal ideals then we may choose x \in R such that x \bmod \mathfrak m' = 0 and x \bmod \mathfrak m = 1 by the Chinese remainder theorem (Lemma 10.15.4). This element x is not invertible and neither is 1 - x which is a contradiction. Thus (4) and (1) are equivalent.
\square
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