Union of Chain of Proper Ideals is Proper Ideal

Theorem

Let $R$ be a ring with unity.

Let $\struct {P, \subseteq}$ be the ordered set consisting of all ideals of $R$, ordered by inclusion.

Let $\sequence {I_\alpha}_{\alpha \mathop \in A}$ be a non-empty chain of proper ideals in $P$.

Let $\ds I = \bigcup_{\alpha \mathop \in A} I_\alpha$ be their union.

Then $I$ is a proper ideal of $R$.

Proof

By Union of Chain of Ideals is Ideal, $I$ is an ideal.

It remains to show that $I \subsetneq R$.

The ideals $I_\alpha$ are all proper, so none of them contain the unity.

Thus $I$ does not contain $1$, which means $I \subsetneq R$.

$\blacksquare$

Also see

• Union of Chain of Ideals is Ideal