Union of Chain of Proper Ideals is Proper Ideal

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Theorem

Let $R$ be a ring with unity.

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

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

Let $\displaystyle I = \bigcup_{\alpha \in A} I_\alpha$ 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 $1$.

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

$\blacksquare$


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