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Does anyone know a simple example of a Boolean ring with a non-principal ideal? Every finitely generated ideal in a Boolean ring is principal, hence such an ideal cannot be finitely generated...

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3 Answers 3

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Depending on what you know, you might consider this is equivalent to another answer, but since it is not spelled out I'll give it a shot.

Take $R=\prod_{i\in I} \mathbb{F}_2$ where $I$ is infinite and $\mathbb{F}_2$ is the field of two elements.

This contains lots of ideals that are not finitely generated (for example,$A =\bigoplus_{i\in I} \mathbb{F}_2$).

This can be shown elementarily, or else if you see that $R$ is not Noetherian, you can recognize it must have non-finitely generated ideals.

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Think small. Any reasonable collection of "small" sets in a Boolean algebra of sets will do. For example, let the ring be the powerset of the reals, and the ideal the collection of countable (this includes finite) subsets. Or, more boringly, think of the powerset of the natural numbers, with ideal the finite sets. Or else think small in the sense of measure.

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Consider the collection of countable and co-countable subsets of the real numbers. The collection of all finite sets is a non-principal ideal; as it the collection of all countable sets.

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