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In the ring $\mathbb{R}^{\infty}$ (with the standard operations of component-wise addition and multiplication), what are the maximal ideals?

It was quite simple to determine that the ideals with a single coordinate fixed at $0$ (i.e. $\{0\} \times \mathbb{R} \times \cdots$, $\mathbb{R} \times \{0\} \times \mathbb{R} \times \cdots$) are clearly maximal ideals, since the quotient ring is clearly isomorphic to $\mathbb{R}$.

However, I've been told there's a maximal ideal that I'm missing, and this likely has something to do with the infinite-dimensional aspect of the ring (otherwise if it was a finite Cartesian product, then any ideal should be a Cartesian product of the ideals of $\mathbb{R}$), though I can't quite put a handle on what it is.

user26857
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bzprules
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2 Answers2

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The maximal ideals are related to ultrafilters.

Let $\def\U{\mathcal U}\U$ be a ultrafilter in $\mathbb N$, and let $I_\U$ be the set of sequences in $\mathbb R^\mathbb N$ such that set indices on which one of them vanishes is an element of $\U$. That is a maximal ideal, and this way you get all of them.

Mariano Suárez-Álvarez
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If you don't know ultrafilters, here is another (but essentially the same as Mariano's) argument:

Consider all sequences of $\mathbb{R}^\mathbb{N}$ which are eventually zero. They form a proper ideal which is not contained in any of the maximal ideals you mentioned. But it must be contained in a maximal ideal...

Dune
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