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Consider the ring $R=\mathbb{Z}^\mathbb{N}$ of integer sequences with the usual componentwise operations and let $I$ be the ideal of sequences that are eventually zero. Questions:

  1. Is there a unique maximal ideal $\mathfrak m \supseteq I$ of $R$ ?
  2. Is $R/\mathfrak m$ isomorphic to a well-known field (like $\mathbb{R}, \mathbb{Q}_p$,...) where $\mathfrak m\supseteq I$ is any maximal ideal ?
tj_
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    It is fairly straightforward to construct an invertible element in an ideal of $R/I$ containing all noninvertibles. Thus $R/I$ is not local. For the sake of illustration: $(0,2,0,2,\dots)+(1,-3,1,-3,\dots)$ is the invertible $(1,-1,1,-1,\dots)$. This makes question 2 more interesting, since there is not such an easy characterization of the elements outside a maximal ideal in $R/I$. – Karl Kroningfeld Aug 26 '13 at 22:20
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    I was talking about this ring just a few days ago, specifically about it's cardinality. It's very big. You can exhibit countably many distinct injective ring homomorphisms $R \hookrightarrow R/I$. As to your question, I have no idea what a maximal ideal in $R$ even looks like, let alone what the quotient might be. I second that this is not off topic, this is an interesting question. – Jim Aug 26 '13 at 23:22

1 Answers1

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No, $m$ is very far from unique. Take any function $p : \mathbb{N} \to \text{primes}$ and any non-principal ultrafilter $U$ on $\mathbb{N}$. Then $R$ admits a quotient map to $\prod_{i=1}^{\infty} \mathbb{F}_{p(i)}$ which in turn admits a quotient map to the corresponding ultraproduct of the fields $\mathbb{F}_{p(i)}$, which is a field, and this map (viewed as a map with domain $R$) contains $I$ in its kernel. (I think this construction describes all possible $m$ but am not sure.)

Generically these fields are pseudo-finite fields and in particular quasi-finite fields. When they are not themselves isomorphic to finite fields (which happens iff there exists some prime $p$ such that $\{ i : p(i) = p \}$ is contained in the ultrafilter $U$, and in particular if $\{ i : p(i) = p \}$ is cofinite), they are unfamiliar.

Qiaochu Yuan
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    Not just unfamiliar but also of characteristic zero (if infinite) and completely difficult to understand. For example if on a large set of indices the $-1$ has a root, then will the product. So we can't even say that in advance. – Asaf Karagila Aug 27 '13 at 08:09
  • Incidentally, you need a weak version of choice (http://en.wikipedia.org/wiki/Boolean_prime_ideal_theorem) to show that non-principal ultrafilters exist. I think it is consistent with ZF that they do not and in this case I think there are no such $m$, so this is an explicit example where you need some amount of choice to furnish maximal ideals containing an ideal. – Qiaochu Yuan Aug 28 '13 at 07:19
  • Yes, Andreas Blass constructed a model in which there are no free ultrafilters on any set. The ultrafilter lemma is strictly stronger than the assertion "every infinite set has a free ultrafilter, but not every filter can be extended to an ultrafilter". – Asaf Karagila Aug 28 '13 at 07:37
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    Incidentally, you can use this idea to motivate the definition of an ultrafilter, starting from the idea that you'd like to write down a homomorphism $\prod K_i \to K$ where $K_i, K$ are all fields, and writing down everything you can figure out about what happens to the idempotents under this map. In fact an ultrafilter on a set $I$ is precisely a homomorphism $\prod_{i \in I} \mathbb{F}_2 \to \mathbb{F}_2$. – Qiaochu Yuan Sep 01 '13 at 21:17
  • Interesting. I never thought about it like that. – Asaf Karagila Sep 01 '13 at 21:25