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Let $I$ be a set of huge cardinality, that is, let $|I|>\mathfrak{c}$. Consider the real product algebra $\mathbb{R}^I$ of all real functions defined on $I$. Can we determine:

1) all algebra homomorphisms $\varphi\colon \mathbb{R}^I\to \mathbb{R}$?

2) all algebra homomorphisms $\varphi\colon \mathbb{R}^I\to \mathbb{C}$, where $\mathbb{C}$ is considered as a real algebra over $\mathbb{R}$?

At least when $|I|\leqslant \mathfrak{c}$, for (1) we can describe the homomorphisms as point evaluations.

Apologies if my questions are trivial. I am not an algebraist and I am just trying to build a general picture of this world.

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$\mathbb{R}^I$ doesn't have any elements squaring to $-1$, so 2) reduces to 1). –  Qiaochu Yuan Jun 18 '12 at 12:49
    
Good point! Thank you very much for this most useful observation! –  A. Carda Jun 18 '12 at 12:50
    
No problem. I should mention that generally speaking it's a bad idea to accept answers this early; having an accepted answer makes it less likely that someone else will come here and offer a different answer (and possibly correct mine). –  Qiaochu Yuan Jun 18 '12 at 13:16
    
See what I mean? I now suspect that my answer is wrong! –  Qiaochu Yuan Jun 20 '12 at 2:47

1 Answer 1

up vote 2 down vote accepted

Edit: Jason DeVito's comment below correctly indicates that this argument is incomplete. It remains to show that the ultraproduct with respect to a non-principal ultrafilter is always nonstandard (in that it contains elements not equivalent to the constant sequences). Equivalently, it remains to show that no ultrafilter $F$ on $I$ can have the property that in any partition of $I$ into at most $|\mathbb{R}|$ disjoint subsets, exactly one subset lies in $F$. (Being an ultrafilter is equivalent to having this property for finite partitions.) I do not actually know if this is true.


You only get evaluation maps.

Let $\varphi : \mathbb{R}^I \to \mathbb{R}$ be an algebra homomorphism. Let $S$ be any subset of $I$. The indicator function $1_S \in \mathbb{R}^I$ is idempotent (it is its own square) and so must be sent to an idempotent of $\mathbb{R}$. The only such idempotents are $0$ or $1$, so this determines a distinguished collection $F$ of subsets of $S$, namely those subsets whose indicator functions map to $1$ under $\varphi$.

$F$ has the following properties. The empty set is never in $F$. Since $\varphi$ is a homomorphism we must have $$\varphi(1_S) = 1, \varphi(1_R) = 1 \Rightarrow \varphi(1_S 1_R) = \varphi(1_{S \cap R}) = 1$$

so $F$ is closed under intersection. Moreover, if $\varphi(1_S) = 1$ and $S \subset R$, then $1_S 1_R = 1_S$, hence $\varphi(1_S 1_R) = \varphi(1_S) = 1$, hence $\varphi(1_R) = 1$, so $F$ is also upward closed. Finally, since $\varphi(1) = 1$, we have $$\varphi(1 - 1_S) = 1 - \varphi(1_S) = \varphi(1_{S^c})$$

hence $S \in F$ if and only if $S^c \not \in F$.

These are precisely the axioms defining an ultrafilter on $I$. Now suppose $f, g \in \mathbb{R}^I$ are two functions such that the set $S$ of $x \in I$ for which $f(x) = g(x)$ lies in $F$. Then $(f - g) = (f - g) 1_{S^c}$ and $\varphi(1_{S^c}) = 0$, hence $\varphi(f) = \varphi(g)$. It follows that $\varphi$ factors through the ultraproduct of $I$ copies of $\mathbb{R}$ defined by the ultrafilter $F$.

This ultraproduct is already a field. If $F$ is a principal ultrafilter, $F$ consists of all sets containing a fixed $x \in I$, and the ultraproduct reduces to $\mathbb{R}$; in this case $\varphi$ is just the evaluation homomorphism at $x$. If $F$ is non-principal, then the resulting ultraproduct is a nonstandard version of $\mathbb{R}$ (it satisfies the same first-order properties but has oddities like infinitesimal and infinitely large elements). The standard version of $\mathbb{R}$ sits properly inside it, and since it is a field it admits no nontrivial quotients, so $\varphi$ cannot exist in this case.

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If $F$ is nonprincipal, is the ultraproduct necessarily nonstandard? I'm aware that it can be nonstandard. –  Jason DeVito Jun 18 '12 at 14:48
    
@Jason: hmm. I hadn't considered that. It looks like there might be problems if the cardinality of $I$ is strictly larger than the continuum. –  Qiaochu Yuan Jun 18 '12 at 20:29

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