## A hands-on description of a “completion” of the free commutative monoid on countably many generators

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This is basically an I'm-weak-at-algebraic-geometry question. I asked it as a warm-up question here, but Ilya N asked me to break that post up into several questions.

Consider the free commutative monoid $X$ on countably many generators. Let $A$ be the algebra of all functions $X \to \mathbb C$. Then $A$ is a commutative and cocommutative bialgebra, and hence a commutative monoid object in $(\text{CAlg})^{\rm op}$, where $\text{CAlg}$ is the category of commutative algebras. A global point of $A \in (\text{CAlg})^{\rm op}$ is an algebra homomorphism $A \to \mathbb C$. The set of global points of $A$ is naturally a commutative monoid that contains $X$ as a submonoid.

What are all the global points of $A$? I'm pretty sure that there are more than just the points of $X$. The form a monoid, because $A$ is a bialgebra: is there a reasonable description of the monoid structure?

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You probably want the monoid to be finite if you are thinking of the bialgebra of functions on it. – Mariano Suárez-Alvarez Jan 6 2010 at 20:20
It's fine as long as each element is the sum of two others in finitely many ways, as is the case here. – Reid Barton Jan 6 2010 at 20:27

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As far as I can tell, the question has nothing to do with the coalgebra structure. You're just asking if an infinite product of fields has exceptional augmentations. – S. Carnahan♦ Jan 6 2010 at 23:18
@Scott: that's totally correct. I meant to imply "what is their monoid structure?" I'll add that. (Man, I've been making lots of edits-to-questions, which annoy me when other people do it. Why can't people ask the questions they mean to begin with ?!? :) ) – Theo Johnson-Freyd Jan 7 2010 at 0:16

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I think that (editing question) kind of works as long as it's an extension of the question. That is, the rule is: when there's already a reply, it should continue to be a valid reply after editing the question. – Ilya Nikokoshev Jan 7 2010 at 0:24

## Preliminaries

First part of your question doesn't use the bialgebra structure. That is, you have a space of functions on countable many points which I'll denote $A = \mathbb C_1\times \mathbb C_2\times\cdots \times\mathbb C_n\times\cdots$ equipped with $+$ and $\times$ pointwise. You would like to classify all algebra maps $\varphi: A\to \mathbb C$.

To start, note that idempotents must go to idempotents, of which $\mathbb C$ has only two: $0$ and $1$. Moreover, $1\times1\times\cdots \mapsto 1$ (unless the whole map is trivial).

## Classification of points

Consider characteristic functions $\lambda_n$ which take value from point labeled $n$. As $\lambda_n^2 = \lambda_n$, we must have $\varphi(\lambda_n) = 0$ or $1$. There are three cases:

• there are two indices $i, j$ such that $\varphi(\lambda_i) = \varphi(\lambda_j) = 1$. This is impossible: $0 = \varphi(\lambda_i\lambda_j) = 1$.
• there exists exactly one index $i$ for which $\varphi(\lambda_i) = 1$. This implies $\varphi(f) = \varphi((1-\lambda_i)f + \lambda_if) = f(i)$.
• all functions $\lambda_i$, and, therefore, all finite sums from $A$, lie in the kernel of $\varphi$.

The maps of the latter type are indeed the "extra" points. Note, however, that they are "wild" and can be easily killed by some extra finitness assumptions, for example the following "limit of zeroes" property: if $\varphi$ restricted to all finite subsets is 0, then $\varphi$ is 0.

## Wild maps

You can construct examples of these wild maps using the axiom of choice. To do that, make the elements of the form $a\times a \times \cdots$ and all their finite modifications to map to $a$; denote those elements $A_0$. Next, select an arbitrary $T_1\in A$ which is transcendental over $A_0$ and map it to arbitrary $t_1\in \mathbb C$. This will fix all elements that lie in algebraic closure $A_1 = A_0(T_1)$. Do that again for $T_2$ and repeat until you have nothing left. At each step you're producing a correct map $A_n \to \mathbb C$, so you get the final map $A\to \mathbb C$ as the limit of those.

Conversely, any wild map can be constructed by the above process, assuming all necessary set-theoretic things. So, the answer is, the wild points are classified by maps from this terribly non-constructive sequence of $T_i$ (of cardinality the same as $A$, that is, continuum) to $\mathbb C$. Those are again in cardinality of continuum.

## Monoid structure

One now is reminded that $A$ came with a natural basis enumerated by numbers $(n_1, n_2, \dots, n_k, \dots) \in \mathbb Z^+\oplus\mathbb Z^+\oplus\cdots\oplus\mathbb Z^+\oplus\cdots$ (which is isomorphic to $\mathbb Z_{>0}^\times$). Therefore, it should be possible to add any two points (denoted $\oplus$). The following properties are clear:

• regular points add normally as $n_k = n_k' + n_k''$;
• adding wild point to either regular or wild point results in a wild point.

Now, although you could write explicitly some wild points, nearly all of them are too hard to describe. The best approximation to the resulting picture is probably this: imagine a set of wild points $W$; the whole space is $W \times \mathbb Z^+\times \mathbb Z^+\times\cdots$. The remaining question, therefore, is what monoid $W$ is equivalent to. For that, you need to settle these questions:

• is it true that you cannot have $w_1 \oplus w_2 \oplus \cdots\oplus w_n = r$ where $r$ is a regular point;
• whether you can subtract them;
• whether and how wild points are divisible by naturals.

## Operations on wild points

I think now is certainly time to post another question :)

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When the field is $k=\mathbb F_2$ instead of $\mathbb C$, you can construct algebra morphisms from the $k$-algebra of functions on an infinite set to $k$ which are not evaluations, by using an ultrafilter.

Let $U$ be a non-principal ultrafilter on $X=\mathbb N$, and let $I$ be the ideal of $k(X)$, the algebra of $k$-valued functions on $X$, such that a function $f:X\to k$ is in $I$ iff $\{x\in X:f(x)=0\}\in U$. This is indeed a proper ideal, as one checks at once, and it is in fact maximal. Moreover, $k(X)/I$ has exactly two elements, so from $U$ we get a morphism $\epsilon:k(X)\to k$. As the ultrafilter $U$ is not principal, this map $\epsilon$ is not an evaluation.

(If you do the same with $\mathbb C$, one cannot prove that $k(X)/I$ is again $\mathbb C$; indeed, it is generaly not!) (On the other hand, the same is true for any finite field, I guess, because saying that the field has $q$ elements is a first order property of the field and Łoś's theorem applies... maybe someone with a firmer grasp on this can confirm this?)

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Yes, you can do the same thing on the algebra of all bounded functions (i.e. you get the Stone-Cech compactification of N) but I don't know what this says about the unbounded case. – Qiaochu Yuan Jan 7 2010 at 0:00
Apparently (mathoverflow.net/questions/3871), maxSpec of C(X) is \beta X, but the residue fields of the extra points can be wild.  But they aren't always wild. – Theo Johnson-Freyd Jan 7 2010 at 0:14
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