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Let $k$ be a commutative ring. Feel free to assume it's a field.

Let $X$ be a set. This question is only interesting when $X$ is infinite.

Write $k^X$ for the $k$-algebra of functions $X \to k$, with the algebra operations defined pointwise.

What are the $k$-algebra homomorphisms $k^X \to k$?

Trivially, for each $x \in X$ there is a projection/evaluation map $k^X \to k$.

Under what circumstances are there any $k$-algebra homomorphisms $k^X \to k$ apart from the projections?

Here are some observations. Observation 2 shows that the question is not entirely trivial, in the sense that there are sometimes nontrivial homomorphisms $k^X \to k$.

  1. When $k$ an integral domain, any homomorphism $\Phi: k^X \to k$ gives rise to an ultrafilter $\mathcal{U}_\Phi$ on $X$. To see this, write $\chi_S \in k^X$ for the characteristic function of a subset $S \subseteq X$. Since $\chi_S$ is idempotent, $\Phi(\chi_S)$ is also idempotent, and is therefore either $0$ or $1$. Write $$ \mathcal{U}_\Phi = \{ S \subseteq X : \Phi(\chi_S) = 1 \}. $$ It's easy to check that $\mathcal{U}_\Phi$ is an ultrafilter on $X$ — in other words, that whenever we write $X = X_1 \amalg \cdots \amalg X_n$, there is precisely one $i$ for which $\Phi(\chi_{X_i}) = 1$.

  2. When $k$ is a finite integral domain — that is, a finite field — the $k$-algebra homomorphisms $k^X \to k$ are in bijection with the ultrafilters on $X$. One direction of this correspondence is given as in (1).

    For the other, start with an ultrafilter $\mathcal{U}$ on $X$. We want to define a homomorphism $\Phi_{\mathcal{U}}: k^X \to k$, so take $\phi \in k^X$. Since $k$ is finite, the fibres $(\phi^{-1}(c))_{c \in k}$ form a finite partition of $X$. So there is precisely one element $c \in k$ such that $\phi^{-1}(c) \in \mathcal{U}$, and we put $\Phi_{\mathcal{U}}(\phi) = c$. It's straightforward to check that $\Phi_{\mathcal{U}}$ is a homomorphism and that the processes $\mathcal{U} \mapsto \Phi_{\mathcal{U}}$ and $\Phi \mapsto \mathcal{U}_\Phi$ are mutually inverse.

  3. When $k$ is an integral domain and $X$ (rather than $k$) is finite, the only homomorphisms $k^X \to k$ are the projections. This follows e.g. from (1) and the fact that ultrafilters on a finite set are principal.

  4. Denote by $X \cdot k$ the $k$-vector space with basis $X$. Then $k^X$, as a $k$-vector space, is isomorphic to the space of linear maps $X \cdot k \to k$. Now any $k$-algebra homomorphism $\Phi: k^X \to k$ is, in particular, a $k$-linear map, so $\Phi$ is an element of the double dual of $X \cdot k$. Hence there can only be nontrivial homomorphisms $k^X \to k$ if there are nontrivial elements of the double dual of $X \cdot k$.

    So my question seems to be closely related to one that's come up a few times here before: how much Choice do we need in order to construct nontrivial elements of the double dual of an infinite-dimensional vector space?

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    $\begingroup$ If $k$ is a field then the ideals in $k^X$ are in bijection with the filters, with the maximal ideals corresponding to the ultrafilters. So the question comes down to: for which ultrafilters is the residue field a trivial extension of $k$? $\endgroup$
    – Tom Goodwillie
    Jun 4, 2015 at 13:20
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    $\begingroup$ If $X$ is huge enough to have nonprincipal ultrafilters stable under infinite countable intersections, then assuming that $k$ is not to big (say $k$ infinite countable, to begin with), you can compute the limit wrt this ultrafilter. Maybe it's reasonable to assume that $X$ has no such ultrafilter unless you have a specific interest in large cardinals. $\endgroup$
    – YCor
    Jun 4, 2015 at 13:33
  • $\begingroup$ @YCor: I don't have a particular interest in large cardinals, thanks, but neither do I particularly want to rule out sets of large cardinality. $\endgroup$
    – Tom Leinster
    Jun 4, 2015 at 13:38
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    $\begingroup$ A given function $\phi:X\to k$ will represent an element outside $k$ in the residue field associated with the ultrafilter $U$ if and only if no level set $\phi^{-1}(c)$ belongs to $U$. So for example if both $X$ and $k$ are countably infinite then there are no more examples of such algebra homomorphisms. $\endgroup$
    – Tom Goodwillie
    Jun 4, 2015 at 13:39
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    $\begingroup$ The JSL 2014 paper by George Bergman might be helpful: arxiv.org/abs/1301.6383 $\endgroup$
    – Avshalom
    Jun 4, 2015 at 19:43

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As explained in the comments, if $k$ is a field, then $k$-algebra homomorphisms $k^X\to k$ are in bijection with $|k|^+$-complete ultrafilters on $X$ (that is, ultrafilters closed under $|k|$-fold intersections, or equivalently, ultrafilters such that whenever you partition $X$ into $|k|$ pieces one of the pieces must be in the ultrafilter). In particular, they are all projections unless (and only unless) there is a measurable cardinal $\kappa$ such that $|k|<\kappa\leq |X|$ (here I count $\aleph_0$ as measurable), since the least cardinal that supports a $|k|^+$-complete ultrafilter is measurable. When $k$ is not a field, the question seems harder.

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    $\begingroup$ Here's an explanation why the ultrafilter is closed under intersections of $\le |k|$ elements. It's equivalent to prove that if $f:k^X\to k$ is the $k$-algebra homomorphism, $(A_x)_{x\in X}$ are subsets of $X$ with union $\bigcup A_x=A$ and $(\forall x,$ $f(1_{A_x})=0)$ then $f(1_A)=0$. Indeed, assume $f(1_A)=1$. Let $u\in k^X$ map $A_x$ to $x$ and the rest to 0. Then $f(u-f(u)1_A)=0$. But $u-f(u)1_A$ is invertible on $A-A_{f(u)}$, so its image by $f$ is also invertible hence nonzero, contradiction. $\endgroup$
    – YCor
    Jun 4, 2015 at 19:34
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    $\begingroup$ If $A$ is a non-local noetherian domain, then, for $X$ countable every $A$-modulo homomorphism $f:A^X\to A$ zero on $A^{(X)}$ is zero. Indeed write $1=a+b$ with $a,b$ both non-invertible. Then we can write $1=a^nu_n+b^nv_n$, hence $f(s)=f((a^nu_ns)_n)+f((b^nv_ns)_n)$. Since $f$ vanishes on finitely supported functions, $f((a^nu_ns)_n)$ belongs to the ideal generated by $a^n$ for all $n$, hence is zero since $A$ is a noetherian domain and $a$ is non-invertible. Similarly the other side is zero, and hence $f(s)=0$. Thus in this case every $A$-algebra homomorphism $A^X\to A$ is a projection. $\endgroup$
    – YCor
    Jun 4, 2015 at 22:10
  • $\begingroup$ @YCor: Nice argument. It seems likely that with a bit of work a similar sort of argument can be made for arbitrary Noetherian rings (either you can reduce to the field case, presumably if $A$ is $0$-dimensional, or else you can make an argument like yours). And of course this argument works more generally if $X$ is smaller than the least measurable cardinal. What is less clear is whether anything can be said when $X$ admits a countably complete ultrafilter but not an $|A|^+$-complete ultrafilter. $\endgroup$
    – Eric Wofsey
    Jun 5, 2015 at 0:50
  • $\begingroup$ it's known that if $\kappa$ is the smallest cardinal with a $\sigma$-complete (=stable under countable intersections) ultr. then it is measurable (has an ultr. stable under $<\kappa$-intersections. For let $U$ be $\sigma$-complete and assuming $\kappa$ not measurable, there exists $s<\kappa$ and $(X_t)_{t\in s}$ disjoint and not in $U$ such that $\bigcup X_t\in U$. Define $U^*=\{B\subset s:\bigcup_{t\in B}X_t\in U\}$. Then $U^*$ is a $\sigma$-complete ultrafilter on $s$, contradicting the minimality of $\kappa$. This at least helps when $k$ is assumed not too big (but $X$ can be big). $\endgroup$
    – YCor
    Jun 5, 2015 at 6:48
  • $\begingroup$ Thanks, Eric. So in particular, whether there exist nontrivial maps $k^X \to k$ depends only on $X$ and the cardinality of $k$, at least when $k$ is a field. $\endgroup$
    – Tom Leinster
    Jun 6, 2015 at 23:42

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