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 $k^X \to k$.

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 $\Phi(\chi_{X_i}) = 1$.

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.

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.

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?