The question I have is more of a curiosity, and that is why I decided to post here instead of Mathoverflow.
Before posing the question, let me set up some background.
Background: Let $\Omega$ be a Boolean algebra. For the purposes of this post a measure $\nu$ on $\Omega$ is a finitely additive map $\Omega\to A$ with values in a linear space $A$. If $A$ is a commutative, unital algebra, $\nu$ is multiplicative or spectral if $\nu(E\cap F)= \nu(E)\nu(F)$ and $\nu(1)= 1$. I will denote by $\mathbf{A}(\Omega, V)$ the linear space of $V$-measures on $\Omega$ and by $\mathbf{AS}(\Omega, A)$ the set of spectral measures.
note: the field of scalars is irrelevant; take the real field for the sake of determinateness.
Theorem 1: The functor $V\mapsto \mathbf{A}(\Omega, V)$ is representable, that is, there is a linear space $S(\Omega)$ and a natural isomorphism $\mathbf{A}(\Omega, V)\cong \mathbf{Vect}(S(\Omega), V)$.
$S(\Omega)$ is the space of simple elements. Denote by $\chi$ the universal measure $\Omega\to S(\Omega)$ and call elements of the form $\chi(E)$ characteristic.
note: By Stone duality, $S(\Omega)$ is isomorphic to the linear space of linear combinations of characteristic functions of clopen sets of the Stone space of $\Omega$. Thus the monikers "simple" and "characteristic elements". Stone duality needs the Boolean prime ideal theorem.
We can introduce a commutative unital algebra structure on $S(\Omega)$ by $\chi(E)\chi(F)= \chi(E\cap F)$ and then extending linearly (or juggling around the universal property). $\chi$ is now spectral and in fact the universal spectral measure.
Theorem 2: The functor $A\mapsto \mathbf{AS}(\Omega, A)$ is naturally isomorphic to $\mathbf{CAlg}(S(\Omega), A)$.
In fact, we have a functor $\mathbf{CAlg}\to \mathbf{Bool}$ that sends a commutative, unital algebra to the Boolean algebra of its idempotents. It is now easy to see that this functor is right adjoint to $S$ and, in particular, $S$ is cocontinuous.
Problem: An easy application of the Boolean prime ideal theorem (BPI for short) implies that if $E\in \Omega$ is non-zero then $\chi(E)$ is likewise non-zero. In particular, $S(\Omega)$ is non-trivial ($0\neq 1$). Conversely, it is not difficult to see that if $1\neq 0$ then if $E\in \Omega$ is non-zero then $\chi(E)$ is non-zero. My question is can the non-triviality of $S(\Omega)$ be proved without recourse to the BPI? Say, in ZF alone? ZF + P with some P weaker than BPI?
note: BPI is weaker than AC by a well-known result of Halpern and Levy.
Here is what I managed to do until now. Theorem 2 guarantees that as long as we have a non-trivial spectral measure on $\Omega$, non-triviality of $S(\Omega)$ drops out. $S$ of the initial Boolean algebra (the non-trivial one constituted solely by the bottom and top elements) is the scalar field, thus, if we have an ultrafilter on $\Omega$ we can prove that $S(\Omega)$ is non-trivial. In particular, if $\Omega$ has even just one atom we are done. But I am at a loss as how to proceed in the case of arbitrary atomless algebras. If we can factor some morphism into a Boolean algebra with non-trivial $S$ then the problem would also be solved, but this is really not an advancement over constructing ultrafilters directly.
At this point, and after not managing to get non-triviality of $S(\Omega)$ without BPI I started wondering if non-triviality of $S(\Omega)$ is actually equivalent to BPI, but the only thing I accomplished is to narrow down to the problem of constructing one algebra morphism from $S(\Omega)$ to the real field -- not much, that is.