Does a power-complete finite pasture exist? Suppose we define a pasture to be an algebraic structure $\langle M, 0, +, \times, \wedge \rangle$ where


*

*$\langle M, 0, +, \times \rangle$ is a ring (not necessarily commutative or unital)

*$\wedge$ distributes over $\times$ on the left: $(a \times b) \wedge c = (a \wedge c) \times (b \wedge c)$

*$\wedge$ distributes $+$ into $\times$ on the right: $a \wedge (b + c) = (a \wedge b) \times (a \wedge c)$


The idea is that a pasture is a bit like a field (in that it consists of a ring with additional structure), but goes off in a slightly different direction (by adding exponentiation instead of division).
Now let's call $x \in M$ a perfect power if $x = y \wedge z$ for some $y, z \in M$. Moreover, let's say that $M$ is power-complete if all of its elements are perfect powers. For example, the trivial pasture $\{0\}$ is clearly power-complete.

Question: Does a nontrivial power-complete finite pasture exist?

I was inspired to ask this question after running a computer search for finite pastures and noticing that they tend to have few perfect powers. In fact, most of the pastures I found had a single perfect power, often (but not always) $0$. If my code is correct, then I have confirmed that no pasture of order $\le 8$ is power-complete, and moreover that no commutative unital pasture of order $\le 10$ is power-complete.
Side note: for $2 \le n \le 10$, the number of non-isomorphic commutative unital pastures of order $n$ is given by $(2, 2, 10, 2, 4, 2, 36, 10, 4)$. This is not a sequence recognized by the OEIS.

Edit: Thanks to a comment by @user60589, I have discovered a bug in my code which invalidates the above results. In fact, there are plenty of examples of power-complete pastures of order $\le 10$.

On any ring $R$ in which all elements are idempotent there is a trivial pasture structure on $R$ defined via
$$ x^y =x $$
for all $x,y \in R$. The left distributivity with multiplication is trivial and the right distributivity with addition is equivalent to the fact that all elements are idempotent.
This pasture structure on $R$ is obviously power-complete.
So for instance for all natural numbers $n$
$$ (\mathbb{Z}/ 2\mathbb{Z} )^n $$
is a power-complete pasture.
