Let $T_R$ be the theory of rings in the language $\{+,\cdot,-,0,1\}$. Let $A$ be the set of one-variable sentences which imply commutativity over $T_R$, and let $B$ be the set of one-variable sentences implied by commutativity over $T_R$. Each set is infinite; for $A$ this is a consequence of Jacobson's theorem (e.g. "$\forall x(x\cdot x=x)$" is in $A$), and for $B$ see this earlier question of mine. Note that "one-variable sentence" allows the variable in question to appear under multiple quantifiers, which is the standard convention in finite-variable logic.
Both $A$ and $B$ are lattices in the usual way; moreover, $A$ has a least element ($\perp$) and $B$ has a greatest element ($\top$). However, beyond this it's not clear to me what $A$ and $B$ "look like." To keep things specific, here is a pair of questions I'm interested in:
Does $A$ have a greatest element?
Does $B$ have a least element?
I'm happy with an answer to either; since I don't know which is easier, I'm putting them both here. I'd also be happy(er!) with a result connecting the two questions, e.g. "$A^{\mathit{op}}\cong B$" wouldn't answer either question but would decrease the number of questions by 1.
It's easy to see that commutativity itself is not equivalent to any sentence in $A$ or $B$. This is because if $X$ and $Y$ are rings with (up to isomorphism) the same one-element-generated subrings then $X$ and $Y$ agree on all one-variable sentences, and so freely adjoining two noncommuting variables to $\mathbb{Z}$ gives a noncommutative ring satisfying all the same one-variable sentences as the commutative ring $\mathbb{Z}[X]$. (Note, however, that a one-variable sentence can still hold of all one-element-generated subrings of a given ring while failing of that ring!) But this doesn't answer the question.
