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Herstein proved that $S$ can be enlarged to the set of all $a^2 p(a) - a$ with $p$ a polynomial (with integer coefficients).

EDIT. Herstein's set may be maximal. The set can't contain any polynomials whose vanishing would be consistent with the ring containing (nonzero) nilpotent elements, so nothing in $S$ can be divisible by $a^2$. The lower degree terms are also highly constrained by the condition that if there is $p$-torsion then no $p^2$-torsion.

EDIT-2. Again, based on my memory of the $x^n=x$ problem, the point was to have conditions that prevent one from replacing $R$ with something like $R \oplus tQ \oplus uQ \oplus tuQ$ with $Q$ noncommutative and $u^2=ut=tu=t^2 = 0$, or adding a noncommutative torsion component, $R \oplus Q$ with $mQ=0$ for some $m > 1$. I think that generating enough examples to show that Herstein's set is maximal should be easier than showing that Herstein's conditions imply commutativity.

Herstein proved that $S$ can be enlarged to the set of all $a^2 p(a) - a$ with $p$ a polynomial (with integer coefficients).

EDIT. Herstein's set may be maximal. The set can't contain any polynomials whose vanishing would be consistent with the ring containing (nonzero) nilpotent elements, so nothing in $S$ can be divisible by $a^2$. The lower degree terms are also highly constrained by the condition that if there is $p$-torsion then no $p^2$-torsion.

Herstein proved that $S$ can be enlarged to the set of all $a^2 p(a) - a$ with $p$ a polynomial (with integer coefficients).

EDIT. Herstein's set may be maximal. The set can't contain any polynomials whose vanishing would be consistent with the ring containing (nonzero) nilpotent elements, so nothing in $S$ can be divisible by $a^2$. The lower degree terms are also highly constrained.

EDIT-2. Again, based on my memory of the $x^n=x$ problem, the point was to have conditions that prevent one from replacing $R$ with something like $R \oplus tQ$ with $Q$ noncommutative and $t^2 = 0$, or adding a noncommutative torsion component, $R \oplus Q$ with $mQ=0$ for some $m > 1$. The general Jacobson problem is not that much different in that the one-generator subrings are so constrained that this lets you draw conclusions about the two-generator ones (which then imply the problem for the whole ring, since all you need is any two elements to generate a commutative subring). I think that generating enough examples to show that Herstein's set is maximal should be easier than showing that Herstein's conditions imply commutativity.

Herstein proved that $S$ can be enlarged to the set of all $a^2 p(a) - a$ with $p$ a polynomial (with integer coefficients).

EDIT. Herstein's set may be maximal. The set can't contain any polynomials whose vanishing would be consistent with the ring containing (nonzero) nilpotent elements, so nothing in $S$ can be divisible by $a^2$. The lower degree terms are also highly constrained by the condition that if there is $p$-torsion then no $p^2$-torsion.

EDIT-2. Again, based on my memory of the $x^n=x$ problem, the point was to have conditions that prevent one from replacing $R$ with something like $R \oplus tQ \oplus uQ \oplus tuQ$ with $Q$ noncommutative and $u^2=ut=tu=t^2 = 0$, or adding a noncommutative torsion component, $R \oplus Q$ with $mQ=0$ for some $m > 1$. I think that generating enough examples to show that Herstein's set is maximal should be easier than showing that Herstein's conditions imply commutativity.

Herstein proved that $S$ can be enlarged to the set of all $a^2 p(a) - a$ with $p$ a polynomial (with integer coefficients).

EDIT. Herstein's set may be maximal. The set can't contain any polynomials whose vanishing would be consistent with the ring containing (nonzero) nilpotent elements, so nothing in $S$ can be divisible by $a^2$. The lower degree terms are also highly constrained.

Herstein proved that $S$ can be enlarged to the set of all $a^2 p(a) - a$ with $p$ a polynomial (with integer coefficients).

EDIT. Herstein's set may be maximal. The set can't contain any polynomials whose vanishing would be consistent with the ring containing (nonzero) nilpotent elements, so nothing in $S$ can be divisible by $a^2$. The lower degree terms are also highly constrained.

EDIT-2. Again, based on my memory of the $x^n=x$ problem, the point was to have conditions that prevent one from replacing $R$ with something like $R \oplus tQ$ with $Q$ noncommutative and $t^2 = 0$, or adding a noncommutative torsion component, $R \oplus Q$ with $mQ=0$ for some $m > 1$. The general Jacobson problem is not that much different in that the one-generator subrings are so constrained that this lets you draw conclusions about the two-generator ones (which then imply the problem for the whole ring, since all you need is any two elements to generate a commutative subring). I think that generating enough examples to show that Herstein's set is maximal should be easier than showing that Herstein's conditions imply commutativity.