Let \(f(x_1,\dots,x_n)\) be a nonzero polynomial without constant term in the free associative ring \(\mathbb{Z}\langle x_1,x_2,\dots\rangle\). A ring \(R\) is called an \(f\)-ring if it satisfies the polynomial identity \(f=0\); it is a virtual \(f\)-ring if for every \(n\) infinite subsets \(X_1,\dots,X_n\) of \(R\) there exist \(r_i\in X_i\) such that \(f(r_1,\dots,r_n)=0\). In the paper under review the authors are interested in the problem whether a virtual \(f\)-ring is an \(f\)-ring. They study the case when \(f\) is nonzero modulo the commutator ideal of \(\mathbb{Z}_m\langle x_1,x_2,\dots\rangle\), where \(m\) is the characteristic of \(R\). Under this condition, the main results give: (1) If \(R\) is a left primitive virtual \(f\)-ring, then it is finite; (2) Under a natural condition on the coefficients of \(f\) modulo the commutator ideal of \(\mathbb{Z}_m\langle x_1,x_2,\dots\rangle\), if \(R\) is an infinite semisimple virtual \(f\)-ring, then \(R\) is an \(f\)-ring; (3) If \(f=x_{i_1}^{\alpha_1}\cdots x_{i_t}^{\alpha_t}\), then every infinite virtual \(f\)-ring is an \(f\)-ring; (4) For \(f(x)=x+a_2x^2+\cdots+a_nx^n\), if \(R\) is a unitary infinite virtual \(f\)-ring, then \(R\) is an \(f\)-ring and is commutative. The authors also study infinite rings with the property that every infinite subset contains a potent element (satisfying \(r^n=r\)). They correct a gap in the proof of a theorem of \textit{A. Abdollahi} and \textit{B. Taeri} [Proc. 31st Iranian Math. Conf., Tehran 2000, Univ. Tehran, 23-27 (2000; Zbl 1003.16031)] which states that an infinite ring which satisfies the above property virtually, does satisfy it, and hence is commutative.