Equivalence between nilpotent and quasi-nilpotent polynomials (and more) Let $A$ be a commutative unital ring, and $A[X]$ its polynomial ring. We know the following facts:

Proposition $1$. Let $f=a_0+a_1X+...+a_nX^n\in A[X]$. Then:
(i) $f$ is nilpotent iff $a_i\in A$ is nilpotent, $\forall i\in \overline{0,n}$;
(ii) $f$ is idempotent iff $f$ is an idempotent constant in $A$;
(iii) $f$ is a zero divisor (by definition, $fg=0$ for some nonzero $g\in A[X]$) iff $fg=0$ for some nonzero constant $g\in A[X]$;
(iv) $f$ is invertible iff $a_0 \in A$ is invertible and $a_i\in A$ is nilpotent, $\forall i\in \overline{1,n}$.


Definition $2$. Consider $P$ a property. We say that $f\in A[X]$ is a quasi-$P$ polynomial, if $f(a)$ has property $P$ in $A, \forall a\in A$ (i.e. every evaluation in elements of $A$ has property $P$).

It is not difficult to show:

Proposition $3$. Let $P\in \{\text{nilpotent}, \text{zero divisor}, \text{invertible}, \text{idempotent}\}$ be a property. Then every $P$-polynomial in $A[X]$ is also quasi-$P$ polynomial. In addition, if a constant polynomial is quasi-$P$, then is also $P$.


QUESTION: What about the converse of Prop. $3$? Maybe I am asking too much, but: can we characterize (as precise as possible) the rings $A$ where the notions of $P$-polynomial and quasi-$P$ polynomial are equivalent? I am interested for $P$ being one of those $4$ aforementioned properties, and these properties to be treated separately.

Any suggestions or further references are welcome. Thank you in advance!
 A: This is a partial answer, showing my effort so far:
It is not surprising to see that the converse of Proposition $3$ is not generally true. I have come up with these examples:
(i) Let $A:=\{x_1,...,x_n\}$ be finite, and $f:=(X-x_1)...(X-x_n)\in A[X]$. Clearly, $f(a)=0, \forall a\in A$, so $f$ is quasi-zero divisor, quasi-nilpotent, quasi-idempotent. But $f$ itself (due to Proposition $1$) is neither zero divisor nor nilpotent nor idempotent. Let $g:=f+1\in A[X]$, then clearly $g(a)=1, \forall a\in A$, so $g$ is quasi-invertible. But $g$ itself is not invertible in $A[X]$. This shows that for finite rings, the notions are not equivalent.
From now on, let $A$ be an infinite commutative unital ring.
(ii) [$\color{blue}{\text{invertible vs. quasi-invertible}}$]

*

*Let $A$ be a field. If $A$ is algebraically closed, then every nonconstant polynomial has a zero in $A$, so it cannot be quasi-invertible. Thus, we reduced to the constant polynomials, and due to Prop. $3$, the notions trivially coincide. If $A$ is not algebraically closed, then there is a nonconstant polynomial with no zeros in $A$, so that polynomial is quasi-invertible, but not invertible; the notions do not coincide!

*If $A$ is a domain with finitely many units, then take $f\in A[X]$ a quasi-invertible. Since $A$ is infinite, $f$ has to take infinitely many times a specific value $u\in U(A)$. So $f-u$ has infinitely many zeros in $A$=domain, so $f=u$, and thus $f$ is invertible, the notions coincide!

(iii) [$\color{blue}{\text{zero divisor vs. quasi-zero divisor}}$]

*

*Let $A$ be a domain. Take $f\in A[X]$ a quasi-zero divisor. Since $0$ is the only zero divisor of $A$, we have $f(a)=0$, $\forall a\in A$. Because $A$ is an infinite domain, it follows that $f=0$, the notions coincide.

(iv) [$\color{blue}{\text{nilpotent vs. quasi-nilpotent}}$]

*

*Let $A$ be a domain. Take $f\in A[X]$ a quasi-nilpotent. Since $0$ is the only nilpotent of $A$, we have $f(a)=0$, $\forall a\in A$. Because $A$ is an infinite domain, it follows that $f=0$, the notions coincide.

(v) [$\color{blue}{\text{idempotent vs. quasi-idempotent}}$]

*

*Let $A$ be a domain. Take $f\in A[X]$ a quasi-idempotent. Since $0,1$ are the only idempotents of $A$, we have $f(a)\in \{0,1\}$, $\forall a\in A$. Because $A$ is an infinite domain, it follows that $f=0 \lor f=1$, the notions coincide!

