Does the statement "the $R$-algebra homomorphisms $A \rightarrow R$ are linearly independent whenever $R$ is a field" admit any generalizations?

Question

The following is well-known.

Proposition. Let $R$ denote a field and $A$ denote an $R$-algebra (not necessarily commutative). Then the set of $R$-algebra homomorphisms $A \rightarrow R$ form a linearly independent subset of the $R$-module of all $R$-module homomorphisms $A \rightarrow R$.

Can this be generalized so that $R$ is only assumed to be a commutative ring, perhaps by putting some conditions on $A$?

Answer 1

It certainly holds if $R$ is a domain.

If $R$ has zero-divisors, say $ab=0$ with $a \neq 0 \neq b$, we have the following counterexample:

Consider $\alpha: R[x] \to R, x \to a$ and $\beta: R[x] \to R, x \to b$ and $\gamma: R[x] \to R, x \to 0$

Then $b\alpha-a\beta+(a-b)\gamma=0$ holds in the $R$-module $\operatorname{Hom}_R(R[x],R)$.

Since $R[x]$ is a very nice $R$-algebra, it should be hard to find any sufficient conditions on $A$, besides trivial cases like $A=R$ or $A$, such that $\operatorname{Hom}(A,R)$ is very small.