In the following, all algebras are associative and unital. Let $J\left(A\right)$ denote the Jacobson radical of an arbitrary algebra $A$. Recall that this is defined as the set of all $a \in A$ such that for each $s \in A$, the element $1 - as \in A$ is invertible.
The following is Dickson's trace criterion for the Jacobson radical (slightly generalized):
Theorem 1. Let $\mathbb{K}$ be a field, and let $n \in \mathbb{N}$. Assume that the integers $1, 2, \ldots, n$ are invertible in $\mathbb{K}$. Let $A$ be a $\mathbb{K}$-algebra that is $n$-dimensional as a $\mathbb{K}$-vector space. Let $a \in A$. Then, $a \in J\left(A\right)$ if and only if every $s \in A$ satisfies $\operatorname{Tr}\left(as\right) = 0$.
Here, the trace $\operatorname{Tr}\left(b\right)$ of an element $b \in A$ is defined as the trace of the endomorphism $A \to A, \ u \mapsto bu$ of the $\mathbb{K}$-vector space $A$.
Note that the "$\Longrightarrow$" direction of Theorem 1 does not require the assumption that $1, 2, \ldots, n$ be invertible; only the other direction requires it.
The definition of $\operatorname{Tr}\left(b\right)$ does not require $\mathbb{K}$ to be a field; it suffices that $A$ is a free $\mathbb{K}$-module. (It even suffices that $A$ is a projective $\mathbb{K}$-module, but I don't want to go that far afield.) Of course, we cannot straightforwardly generalize Theorem 1 to arbitrary commutative rings $\mathbb{K}$, since it would fail even for $A = \mathbb{K}$. But here is an attempt at a generalization that would work:
Conjecture 2. Let $\mathbb{K}$ be a commutative ring, and let $n \in \mathbb{N}$. Assume that the integers $1, 2, \ldots, n$ are invertible in $\mathbb{K}$. Let $A$ be a $\mathbb{K}$-algebra that is a free $\mathbb{K}$-module of rank $n$. Let $a \in A$. Then, $a \in J\left(A\right)$ if and only if every $s \in A$ satisfies $\operatorname{Tr}\left(as\right) \in J\left(\mathbb{K}\right)$.
Question. Is this conjecture correct? If not, is at least one of its two directions correct?
Almost nothing in rschwieb's proof of Theorem 1 generalizes easily to this setting; Artinianity breaks down, and elements of Jacobson radicals don't have to be nilpotent. I'm also not exactly swamped by good examples of Jacobson radicals, so there might be a counterexample much shorter than this post.
The only thing I was able to do is make some questionable headway into the "$\Longleftarrow$" direction of Conjecture 2. Namely, assume that every $s \in A$ satisfies $\operatorname{Tr}\left(as\right) \in J\left(\mathbb{K}\right)$. Consider the commutative ring $\overline{\mathbb{K}} := \mathbb{K} / J\left(\mathbb{K}\right)$, which is well-known to satisfy $J\left(\overline{\mathbb{K}}\right) = 0$. Let $\overline{A}$ be the $\overline{\mathbb{K}}$-algebra $\overline{\mathbb{K}} \otimes_{\mathbb{K}} A$; this is a free $\overline{\mathbb{K}}$-module of rank $n$. Now, consider the projection $\overline{a}$ of $a$ onto $\overline{A}$. Then, our assumption yields that every $s \in \overline{A}$ satisfies $\operatorname{Tr}\left(\overline{a}s\right) \in J\left(\overline{\mathbb{K}}\right) = 0$ (here we are using the fact that surjective ring homomorphisms induce homomorphisms between the Jacobson radicals). Now, a well-known fact from linear algebra (e.g., Corollary 4.1 (b) in my note The trace Cayley-Hamilton theorem) yields that for each $s \in \overline{A}$, the element $\overline{a}s$ is nilpotent, whence the element $1 - \overline{a}s$ is invertible; hence, $\overline{a} \in J\left(\overline{A}\right)$. It sounds reasonable to expect that this entails $a \in J\left(A\right)$, although the exact mechanics of how this should follow eludes me.
