Ring and $A$-module, but not $A$-algebra Suppose $\mathbb A_i := (A_i, +_i, -_i, \cdot_i, 0_i, 1_i)$, $i\in \{1, 2\}$, are commutative unital rings. Let $\star \colon A_1 \times A_2 \to A_2$ be a map such that $\mathbf A_2:=(A_2, +_2, -_2, 0_2, \star)$ is a $\mathbb A_1$-module.
Can yuo provide an example of $\mathbb A_1, \mathbb A_2, \mathbf A_2$ such that $\mathcal A_2:=(A_2, +_2, -_2, \cdot_2, 0_2, 1_2, \star)$ is not a $\mathbb A_1$-algebra?
 A: You are essentially asking for an example of an abelian group $A$ that can be given the structure of a commutative, unital ring in two different ways $A_1$ and $A_2$ such that the underlying additive group of each one is equal to $A$.  If we have this, then the ring structure $A_1$ makes the additive group $A$ an $A_1$-module, and thus a fortiori makes the ring $A_2$ an $A_1$-module, but not an $A_1$-algebra, since by definition the product in $A_1$ differs from that in $A_2$.
Here is a simple example: let $A = \mathbb{R}^2$ be the indicated additive group, and give it two ring structures:
$$A_1 = \mathbb{C} \cong \mathbb{R}[x]/(x^2 + 1) \qquad
  A_2 = \mathbb{R}[x]/(x^2).$$
Then $A$ is an $A_1$-module (the complex vector space $\mathbb{C}$) but this module structure does not make $A_2$ an $A_1$-algebra, since in fact the element $x \in A_1$ (otherwise known as $i \in \mathbb{C}$) maps to $x \in A_2$, where $x^2 = -1$ before and $x^2 = 0$ after.
The moral is that the structure of an $A_1$-module is far from sufficient to determine a ring structure.
