Remark.  [ca-0018]
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    For a commutative ring \(R\), the algebra structure is equivalently an \(R\)-module structure on an associative ring \(A\) with identity, with the same addition, such that \[ r \bullet (x * y) = (r \bullet x) * y = x * (r \bullet y) \qquad (r \in R,\ x,y \in A). \] This is the formulation of [jadczyk2019notes, Definition 1.6, p. 4]. Both formulations require associativity and an identity.

    Proof. 

      Given the central ring homomorphism \(\mathit {1}\), define \(r \bullet x=\mathit {1}(r)*x\). The ring laws and the preservation of addition, multiplication and identity by \(\mathit {1}\) give \[ \begin {aligned} r \bullet (x+y)&=\mathit {1}(r)*x+\mathit {1}(r)*y,\\ (r+s)\bullet x&=\mathit {1}(r)*x+\mathit {1}(s)*x,\\ r\bullet (s\bullet x)&=(\mathit {1}(r)*\mathit {1}(s))*x=(r*s)\bullet x,\\ 1_R\bullet x&=1_A*x=x. \end {aligned} \] Thus the action defines an \(R\)-module. Associativity gives \(r\bullet (x*y)=(r\bullet x)*y\); centrality of \(\mathit {1}(r)\) then gives \((r\bullet x)*y=x*(r\bullet y)\).

      Conversely, given the module formulation, put \(\mathit {1}(r)=r\bullet 1_A\). The module axioms give \(\mathit {1}(r+s)=\mathit {1}(r)+\mathit {1}(s)\) and \(\mathit {1}(1_R)=1_A\). They also give \(\mathit {1}(0_R)=0_A\), by cancellation in \(0_R\bullet 1_A=(0_R\bullet 1_A)+(0_R\bullet 1_A)\). Compatibility with multiplication implies \[ \begin {aligned} \mathit {1}(r)*\mathit {1}(s) &=(r\bullet 1_A)*(s\bullet 1_A)\\ &=r\bullet (1_A*(s\bullet 1_A))\\ &=(r*s)\bullet 1_A=\mathit {1}(r*s),\\ \mathit {1}(r)*x&=r\bullet x=x*\mathit {1}(r). \end {aligned} \] Hence \(\mathit {1}\) is a central, identity-preserving ring homomorphism, and its induced action is the original one. Starting with a central ring homomorphism instead recovers it because \(\mathit {1}(r)*1_A=\mathit {1}(r)\).

      The corresponding Mathlib constructions are Algebra.toModule and Algebra.ofModule. Mathlib also permits semirings; under the ring hypotheses here, the two formulations describe the same structures.