A $k[[t]]$-algebra is free and finitely generated independently of the $k[[t]]$-algebra structure

Question

Let $k$ be a field and $R=k[[t]]$ be the formal power series ring. Let $A$ be a ring and let $\phi,\psi:R\to Z(A)$ be ring homomorphisms, where $Z(A)$ is the centre of $A$, so that $A$ is an $R$-algebra in two different ways. Then these ways induce two different structures of an $R$-module on $A$: $$ r\cdot_{\phi} a = \phi(r)a, $$ $$ r\cdot_{\psi} a = \psi(r)a. $$ Is it true that if $A$ is finitely generated and free as a module in the first way, then it is also finitely generated and free as a module in the second way?

Answer 1

No. Take $A = k[[t]]$, $\phi = \operatorname{id}$, and $\psi: t \mapsto 0$. Clearly $A$ is a finitely generated free $A$-algebar via $\phi$, but it's not finitely generated wrt $\psi$.