If $R_3$ is a finitely generated $R_2$-module and $R_2$ is a finitely generated free $R_1$-module, can we say that $R_3$ is a free $R_1$-module?

Question

Let $R_1 \subseteq R_2 \subseteq R_3$ be Noetherian rings. If $R_3$ is a finitely generated $R_2$-module and $R_2$ is a finitely generated free $R_1$-module, then $R_3$ is a finitely generated $R_1$-module. I was wondering if we can say that $R_3$ is a free $R_1$-module? if not (as I think so), is there any condition we can add to the statement to make it true?

Answer 1

The answer to the question "can [we] say that $R_3$ is a free $R_1$-module?" is No. Let $R_1=\mathbb Z$, $R_2 = \mathbb Z[x]/(x^2)$, $R_3 = \mathbb Z[x,y]/(x^2,xy,y^2,2y)$.

As for the question "is there any condition we can add to the statement to make it true?" the answer is basically "Nothing nonobvious will work", as one sees by considering what happens in the case where $R_1=R_2$.