# 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?

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?

• The condition can be that $R_2$ is a free $R_1$ module and $R_3$ is a free $R_2$ module – Xuqiang QIN Oct 29 '16 at 23:00
• If $R_2$ is a free $R_1$ module and $R_3$ is a free $R_2$ module, then you have your result.. – RKD Oct 29 '16 at 23:01
• Thank you. I know that but I was wondering if there is another condition as I cannot prove that $R_3$ is free over $R_2$. – user279941 Oct 30 '16 at 1:12

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 $R_1=R_2$.
• The last case $R_1=R_2$ is enough! Thank you. – user279941 Oct 30 '16 at 16:07