Proposition 7.8 Atiyah Macdonald says that:

Let $A\subseteq B\subseteq C$ be rings. Suppose that $A$ is Noeterian, that $C$ is finitely generated as an $A$-algebra and that $C$ is either (i) finitely generated as a B-module or (ii) integral over $B$. Then $B$ is finitely generated as an $A$-module.

I try to solve it and here is my solution: It is easy to see $C$ is Noetherian as an $A$-module. On the other hand, $B$ is also an $A$-module (because $B$ is subring of $C$). So $B$ is Noetherian as an $A$-module and we have done.

I know it is wrong but I can not realise where it is. Can you help me point it out?

  • $\begingroup$ By the way, the proposition you were trying to prove is false. You have changed the proposition in transferring it from Atiyah to this post, and judging by your attempted proof, it seems like you were trying to prove this one, not the one in Atiyah. The error you made in transferring is that you wrote "Then $B$ is finitely generated as an $A$-module" instead of "Then $B$ is finitely generated as an $A$-algebra." It is easy to see that your version is false; for example, the rings $k \subseteq k[x] \subseteq k[x]$ with $k$ a field. $\endgroup$
    – user379719
    Oct 21, 2016 at 17:56

1 Answer 1


In general $C$ is not a Noetherian $A$-module. For example $C = \mathbb{Z}[X]$ is a finitely generated $A$-algebra for $A = \mathbb{Z}$ but it is not a Noetherian $A$-module, as $$ \{ \text{polynomials of degree } \leq 1\} \subsetneq \{ \text{polynomials of degree } \leq 2 \} \subsetneq \dots $$ is a strictly increasing sequence of $A$-submodules of $C$.

  • $\begingroup$ If B be a finitely generated A-algebra and A is Noetherian then B is Noetherian. It does not mean that B is A-module Noetherian, so B is noetherian in what meaning? $\endgroup$
    – Soulostar
    Oct 12, 2016 at 11:30
  • $\begingroup$ It means $B$ is Noetherian as a ring (equivalently, as $B$-module). $\endgroup$
    – Dune
    Oct 12, 2016 at 11:33
  • $\begingroup$ Yes, thank you so much. I got it. $\endgroup$
    – Soulostar
    Oct 12, 2016 at 11:34
  • $\begingroup$ Sorry, I have one trouble which is relevent to this one. B is Noetherian as B module then B is finite? Is it true? $\endgroup$
    – Soulostar
    Oct 19, 2016 at 12:55

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