# transitivity of finitely generated condition

Let $A \subseteq B \subseteq C$ be rings. I know that if $B$ is a finitely generated $A$-module and $C$ is a finitely generated $B$-module, then $C$ is a finitely generated $A$-module. (Proof is in Atiyah or here)

But I have some silly questions, which are some variances of the above. Let $A \subseteq B \subseteq C$ be rings. What can we say about $C$ ($C$ is a finitely generated $A$-module / $C$ is a finitely generated $A$-algebra / others) in the following cases?

1. $B$ finitely generated $A$-module and $C$ finitely generated $B$-algebra
2. $B$ finitely generated $A$-algebra and $C$ finitely generated $B$-module
3. $B$ finitely generated $A$-algebra and $C$ finitely generated $B$-algebra

And in the same situation, is the follwing true?

a. $C$ is a finitely generated $A$-module implies $B$ is a finitely generated $A$-module and $C$ is a finitely generated $B$-module?

b. $C$ is a finitely generated $A$-algebra implies $B$ is a finitely generated $A$-algebra and $C$ is a finitely generated $B$-algebra?

Edit: I noticed that $A\subseteq B$ rings and $B=\sum_{i=1}^n A b_i$ is a finitely generated $A$-module implies $B$ is a finitely generated $A$-algebra, since $B=\sum_{i=1}^n A b_i=A[b_1,\cdots,b_n]$ by checking left, right inclusions. And I also solved (3), since if we let $B=A[b_1,\cdots,b_n], C=B[c_1,\cdots,c_m]$, then $C=(A[b_1,\cdots,b_n])[c_1,\cdots,c_m]=A[b_1,\cdots,b_n,c_1,\cdots,c_m]$ by checking both inclusions.

From this two facts, in questions (1), (2) we can deduce that $C$ is at least a finitely generated $A$-algebra. By using Amitesh's answer of (a), in (1), (3) $C$ need not be a finitely generated $A$-module otherwise $C$ should be a finitely generated $B$-module.

Then the left question is that:
In (2), what is the example of $C$ not being a finitely generated $A$-module?
In (a), decide whether or not $B$ needs to be a finitely generated $A$-module(or algebra).

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-1 The abbreviations are confusing since they make the question harder to read. Also, you have not shown any attempt(s) to solve the question. I agree that a. and b. are very good questions; however, 1-3 are elementary and I think you should have at least tried to solve them before asking here. – Amitesh Datta Jun 5 '11 at 9:56
@Amitesh I edited the abbreviations. And I'm sorry for the easy questions. I didn't know whether it would be easy or hard, so I questioned first. Next time, I will try some more and then ask questions. – Gobi Jun 6 '11 at 2:31
@Amitesh I added what I did with your answers. The problems are fairly reduced, and I'm working on the left. – Gobi Jun 6 '11 at 3:02
Dear Gobi, thank you very much! I should make it clear, however, that there are absolutely no problems with asking easy questions. Virtually all mathematicians have been stumped by an easy question at least once. I just felt that it would have been better for you to think about these questions a little more. I have upvoted your answer now that it is edited. – Amitesh Datta Jun 6 '11 at 4:27
Let me also add it is excellent that you are thinking about these questions. It is always important to play with definitions like this and certainly a few of these questions are non-trivial. – Amitesh Datta Jun 6 '11 at 4:29

I will state the answers (without proof) below. Please try to prove them on your own (they are mostly trivial exercises with the definitions). If you get stuck, please feel free to comment and I (or someone else) will assist you.

1. $C$ is a finitely generated $A$-algebra
2. $C$ is a finitely generated $A$-algebra
3. $C$ is a finitely generated $A$-algebra

a. If $C$ is a finitely generated $A$-module, then $C$ is a finitely generated $B$-module. I leave it as a (little tricky) exercise to decide whether or not $B$ needs to be a finitely generated $A$-module.

b. If $C$ is a finitely generated $A$-algebra, then $C$ is a finitely generated $B$-algebra. However, it is not necessarily true that $B$ is a finitely generated $A$-algebra. For example, we consider the chain $k\subseteq k[x_1x_2,x_1x_2^2,x_1x_2^3,\cdots]\subseteq k[x_1,x_2]$.

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On the other hand, if $C$ is a finitely generated $A$-algebra and if $C$ is also a finitely generated $B$-module, then $B$ is a finitely generated $A$-algebra. In fact, this is proved in the beginning of chapter $7$ of Atiyah and Macdonald. – Amitesh Datta Jun 5 '11 at 9:48
The case is proved with the assumption that $A$ is Noetherian. Is this assumption can be removed? When I checked the proof, if (a) were true ($C$ is a finitely generated $A$-module implies $B$ is a finitely generated $A$-module.) then the proof seemed to remain true. – Gobi Jun 6 '11 at 3:14
Indeed, I forgot the Noetherian assumption. Thank you very much for pointing this out! – Amitesh Datta Jun 6 '11 at 4:26