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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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    @Amitesh$I$know that kind of easy quotient of a polynomial ring, but since I failed to find the example of (a), I thought my understanding was short. I tried to mod out $k \subseteq k[x,xy,xy^2,\cdots] \subseteq k[x,y]$ with ideals generated by some of $xy, x^2, y^2, x^2 y, y-x^2$, etc. But it seems to be no answer in these.2011-06-09

1 Answers 1

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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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    Ind$e$ed, I forgot the Noetherian assumption. Thank you very much for pointing this out!2011-06-06