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Let $k$ be a field (if necessary assume $k$ to be algebraically closed). Let $A$ be a finitely generated $k$-algebra and let $B$ be a subalgebra of $A$. Remark that $B$ doesn't have to be noetherian, let alone finitely generated (consider $A=k[x,y]$, $B=k[x,xy,xy^2,\dotsc]$).

Question 1. Does it follow that $\mathrm{dim}(B) \leq \dim(A)$?

This is true when $B$ is also finitely generated and $A$ is an integral domain. Because then we may use the expression of the dimension as the transcendence degree of the field of fractions. Is it also true when $B$ is not assumed to be finitely generated, or if $A$ is not an integral domain?

Question 2. Assume that $B$ is noetherian. Does it follow that $B$ is finitely generated?

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    Sure, otherwise the question is boring and my remarks would be nonsense.2012-12-05

1 Answers 1

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$1.$ As I've mention here if $A$ is a polynomial ring the answer to your question is yes.

If $A$ is an integral domain and $B$ is a $k$-subalgebra (not necessarily finitely generated) the answer is still yes. See here.

If $A$ is not an integral domain, I can't see any reason to be true.

$2.$ In this paper the authors claim that for $\dim B>1$ noetherianity is not enough to imply that $B$ is affine. (But you know, they are japanese and japanese are very good in (counter)examples, so we can trust them.)

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    Oh sorry, I meant question 2.2013-01-04