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Let $k$ be a field and $A$ be a finitely generated (commutative) algebra over $k$. If $A_1$ and $A_2$ are finitely generated $k$-subalgebras of $A$, is it true that $A_1 \cap A_2$ is also finitely generated (as an algebra) over $k$? What if $A$ is a polynomial ring?

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I think you can prove this using some version of Steinitz exchange. –  Qiaochu Yuan Jan 29 '10 at 5:03

2 Answers 2

up vote 5 down vote accepted

Thomas Bayer has found a counter-example using rings of invariants inside polynomial rings.

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I am choosing this as an answer because it answers both questions simultaneously. It seems that Bayer's example requires the dimension of the polynomial ring to be at least 10. I wonder what happens in lower dimensions! –  auniket Jan 29 '10 at 22:23

It is enough to show that the intersection of two finitely generated semigroups inside a finitely generated commutative semigroup is not necessarily finitely generated, for then you can consider the semigroup algebras.

So let $A$ be freely generated by $\{y,z\}\cup\{x_n:n\geq1\}$ subject to the relations $yx_n=x_{n+1}$ and $zx_n=x_{n+1}$ for all $n\geq 1$, and $x_nx_m=x_{nm}$ for all $n,m\geq1$ (notice that $A$ in fact coincides with the given set of generators...). Let $A_1$ be the subsemigroup generated by $y$ and $x_1$, and let $A_2$ be the subsemigroup generated by $z$ and $x_1$. Then $A$, $A_1$ and $A_2$ are finitely generated and commutative, yet the intersection $A_1\cap A_2$ is the subsemigroup of $A$ generated by $\{x_n:n\geq1\}$, which is isomorphic to $\mathbb N$ under the product. This is not finitely generated.

Later: Yemon asks in a comment if one can change this so that the containing algebra is a domain. I think this works: let $A$ be the algebra generated by $\{y,z,u\}\cup\{x_n:n\geq2\}$ subject to the relations $yx_n=x_{n+1}$ and $zx_n=x_{n+1}+u$ for all $n\geq 2$, and $x_nx_m=x_{nm}$ for all $n,m\geq1$, let $A_1$ be generated by $y$ and $x_2$, and let $A_2$ be generated by $z$, $u$ and $x_2$. (I have to remove $x_1$ for otherwise $x_1(x_1-1)=0$)

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Of course, the algebra generated by A has an element y-z which annhilates a large part of the algebra... out of curiosity, is there a way to modify your example so as to get an integral domain for the containing algebra? –  Yemon Choi Jan 29 '10 at 8:33
The modified example is not a domain either! It satisfies u = yu. In any case, thanks for the simple example. –  auniket Jan 29 '10 at 22:04

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