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?

Update (for the sake of completeness, April 1, 2017): This paper (disclaimer: it's mine) describes the smallest dimensional counterexample to the second question in zero characteristic. In positive characteristic the answer seems to be unknown.

  • I think you can prove this using some version of Steinitz exchange. – Qiaochu Yuan Jan 29 '10 at 5:03
up vote 6 down vote accepted

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

  • 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 $x_1(x_1-1)=0$)

  • 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
  • The parethentical remark "notice that $A$ in fact coincides with the given set of generators" appears to be wrong: for example, $y^2$ and $yz$ are elements of $A$ that are not in the given set of generators. And in fact the whole example is a bit odd: expanding $yx_nx_m$ in two different ways (associativity!), we get $$(yx_n)x_m=x_{n+1}x_m=x_{(n+1)m}=x_{nm+m},$$ while $$y(x_n x_m)=yx_{nm}=x_{nm+1},$$ so $x_{nm+m}=x_{nm+1}$ for all $n,m$. – Victor Protsak Apr 2 '17 at 5:45
  • (Cont'd) Setting $n=1, m=2$, it follows that $x_3=x_4$, hence $x_4=yx_3=yx_4=x_5$, etc, and the whole thing collapses from there. – Victor Protsak Apr 2 '17 at 5:53
  • I have posted a modification of your example that does work as an answer. – Victor Protsak Apr 2 '17 at 22:02

Here is a variant of Mariano's construction that answers the question. We shall construct a commutative semigroup $S$ and its subsemigroups $S_1,S_2$ such that

(*) $S,S_1,S_2$ are finitely generated, but $S_1\cap S_2$ is not.

Then passing to the semigroup algebras $A=k[S]$, $A_1=k[S_1]$, $A_2=k[S_2]$, which satisfy $A_1\cap A_2=k[S_1\cap S_2]$ and have the same finite generation properties as the corresponding semigroups, would yield a counterexample to the question.

Let $T$ be the commutative subsemigroup of $\Bbb{Z}^2$ generated by $x_i=(i,1)$ for $i\geq 0$. Thus $$T=\{(a,b)\in\Bbb{Z}^2: a\geq 0, b\ge 1 \text{ or } a=b=0\}$$ consists of the integer points in the semiopen real cone $\{(a,b)\in \Bbb{R}^2: a\geq 0, b>0\}$ together with the origin. It is easy to see that $T$ is not f.g. Let $S$ be formed by adjoining commuting elements $y,z$ to $T$ subject to the relations $yx_i=x_iy=x_{i+1}$ and $z x_i=x_i z=x_{i+1}$. Set $S_1=\langle T,y\rangle$ and $S_2=\langle T,z\rangle$. Clearly, $S_1$ is generated by $x_0$ and $y$, $S_2$ is generated by $x_0$ and $z$ and $S$ is generated by $x_0, y$ and $z$. Moreover, $S_1\cap S_2=T$ is not f.g., meaning (*) holds.

Where does this come from? First, $T$ is the standard example of a non-f.g. commutative semigroup embedded into an f.g. one. In fact, both $S_1$ and $S_2$ are naturally isomorphic to $\Bbb{Z}_{+}^2$, with $y$, resp. $z$, corresponding to $(1,0)$, and $S$ is the amalgamated product of $S_1$ and $S_2$ over $T$ in the category of commutative semigroups. However, I do not see how to modify this construction to make $S$ an affine semigroup (so that $A=k[S]$ is a domain). In particular, it appears impossible to get a counterexample with $A$ a polynomial ring in this fashion.

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