Let $\{a_{ij}\}$ for $i=1,2,3$, and $j=1,...,16$ be algebraically independent elements over some prime field. Let $k$ be a field containing all $a_{ij}$. Then consider $k^{16}$ as $k$-vector space and $U$ be the linear subspace of all vectors orthogonal to the $(a_{i1},...,a_{i16})$. Now let $G\subset GL_{32}(k)$ be the set of transformations of $k[x_1,...,x_{16},t_1,...,t_{16}]$ (the $x_i$ and $t_i$ be indeterminates) that fix $t_i$ for all $i$, and map $x_i$ to $x_i+b_it_i$ for some $(b_1,...,b_{16})\in U$. Claim:

$k[x_1,...,x_{16},t_1,...,t_{16}]^G$ is not finitely generated.

This is the famous first counterexample to Hilbert's conjecture known as the fourteenth problem (of his 23 published problems). I'm trying to understand the proof that this actually works, and I'm already a little confused with some arguments / steps in the first some sentences. Maybe you can help me out there.

We set $u=t_1\cdots t_{16}$, $v_j=u/t_j$, $w_j=v_jx_j$, and finally $y_i=\sum a_{ij}w_j$ for $i=1,2,3$. Of course the $t_i$ are invariant under $G$, and the $y_i$ are also. Now he writes:

"Since $k[w_1,...,w_{16}]=k[y_1,y_2,y_3,w_4,...,w_6]$, we have $$k(x_1,...,x_{16},t_1,...,t_{16})=k(y_1,y_2,y_3,x_4,...,x_{16},t_1,...,t_{16}),$$ and $G$ is the set of linear transformations $\sigma$ of $k[y_1,.y_2,y_3,x_4,...,x_{16},t_1,...,t_{16}]$ such that $\sigma(y_i)=y_i$, $\sigma(t_j)=t_j$ and $\sigma(x_l)=x_l+b_lt_l$ with arbitrary elements $b_4,...,b_{16}$ of $k$."

I don't really get where that "Since" is coming from. I can show both equalities, but I'm not really using the first one to show the 2nd one except for that it tells me that $w_1,w_2,w_3$ are in the right hand side. So I think there maybe is a "direct argument" other than calculation here to get the 2nd equation. Maybe a "direct relation" between those two equations. I thought about taking the quotient field on both sides, but that didn't seem to work for me, either.

It also made me think about the way I tried to prove equality of two rings / fields of the form $k[...]=k[,,,]$, or $k(...)=k(,,,)$. For the second type, i.e. in this case the 2nd equation above, I just showed that $x_1,x_2,x_3$ is in the rhs, and that $y_i$ are in the lhs. But that's exactly the same thing I would do were there no round parentheses but []. Am I doing anything wrong here?

Edit: To make this more precise, I think what I'm asking here is: Let $f_1,...,f_r,g_1,...,g_s\in k[x_1,...,x_n]$. What would I need to do to show that $k[f_1,...,f_r]=k[g_1,...,g_s]$, and in comparison, what would be needed to have $k(f_1,...,f_r)=k(g_1,...,g_s)$? Is there a fundamental difference between those 2 methods?

Also, I don't quite get why we can use arbitrary $b_i$ here now, but I didn't really think about that yet, so I'm gonna do that first and try stuff myself there, too :)

Thank you in advance for reading this wall of text, I'm sorry it got so long.


1 Answer 1


First, the theorem is that the invariant ring is Not finitely generated.

"Since" will help you, because Assuming $k[w_1,...,w_{16}]=k[y_1,y_2,y_3,w_4,...,w_{14}]$, we will get $k(x_1,\ldots,x_{16},t_1,\ldots,t_{16})=k(w_1,...,w_{16},t_1,...,t_{16})$. Now use first equality to get $k(y_1,y_2,y_3,w_4,...,w_{16},t_1,...,t_{16})$ which equal the desired field.

Any choice of vector $(b_1,...,b_{16})$ in U will work and give you that $y_i$ is G-invariant.

  • $\begingroup$ Hi, of course you're right, that was a huge typo up there. $\endgroup$ Feb 9, 2013 at 9:13

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