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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 finitely generated.

This is the famous first counterexample to Hilbert's question/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 $i=1,2,3$, and $j=1,...,16$. 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. (Of course that "since" does make sense after all, since I use something it provides to show the 2nd equality, but somehow I think that's not what he meant there perhaps). 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?

To make this more precise, I think what I'm asking as second part 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?

I know that this question is not really on a mathoverflow level of mathematics, but I posted it on MSE here without answers or comments so far, so I hoped one of you could help me out here.

Thank you in advance!

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Q2: Showing equalities is quite similar for $[]$ and $()$. For, if $F$ is a $k$-algebra and $S,T$ are subsets of $F$, then $k[S]$ is the least $k$-subalgebra of $F$ that contains $S$ (i.e. $k[S]$ is the intersection of all $k$-subalgebras of $F$ containing $S$). Hence $S \subseteq k[T]$ implies $k[S] \subseteq k[T]$. Similarly, if $F$ is a field, then $k(S)$ is the least subextension of $F|k$ that contains $S$. Hence, $S \subseteq k(T)$, also implies $k(S) \subseteq k(T)$. – Ralph Dec 15 '12 at 10:02
@Ralph: Thanks, I was already becoming confused, so at least there's nothing wrong in the stuff I did to show these equalities. – InvisiblePanda Dec 15 '12 at 10:09
@Ralph: What I forgot to say and just remembered: feel free to post this as an answer, since I'm fine now with the other question about a "more direct way", and I'll accept it! – InvisiblePanda Dec 23 '12 at 21:20

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