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Let $a_1,a_2 \in k[x_1,x_2]$, $k$ is a field of characteristic zero. There is a result (that can be found in a paper of Nagata, Corollary 1.3) saying that if $a_1$ and $a_2$ are algebraically dependent over $k$, then there exists a polynomial $h \in k[x_1,x_2]$ such that $a_1=u_1(h)$ and $a_2=u_2(h)$, where $u_1(t), u_2(t) \in k[t]$.

My question is what happens in higher dimensions, namely:

Let $a_1,\ldots,a_n \in k[x_1,\ldots,x_n]$, $k$ is a field of characteristic zero. Is it true that if $a_1,\ldots,a_n$ are algebraically dependent over $k$, then there exist $n-1$ polynomials $h_1,\ldots,h_{n-1} \in k[x_1,\ldots,x_n]$ such that $a_i=u_i(h_1,\ldots,h_{n-1})$, where $u_i(t_1,\ldots,t_{n-1}) \in k[t_1,\ldots,t_{n-1}]$, $1 \leq i \leq n$.

For example, $n=3$, $a_1=a_2=xyz$, $a_3=z$, so in this case $h_1=xyz$ and $h_2=z$.

Remarks: (1) In $k[x_1,\ldots,x_n]$, $n$ polynomials are algebraically dependent iff their Jacobian is zero. (2) Perhaps there is a connection between my above question and the kernel conjecture?

Thank you very much for any help!

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  • $\begingroup$ Though I do not have a specific example, I suspect this false. Similar question with fields ($a_i\in k(x_i)=K$, algebraically dependent does not mean you have $h_i\in K$ and $u_i\in k(t_1,\ldots, t_{n-1})$ with $a_i=u_i(h_j)$). The field question for $n\geq 4$ is false, because there are unirational non-rational varieties of dimension $\geq 3$. For $n=3, k=\mathbb{C}$, this is true, since unirational surfaces are indeed rational over complex numbers. $\endgroup$ – Mohan Aug 6 '17 at 0:13
  • $\begingroup$ Thanks! Perhaps, according to your comment, my question has a positive answer in the following cases: (i) $n=2$, $k$ is a field of characteristic zero (by Nagata's Corollary 1.3). (ii) $n=3$, $k=\mathbb{C}$. $\endgroup$ – user237522 Aug 6 '17 at 0:18
  • $\begingroup$ $n=2$, you do not need characteristic zero. $\endgroup$ – Mohan Aug 6 '17 at 0:20
  • $\begingroup$ Any reference? Actually, in Nagata's paper he assumes that $k$ is algebraically closed of characteristic zero, not just of characteristic zero. $\endgroup$ – user237522 Aug 6 '17 at 0:22
  • $\begingroup$ Any suggestions what should be the right statement (and how to prove it)? Is this known and appears in the literature? $\endgroup$ – user237522 Aug 6 '17 at 0:28

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