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Let $\Omega$ be an algebraically closed field of characteristic $0$, $k$ a subfield such that $\mathrm{tr.deg}(\Omega/k)=\infty$. Let $u_1,\dots,u_n,u_{n+1}\in \Omega$ be algebraically independent over $k$, $P$ be a prime(not maximal) ideal of $k[X_1,\dots,X_n]$. Does there exists $x_1,\dots,x_n\in \Omega$ such that the following conditions are satisfied:

$$\begin{aligned}&(1)u_{n+1}=u_1 x_1+\cdots +u_n x_n,\\ &(2)\text{The image of }k(u_1,\dots,u_n)[X]\rightarrow \Omega, X_i\mapsto x_i\text{is isomorphic to } k(u_1,\dots,u_n)[X]/P \end{aligned}$$ where we still denote by $P$ the ideal $P k(u_1,\dots,u_n)[X] $ which is prime since a purely transcendental extension is regular.

The second condition is easy to be satisfied. Actually, if we forget $(1)$ and set $u_{n+1}’:=u_1x_1+\cdots u_nx_n$, then we can prove that $u_{n+1}’$ is transcendental over $k(u_1,\dots,u_n)$(see p212, Introduction to algebraic geometry, S. Lang). But I am not sure whether the converse is true or not. Thanks in advance if anyone could offer some help.

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It seems not possible in general to me. If $P$ is the prime ideal generated by the irreducible polynomial $f=u_1X_1+...u_nX_n\in k(u_1,...,u_n)[X_1,...,X_n]$, then condition (2) would imply $u_1x_1+...+u_nx_n=0$. Am I right?

Edit: I'll try to explain it better as requested: Having $P$ as above (principal ideal defined by that polynomial), condition (2) means, by Noether Isomorphism Theorem, that $P$ is the kernel. This means that the image of $f$, namely $u_1x_1+...+u_nx_n$, must be zero, which is not algebraically independent of $u_1,...,u_n$.

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  • $\begingroup$ Please provide additional details in your answer. As it's currently written, it's hard to understand your solution. $\endgroup$
    – Community Bot
    Aug 30 at 7:52
  • $\begingroup$ Sure, you are right. I have modified the question. Thank you for pointing it out. $\endgroup$
    – Makimura
    Aug 30 at 8:14
  • $\begingroup$ In that case, it seems that the ideal $P\Omega+(u_1X_1+...+u_nX_n-u_ {n+1})\Omega$ shoud have a tuple $(x_1,...,x_n)\in \Omega^n$ satisfying all equations, but this does only guarantee $P\mapsto\{0\}$. The kernel could be larger... I don't know yet. I would vote the question if I could. $\endgroup$ Aug 30 at 9:13
  • $\begingroup$ I cannot comment in your answer due to low reputation, so I'll write here. It's beautiful! By the way, maybe you have to write $u_{n+1}'$ instead $u_{n+1}$ in the definition of $k(u')$. $\endgroup$ Sep 1 at 9:42
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I found the answer from Serge Lang’s book. As I mentioned in the description of the question, take arbitrary $x_1,\dots,x_n\in \Omega$ such that $P$ is the kernel of $k(u_1,\dots,u_n)[X]\rightarrow \Omega, X_i\mapsto x_i$. Let $u_{n+1}’:=u_1x_1+\cdots+u_nx_n$, then $u_{n+1}’$ is algebraically independent over $k(u_1,\cdots,u_n)$. Denote that $k(u’):= k(u_1,\dots,u_n,u_{n+1}’ )(\text{reps. } k(u):= k(u_1,\dots,u_n,u_{n+1}))$. Consider an isomorphism $$\delta:k(u’)\rightarrow k(u),u_{n+1}’\mapsto u_{n+1}.$$ We extend $\delta$ to an isomorphism of $k(u’,x_1,\dots,x_n)$, then the image of $x_1,\dots,x_n$ is the tuple satisfying our two conditions.

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