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.



Existence of generic zeros
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.


