Existence of generic zeros

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.

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 $$u_1,...,u_n$$.
• 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. Aug 30 at 9:13
• 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 $k(u')$. Sep 1 at 9:42
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.