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Let $u,v \in \mathbb{C}[x]$.

Assume that $\mathbb{C}(u,v)=\mathbb{C}(x)$. Then by this answer, there exists $\alpha,\beta,\gamma \in \mathbb{C}$ such that $\langle u-\alpha,v-\beta \rangle = \langle x-\gamma \rangle$, where $\langle * \rangle$ is the ideal generated by $*$.

Is this result can be generalized to $\mathbb{C}[x,y]$?

First attempt Let $s,t,u,v \in \mathbb{C}[x,y]$. Assume that $\mathbb{C}(s,t,u,v)=\mathbb{C}(x,y)$. Is it true that there exist $\alpha,\beta,\gamma,\delta,\epsilon_1,\epsilon_2 \in \mathbb{C}$ such that $\langle s-\alpha,t-\beta,u-\gamma,v-\delta \rangle = \langle x-\epsilon_1,y-\epsilon_2 \rangle$?

Second attempt: Let $u,v \in \mathbb{C}[x,y]$. Assume that $\mathbb{C}(u,v)=\mathbb{C}(x,y)$. Is it true that there exist $\alpha,\beta,\gamma,\delta \in \mathbb{C}$ such that $\langle u-\alpha,v-\beta \rangle = \langle x-\gamma,y-\delta \rangle$?

Are there counterexamples for both attempts?

Thank you very much!

Edit: Is the converse also true? Namely: Let $u,v \in \mathbb{C}[x,y]$ and assume that there exist $\alpha,\beta,\gamma,\delta \in \mathbb{C}$ such that $\langle u-\alpha,v-\beta \rangle = \langle x-\gamma,y-\delta \rangle$ ($u$ and $v$ generate a maximal ideal). Then $\mathbb{C}(u,v)=\mathbb{C}(x,y)$.

I guess the answer to the edit is positive, adjusting this one-dimensional answer to our two-dimensional case.

Remarks about the edit:

(1) A generalized edit question: Let $u_i \in \mathbb{C}[x,y]$, $1 \leq i \leq n$, $2 \leq n \in \mathbb{N}$, and assume that there exist $\alpha_i, \gamma, \delta \in \mathbb{C}$ such that $\langle u_i-\alpha_i \rangle_{i=1}^{n} = \langle x-\gamma,y-\delta \rangle$ (the $u_i$'s generate a maximal ideal). Then $\mathbb{C}(u_1,\ldots,u_n)=\mathbb{C}(x,y)$.

(2) A special case of the edit: Let $u,v \in \mathbb{C}[x,y]$ and assume that there exist $\alpha,\beta,\gamma,\delta \in \mathbb{C}$ such that $\langle u-\alpha,v-\beta \rangle = \langle x-\gamma,y-\delta \rangle$ ($u$ and $v$ generate a maximal ideal). Assume, in addition, that the Jacobian of $u$ and $v$ is a non-zero scalar, namely, $Jac(u,v):=u_xv_y-u_yv_x \in \mathbb{C}^{\times}$. Then $\mathbb{C}[u,v]=\mathbb{C}[x,y]$. Indeed, $(u,v)$ is a Jacobian pair, then $(u-\alpha,v-\beta)$ is clearly also a Jacobian pair and, by assumption, $\langle u-\alpha,v-\beta \rangle$ is a maximal ideal of $\mathbb{C}[x,y]$, hence by a known theorem $(u-\alpha,v-\beta)$ is an automorphic pair, and then $(u,v)$ is an automorphic pair, $\mathbb{C}[u,v]=\mathbb{C}[x,y]$.

The later assumption on the invertibility of the Jacobian of $u$ and $v$ does not follow from the former assumption that $u$ and $v$ generate a maximal ideal; for example, $u=x$ and $v=y+xy$. From $y=v-xy=v-uy$, we see that $\langle u,v \rangle = \langle x,y \rangle$, and the Jacobian of $u$ and $v$ is not invertible (not a scalar). Notice that $\mathbb{C}(u,v)=\mathbb{C}(x,y)$ because, $y=\frac{v}{1+u}$, so the question in the edit may have a positive answer.

Any comments about the edit are welcome!

Xander Henderson
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user237522
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  • What does the angle bracket $\langle \rangle$ notation mean? – EffingLoveMath Jun 22 '23 at 17:49
  • @DanielDonnelly, thank you for your comment. It means the ideal generated by the elements in that $\langle \rangle$. I will edit it now. – user237522 Jun 22 '23 at 17:51
  • Adapting the proof will be tricky since sets of zeros of a polynomial or polynomials can be closed algebraic curves. – julio_es_sui_glace Jun 22 '23 at 18:26
  • @JulesBesson, thank you. Please, do you think that the proof of the one-dimensional case can be adapted to at least one of my attempts? – user237522 Jun 22 '23 at 18:43
  • Both statements are correct. – Mohan Jun 22 '23 at 18:51
  • You could try replacing $\mathbb{C}$ by the field $\mathbb{C}(x)$ sometimes this works – julio_es_sui_glace Jun 22 '23 at 18:55
  • @Mohan, thank you very much!! – user237522 Jun 22 '23 at 19:19
  • @JulesBesson, thank you very much, too!! You probably hinted in your first comment that both attempts are correct? Does your second comment apply only for the second attempt? – user237522 Jun 22 '23 at 19:21
  • No I'm not sure to be honest it was only an idea, @Mohan seems to know he is probably more competent than me – julio_es_sui_glace Jun 22 '23 at 19:25
  • @JulesBesson, thanks. Mohan knew how to prove the result in the link, he mentioned that result in https://math.stackexchange.com/questions/2803632/a-sufficient-and-necessary-condition-for-mathbbcfx-gx-mathbbcx?noredirect=1&lq=1 – user237522 Jun 22 '23 at 19:28
  • @Mohan, please: (1) Are there infinitely many such $\alpha,\beta,\ldots$? I think yes. (2) Is the following statement also true: Let $u,v,w \in \mathbb{C}[x,y]$. Assume that $\mathbb{C}(u,v,w)=\mathbb{C}(x,y)$. Is it true that there exist $\alpha,\beta,\gamma,\delta,\epsilon \in \mathbb{C}$ such that $\langle u-\alpha,v-\beta,w-\gamma \rangle = \langle x-\delta ,y-\epsilon \rangle$? I think yes. – user237522 Jun 22 '23 at 19:57

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Let $A\subset B$ be integral domains with $B$ a finitely generated $A$- algebra. Assume that the fraction fields are the same. Then one can find an $s\neq 0$ in $A$ such that $A_s=B_s$. Now you can easily check that for any maximal ideal $M$ of $A$ not containing $s$, $MB$ is a maximal ideal of $B$. In your case, further use Hilbert Nullstellensatz to get what you want.

Mohan
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  • Thank you very much! I think I understand most of your proof and it is nicer than the proof in the link. What I do not understand is why one can find $s \neq 0$ in $A$ such that $A_s=B_s$. I guess this is a well-known property for a ring extension $A \subseteq B$ as described above (integral domains, $B$ f.g. $A$-algebra, same fraction fields). – user237522 Jun 22 '23 at 21:56
  • @user237522 Let $B=A[y_1,y_2,\ldots y_n]$. Fraction fields are same says $y_i=a_i/b_i$ for $a_i,b_i\in A$. Take $s=\prod b_i$. – Mohan Jun 23 '23 at 01:47
  • Thank you, I now understand. Nice pure algebraic proof, as I like. – user237522 Jun 23 '23 at 11:00
  • Mohan, please what about the other direction? Namely, let $u_1,\ldots,u_n \in \mathbb{C}[x,y]$, $n \geq 2$, generate a maximal ideal in $\mathbb{C}[x,y]$. Then, $\mathbb{C}(u_1,\ldots,u_n)=\mathbb{C}(x,y)$? – user237522 Aug 23 '23 at 17:19
  • Is it possible to adjust your one-dimentional proof? https://math.stackexchange.com/questions/2803632/a-sufficient-and-necessary-condition-for-mathbbcfx-gx-mathbbcx?noredirect=1&lq=1 It seems to work here also, with the right adjustments. I will try to write the details. – user237522 Aug 23 '23 at 17:24
  • Please, would you take a look at my new edit? – user237522 Mar 30 '24 at 22:42