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Let $f=f(x,y),g=g(x,y) \in \mathbb{C}[x,y]$, each of degree $\geq 1$, and $f,g$ are algebraically independent over $\mathbb{C}$ (= their Jacobian $\in \mathbb{C}[x,y]-\{0\}$).

(1) Is there a sufficient condition that will guarantee that $\mathbb{C}(f,g)=\mathbb{C}(x,y)$?

Perhaps it would help if we will consider $f$ and $g$ as polynomials in one variable $y$ over $\mathbb{C}(x) \subset \bar{\mathbb{C}(x)}$?

A related question I have asked is this question, but I have not got an answer for it, so I decided to ask the above related question here, hoping to get some additional help.

A remark: I have just found three relevant papers: 1, 2 and 3, where the last one seems to answer my question (Theorem 1.10, or more accurately, since here the base field is of characteristic zero, Theorem 1.10 in characteristic zero is a result of K.P. Russell "Field generators in two variables", J. Math. Kyoto Univ., 1.

(2) Does it make a difference if we allow $f,g \in \mathbb{C}(x,y)$?

For example, $f=x$ and $g=\frac{1}{y}$ (what is the analog notion of being algebraically independent?).

Any comments are welcome! Thank you.

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  • $\begingroup$ Is there a sufficient condition? sure there are plenty, including tautological ones. One would rather ask if there a necessary and sufficient condition checkable in terms of $f,g$ (say, which would be algorithmic when $f,g$ have rational coefficients). $\endgroup$ – YCor May 9 '18 at 23:14
  • $\begingroup$ Thanks for your comment. Please, could you elaborate on what exactly that condition is (details/ideas not mentioned in the paper of Russell that may help in practice). $\endgroup$ – user237522 May 10 '18 at 5:23
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The two polynomials that you are giving provide a morphism $\tau\colon\mathbb{C}^2\to \mathbb{C}^2$, given by $(x,y)\mapsto (f(x,y),g(x,y))$.

This map $\tau$ is dominant if and only if $f$ and $g$ are algebraically independent (which seems what you already ask).

This map $\tau$ is birational if and only if $\mathbb{C}(f,g)=\mathbb{C}(x,y)$. This is also equivalent to ask that is is generically injective, which means that the preimage of a general point consists of one single point.

In practical, you can extend your morphism to a birational map from $\mathbb{P}^2$ to $\mathbb{P}^2$ by homogenising, and use Bézout Theorem: you find the common roots of the two homogenised polynomials in $\mathbb{P}^2$ (all these are on the line at infinity because the map is a morphism on $\mathbb{C}^2$). You then blow-up these and count the multiplicities of all base-points, including infinitely near ones. Then the map is birational if and only if $d^2-\sum m_i^2=1$, where $d$ is the degree (max of the degree of $f$ and $g$) and the $m_i$ are the multiplicities.

The article you mention are dealing with a related question which takes only one polynomial $f$ and try to find if it is a variable, i.e. if there is $g$ such that $(f,g)$ is birational.

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  • $\begingroup$ Thank you very much! Could you please apply in a specific example what you suggested? In particular, I am curious to know when $f$ and $g$ of the following forms generate $\mathbb{C}(x,y)$: $f=a_{2n+1}y^{2n+1}+a_{2n-1}y^{2n-1}+\cdots+a_3y^3+a_1y$ and $g=b_{2m+1}y^{2m+1}+b_{2m-1}y^{2m-1}+\cdots+b_3y^3+b_1y$, $a_j,b_j \in \mathbb{C}[x]$, namely, $f$ and $g$ are skew-symmetric w.r.t. the involution $(x,y) \mapsto (x,-y)$. $\endgroup$ – user237522 Jun 2 '18 at 19:11
  • $\begingroup$ Actually, I meant to ask when $f=f_{2n+1}+f_{2n-1}+\cdots+f_3+f_1$ and $g=g_{2m+1}+g_{2m-1}+\cdots+g_3+g_1$ generate $\mathbb{C}(x,y)$, where each $f_i,g_i$ is $(1,1)$-homogeneous of degree $i$. (For the above form it can be shown, thanks to Proposition 3.6 of Nagata's paper repository.kulib.kyoto-u.ac.jp/dspace/bitstream/2433/212692/1/…, that $\mathbb{C}[f,g]=\mathbb{C}[x,y]$ so $\mathbb{C}(f,g)=\mathbb{C}(x,y)$). $\endgroup$ – user237522 Jun 19 '18 at 19:30
  • $\begingroup$ The maps that you give (in both cases) are much too general to be able to compute it. You should give an example in practice to check. $\endgroup$ – Jérémy Blanc Jun 22 '18 at 7:28
  • $\begingroup$ Thank you. Is it possible to compute the case $\deg(f)=3$, $\deg(g) \geq 3$? Are there 'general' cases for which a computation is possible? $\endgroup$ – user237522 Jun 24 '18 at 21:20
  • $\begingroup$ If $deg(f)=3$, then the zero set is a cubic. If this one is rational (irreducible with one cusp or a node), then there exists an element $h$ such that $(f,h)$ is birational. After applying the inverse of this, you reduce to the case where $f=x$, which is then easy: $(x,g)$ is birational if and only if $g$ is of the form $\frac{ay+b}{cy+d}$ where $a,b,c,d\in \mathbb{C}(x)$ are such that $ad-bc≠0$. If $f$ is irreducible and not rational, then $(f,g)$ is never birational. If $f$ is not irreducible, you have to decide if you can send $f$ onto a line by a Cremona transformation. $\endgroup$ – Jérémy Blanc Jul 15 '18 at 22:04

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