6
$\begingroup$

What could be conditions on $k\in\mathbb{C}[x,y,z]$ that would ensure that any polynomial $f\in\mathbb{C}[x,y,z]$ that is algebraically dependent of $k$ is indeed a polynomial in $k$, ie $f\in\mathbb{C}[k]$. References?

$\endgroup$

1 Answer 1

11
$\begingroup$

$\def\CC{\mathbb{C}}$A necessary and sufficient condition is that $k$ cannot be written as $h(\ell(x,y,z))$ for $h \in \CC[t]$ of degree $>1$ and $\ell \in \CC[x,y,z]$. Clearly, this is a necessary condition since, if $k = h(\ell)$, then $k$ and $\ell$ are integrally dependent. We now prove sufficiency.

Let $A = \CC[k]$ and let $B$ be the integral closure of $A$ in $\CC[x,y,z]$. I claim that $B \cong \mathbb{C}[\ell]$ for some $\ell \in \mathbb{C}[x,y,z]$. First of all, $B$ is finite over $A$ and so finitely generated and one-dimensional. Also, it is normal. So $\mathrm{Spec}(B)$ is a smooth curve of some genus and some number of punctures. Since $B \subset \CC(x,y,z)$, the curve is unirational which, for curves, is the same thing as rational. So $\mathrm{Spec}(B)$ is genus zero. Also, $B \subset \CC[x,y,z]$ so $B$ has no units other than $\CC^{\times}$ and we see that $\mathrm{Spec}(B)$ has only one puncture. In short, $\mathrm{Spec}(B) \cong \mathbb{A}^1$ and $B \cong \CC[\ell]$.

In particular, we have $k=h(\ell)$ for some $h \in \CC[t]$. If $h$ has degree $\geq 2$, as noted above, than $k$ and $\ell$ are algebraically dependent and $\ell \not \in \CC[k]$.

Conversely, suppose that $h$ has degree $1$, in which case $B = A$. I claim that, if $f$ and $k$ are algebraically dependent, then $f \in \CC[k]$. Proof: If $f$ and $k$ are algebraically dependent then $p(k) \cdot f$ is integral over $\CC[k]$ for some polynomial $p$. So $p(k) f \in B = \CC[k]$ and we deduce that $f \in \mathrm{Frac} \ \CC[k]$. If $f$ is in $\mathrm{Frac}\ \CC[k]$ but not $\CC[k]$, then the ring $\CC[k,f]$ will contain a unit not in $\mathbb{C}^{\times}$, contradicting that $k$ and $f$ are both in $\CC[x,y,z]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.