2
$\begingroup$

Suppose we have an integral (reduced and irreducible) affine curve $X$ over an algebraically closed field $k$. Suppose $t=\frac{f}{g}$ is a rational function on $X$ such that there exists an $n\in \mathbb{N}$ such that $t^n=(\frac{f}{g})^n$ is a regular function, i.e., $t^n\in \Gamma(X,\mathcal{O}_{X})$ (the coordinate ring or ring of regular functions on $X$). Does this imply that $t$ itself is a regular function?

Note that if the curve is non-singular, this is immediate, as in this case $\Gamma(X,\mathcal{O}_{X})$ is a UFD and hence $b^n|a^n$ implies $b|a$ in $\Gamma(X,\mathcal{O}_{X})$. But is this true in general for an integral affine curve?

$\endgroup$
1
  • $\begingroup$ Why is $\Gamma(X, \mathcal O_X)$ a UFD in the non-singular setting? $\endgroup$ Commented Feb 25, 2017 at 16:24

1 Answer 1

3
$\begingroup$

What about $X = V(y^2 - x^3) \subset \mathbb A^2$ and $t = y/x$? Then $t^2 = x$ is regular, but (unless I'm being dumb) $t$ isn't regular itself.

$\endgroup$
2
  • $\begingroup$ No I think your example is correct. Thanks! $\endgroup$ Commented Feb 25, 2017 at 11:09
  • $\begingroup$ Indeed $(0,0)\in X$ so $t$ is not regular, and $t^2=x/1$ is clearly regular. (+1) $\endgroup$ Commented Feb 25, 2017 at 12:54

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.