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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?

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    Why is $\Gamma(X, \mathcal O_X)$ a UFD in the non-singular setting? – 2017-02-25

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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.

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    No I think your example is correct. Thanks! – 2017-02-25
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    Indeed $(0,0)\in X$ so $t$ is not regular, and $t^2=x/1$ is clearly regular. (+1) – 2017-02-25