Are there non-continuable functions that become continuable when raised to some power?

Question

Let $X$ be a complex algebraic variety (integral, separated scheme of finite type over $\mathbb C$) and $U\hookrightarrow X$ an open subvariety. I will say that $f\in\mathscr O_X(U)$ is continuable if there exists some $g\in \mathscr O_X(X)$ with $g|_U=f$. My question is simply: Can it happen that $f$ is not continuable, but there exists a $k\in\mathbb N^+$ with $f^k$ continuable?

I was debating with a friend who thinks it is possible but could not give an example, while I thought that it is impossible but wasn't able to give a satisfying proof. So, I'd be really glad if you could provide one of the two (that is, a simple example of such a phenomenon or a proof that it can not happen).

Answer 1

What about the following: $X = Spec \mathbb{C}[x_1,x_2]/(x_1^3+x_2^2)$, and $U=D(x_1)$, i.e $Spec \mathbb{C}[x_1^{\pm 1},x_2]/(x_1+(x_2/x_1)^2)$. Take $f =(x_2/x_1)$ on it. It is not a global section, but surely $(x_2/x_1)^2 = -x_1$ is.

Answer 2

On the other hand, for smooth curves it is true: If $K$ is the function field, then $f \in K$ is continuable iff $f \in \mathcal{O}_X(X)$ iff $\mathrm{ord}_P(f) \geq 0$ for all closed points $P \in X$, where $\mathrm{ord}_P$ denotes the valuation at $P$. Since $\mathrm{ord}_P(f^k)=k \cdot \mathrm{ord}_P(f)$, it follows that $f$ is continuable iff $f^k$ is continuable.

What about smooth varieties in higher dimensions?

Answer 3

This answer is a generalization of Martin's answer.

Claim: If $X$ is a normal complex algebraic variety, then any function $f\in \mathcal{O}_X(U)$ with $f^k$ continuable must be continuable too.

Proof:

Notice first that wlog we can choose $U:=\{x\in X:f\in\mathcal{O}_{X,x}\}$ to be the maximal open subset of $X$ such that $f|_U$ is regular. If $U\neq X$, by algebraic Hartog theorem $X\setminus U$ is of purely codimension $1$, which means $X\setminus U$ is a finite union of distinct prime Weil divisors $D_i$.

Since $f^k$ is regular on $X$, we have $\operatorname{ord}_{D_i}(f^k)\geq 0$, hence $\operatorname{ord}_{D_i}(f)=\operatorname{ord}_{D_i}(f^k)/k\geq 0$, which means $f\in \mathcal{O}_{X,\zeta_i}$, where $\zeta_i$ is the generic point of $D_i$. Therefore, $\zeta_i \in U\cap D_i=\emptyset$, we get a contradiction.

Edit: Actually here is a much simpler proof:

Suppose $X$ is covered by affine open subsets $U_j$, then $\mathcal{O}_X(U_j)$ is integrally closed by the normality of $X$. Since $f$ is regular at the generic point of $X$, $f$ belongs to the quotient field of $\mathcal{O}_X(U_j)$ for every $j$. Since $\mathcal{O}_X(U_j)$ is integrally closed and $f^k$ is regular on every $U_j$, so $f$ is also regular on every $U_j$, hence $f$ is regular on $X$.