For compact and connected $K\subset\mathbb{C}$, let $H(K)$ be the set of functions that extend to a holomorphic function over an open neighborhood of $K$. This answer states that $H(S^1)$ is a UFD (and it seems that it is even a PID). There is a comment stating that the result cannot be generalized to the case for general $K$. I don't understand why, because it seems to me that $H(K)$ is always a PID.
My thoughts: pick arbitrarily $f\in H(K)$, then $f$ is analytic over an open neighborhood $U$ of $K$. We can suppose that $U$ is connected (see $\star$ below). Let $Z$ be the set of zeros of $f$ in $U$, then $Z\setminus K$ is a closed in $U$, and $Z\cap K$ is finite: either statement being not true would imply that $Z$ has an accumulation point in $U$, so $f\equiv 0$ by connectivity of $U$. Dividing $f$ by linear polynomials yields an analytic function $f_0$ that has no zero in $K$. Since $\dfrac{1}{f_0}$ is analytic over the open neighborhood $U\setminus(Z\setminus K)$ of $K$, $f_0$ is a unit in $H(K)$. We see that the only irreducible elements are $\{z-\alpha:\alpha\in K\}$, and it's easy to see that $H(K)$ is a UFD.
Note that, by the Bézout identity for $\mathbb{C}[X]$, $H(K)$ is a Bézout domain, hence by this question, $H(K)$ is a PID.
($\star$) Suppose that $X$ is a locally connected space, $A\subset X$ is connected, and $U$ is an open neighborhood of $A$. Then there exists a connected open neighborhood of $A$ that is contained in $U$.
Proof: For each $x\in A$, pick a connected open neighborhood $U_x$ of $x$ that is contained in $U$. We claim that $\displaystyle\bigcup_{x\in A} U_x$ is connected. Suppose that $\displaystyle\bigcup_{x\in A} U_x = C\cup D$ with $C,D$ open and nonintersecting. By connectivity of $A$, we have $A\subset C$ or $A\subset D$. Suppose that $A\subset C$. For each $x\in A$, we must have $U_x\subset C$ or $U_x\subset D$ by connectivity of $U_x$, so $U_x\subset C$ since $x\in C$. This shows that $C = \displaystyle\bigcup_{x\in A} U_x$.
But the comment to the answer cited in the first paragraph tells me that I'm too optimistic. The result stated in the answer itself says that $H(K)$ is noetherian if $K$ is real semi-analytic (whose meaning, unfortunately, is unclear to me because the link there is broken). In short, $H(K)$ cannot be always PID, or even UFD.
I was wondering that what are the parts where my thoughts go wrong. Thank you for your help.
