Ring of entire functions is integrally closed or not? 
Is the ring $\mathscr{O}(\mathbf{C})$ of entire functions integrally closed (in its field of fractions, the meromorphic functions)? 

I know it's not factorial, but this doesn't exclude the possibility of it being integrally closed.
 A: Yes, $\mathscr{O}(\mathbb{C})$ is a GCD domain, and as such it is integrally closed.
You can also see it directly, if $f \in \mathscr{M}(\mathbb{C})$ satisfies
$$\sum_{k=0}^n a_k(z) f(z)^k \equiv 0$$
with $a_k \in \mathscr{O}(\mathbb{C})$ and $a_n = 1$, if it had a pole of order $m$ in $z_0$, you would have
$$1 + \sum_{k=0}^{n-1} a_k(z)f(z)^{k-n} \equiv 0$$
in a punctured neighbourhood of $z_0$, but
$$\lim_{z\to z_0} \sum_{k=0}^{n-1} a_k(z)f(z)^{k-n} = 0$$
since the $a_k$ are bounded.
A: 
Let $A$ be the ring of entire functions $\mathbb{C} \to \mathbb{C}$. Then $A$ is integrally closed.

First, observe that the field $K$ of meromorphic functions, defined everywhere minus precisely the (isolated) poles, contains $A$, and its elements can all be written as $f/g$ with $f$, $g \in A$ (formal fractions in some sense), so we may identify $K = \text{Frac}(A)$.
Suppose $h \in K$ satisfies the monic $A$-polynomial$$x^n + a_{n-1}x^{n-1} + \dots + a_0,$$with $a_i \in A$ entire functions. Pick arbitrary representatives $f$, $g \in A$ (with $g \neq 0$ not identically zero) with $h = f/g$ formally.
For any zero $z_0$ of $g$, say of finite order $n \ge 1$ since $g \neq 0$, we show that $z_0$ is a zero of $f$ of order at least $n \ge 1$, possibly $\infty$ if $f = 0$. Indeed, if not, then $f/g = F/G$ (formally, i.e. $fG - gF = 0$ as entire functions) for some $F$, $G \in A$ with $G(z_0) = 0$ but $F(z_0) \neq 0$, and we have$$0 = F^n + a_{n-1} F^{n-1} G + \dots + a_0 G^n$$as entire functions, so plugging in $z_0$ gives $0 = F(z_0)^n$, contradiction (this is similar to the proof of the Rational Root Theorem).
It follows that $f(z)/g(z)$ extends holomorphically to the whole plane (since the zeros of $g$ are isolated), say to an entire function $H \in A$: then clearly $f - gH = 0$ identically, so$$h = {f\over{g}} = {H\over1} \in A$$formally, establishing that $A$ is indeed integrally closed.
Alternatively, a meromorphic function $f$ lies in $A$ if and only if $\text{ord}_{z = c} f \ge 0$ for all $c \in \mathbb{C}$. Thus, $A$ is an intersection of DVRs inside $\text{Frac}(A)$. Each DVR is integrally closed, so $A$ is integrally closed.
