What are irreducible and prime elements of the ring of holomorphic functions? 
Is $H(\mathbb C)$, the ring of all holomorphic functions in $\mathbb C$, a UFD? What are the irreducible and prime elements in it?

Answer:
If $f(z)= z-a$ where $a \in \mathbb C$ then $f(z)$ is irreducible and if $\deg(f)\geq 2$ then $f(z)$ is reducible, so probably polynomials like $f(z)$ mentioned above are the only irreducible elements. Not sure of the prime elements?
$e^z \in H(\mathbb C) $ and $e^z \neq p_1\cdots p_n$ (cannot be written as finite product of irreducible elements) so $H(\mathbb C) $ not a UFD. 
Are these observations correct?
 A: A nonzero holomorphic function has a well-defined multiset of zeros, and $f/g$ gives a well defined holomorphic function if and only if the multiset of zeros of $g$ is contained in the multiset of zeros of $f$ (in the sense of every complex number having at least the same multiplicity as zero of $f$ as it has as zero of $g$). In this case $g$ divides $f$ in $H(\Bbb C)$, and in particular the invertible holomorphic functions (divisors of the constant function $1$) are precisely those that have no zeros at all.
Every $f$ with a zero in $a$ is divisible by $x\mapsto x-a$, which is irreducible (for any divisor $g$ either $f/g$ is invertible, if $g(a)=0$, or $g$ itself is invertible, if $g(a)\neq0$). So up to invertible factors, the set of irreducible elements is the set of these degree $1$ polynomial functions $x\mapsto x-a$. But then any holomorphic function with infinitely many zeros, like indeed $\sin$, cannot be written as product of irreducibles, and $H(\Bbb C)$ is not an UFD. It is however a GCD-domain and even a Bézout domain.
