TITLE: Why is the ring of holomorphic functions not a UFD?
QUESTION [22 upvotes]: Am I correct or not? I think that a ring of holomorphic functions in one variable is not a UFD, because there are holomorphic functions with an infinite number of $0$'s, and hence it will have an infinite number of irreducible factors! But I am unable to get a concrete example. Please give some example.

REPLY [26 votes]: You are perfectly right: the ring of entire functions $\mathcal O(\mathbb C)$ is not a UFD. Here is why:  
In a UFD a non-zero element has only finitely many irreducible (=prime) divisors and this does not hold  for our ring $\mathcal O(\mathbb C)$.
Indeed the only primes in $\mathcal O(\mathbb C)$ are the affine functions $z-a$ and on the other hand the function $\sin(z)$ is divided by the infinitely many primes  $z-k\pi\; (k\in \mathbb Z) $. 
The same proof shows that for an arbitrary domain $D$ the ring $\mathcal O(D)$ is not a UFD, once you know Weierstrass's theorem which implies that there exist non identically zero holomorphic functions in $D$ with infinitely many zeros. 
NB
It is sufficient for the proof above to show that the $z-a$ are irreducible in $\mathcal O(\mathbb C)$ (you don't need that there are no other irreducibles). And that is easy: if $a=0$ for example, write $z=fg$ and you will see that $f$ (say) has no zero and is thus a unit $f\in \mathcal O(\mathbb C)^*$.