Is the ring of entire functions on $\mathbb C$ a UFD?

Question

Is the ring of entire functions on $\mathbb C$ a UFD ?

I know the irreducible elements are of the form $f(z)=(z-\alpha)g(z),$ where g is a unit in the ring.But I have no idea how to prove or disprove it? Please help.

Answer 1

If you know that the irreducible elements are associates of $z-\alpha$, then the only functions that can be decomposed as a product of irreducible elements are those with finitely many zeros.

If follows that the ring of entire functions is not a UFD, because there are entire function with infinitely many zeros, such as $\sin(z)$.

The ring thus fails to be a UFD because it is not even a factorization domain, let alone have uniqueness of factorization.

Answer 2

See this paper http://www2.math.uu.se/research/pub/Bjermo.pdf by Jonas Bjermo. The punchline is that the primes of your ring are the elements of the form $(z-c)$ for some constant $c$. Thus any function with infinitely many zeros cannot be written as a finite product of primes.