Why is the ring of holomorphic functions not a UFD?

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.

Solution 1:

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 $f\in \mathcal O(\mathbb C)^*$.

Solution 2:

Consider $f(z):=\sin(\pi z)=\pi z\prod_{j=1}^{+\infty}\left(1-\frac{z^2}{j^2}\right)$. It's an holomophic function, and the zeroes are the integers. If $f$ could be written as a product of a unit and a finite product of irreducible elements of $\mathcal O(\Bbb C)$, then one of these irreducible elements would have infinitely many roots. It's not possible, for example if $r_n$ are the roots of this elements $g_1$, then $g_1(z)=(z-r_1)h_1(z)$, where $h_1(z)=\prod_{j=2}^{+\infty}(z-r_j)a_j$ and none of these two elements is an unit.

Solution 3:

Look at $f(z)=\sin(z)$ for example...