TITLE: What are irreducibles in the ring $R$ of entire functions?
QUESTION [5 upvotes]: Here's how far I have gone:
I first find the units in $R$. Let $f$ $ \in $ $R$ be a unit. Then $\exists$ $g$ $\in$ $R$ such that $fg=1$, i.e $f(z)g(z)=1$ $\forall$ $z$ $\in$ $\Bbb C$.
Then, $f(z)$ $=$ $\frac 1{g(z)}$ $\forall$ $z \in \Bbb C$. Hence $f(z),g(z) \neq 0$ $\forall$  $z \in \Bbb C$.
Now to find the irreducibles, let $f \in R$ be an irreducible. Hence whenever 
$f(z)=g(z)h(z)$ for $g,h \in R$, then $g$ or $h$ is a unit.
Let $h$ be the unit and $\alpha$ be a zero of $f$. Then $f(\alpha)=g(\alpha)h(\alpha)=0$ $\Rightarrow$ $g(\alpha)=0$. Hence $z-\alpha$ is a factor of $g$. So $g(z)=(z-\alpha)^ng_1(z)$ for some $n \in \Bbb N$, where $g_1$ is unit. Since $f$ is irreducible, hence $n=1$.
I'm stuck here. What can be my next step?

REPLY [5 votes]: Conclude by saying that
$$f(z)=(z -\alpha) g_1(z) h(z)$$
Since $(z-\alpha)$ is not a unit, you must have that $g_1(z) h(z)$ is a unit.
So irreducibles are exactly elements of the form $(z-\alpha)u(z)$ with $\alpha \in \Bbb{C}$, and $u$ an entire non-vanishing function.