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Show that the ring $R$ of entire functions does not form a Unique Factorization Domain (U.F.D)

My try:

I will first check whether $R$ forms an Integral Domain then check whether it is Factorization Domain and ultimately a U.F.D.

I.D. Let $f,g\in R$ be such that $f.g=0$ To check whether $f\equiv0 ;g\equiv 0$ .If I assume that neither $f=0$ or $g=0$ then let $f(x_1)\neq 0;g(x_2)\neq 0$ for some $x_1,x_2\in \mathbb C$.Now since zeros of an analytic function are isolated there will exist open balls $B_1,B_2$ such that $f\neq 0 ,g\neq 0$ on $B_1,B_2$

How to conclude from here that $fg\neq 0$ because it may happen that $B_1\cap B_2=\emptyset $?

How to conclude whether $R$ is UFD or not?

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  • Which entire functions are irreducible? – Igor Rivin May 20 '15 at 04:09
  • It's a basic theorem in complex analysis that for any open connected subset $\Omega$ in $\mathbf C$, the holomomorphic functions on $\Omega$ are an integral domain. In particular, this is true when $\Omega = \mathbf C$. This should be discussed in lots of complex analysis books. – KCd May 20 '15 at 04:11

1 Answers1

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As you note, there is an open set where $f \neq 0.$ On that set, $g\equiv 0,$ and therefore $g \equiv 0$ everywhere. For UFD, you need to know what the irreducibles are. In fact, they are clearly linear polynomials (prove), and since for a UFD factorization you need a FINITE number of factors, functions such as $\sin z$ are NOT uniquely factorizable into irreducibles.

Igor Rivin
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