# Proving that Ring of Complex Entire functions is neither Artinian nor Noetherian

Question: To prove that the Ring of Complex Entire functions is neither Artinian nor noetherian.

Proof: Clearly $R$ is not Artinian because it is a commutative integral domain which is not a field, and $R$ is not noetherian because its not a factorisation domain.

Is there a proof of this theorem using the Ascending Chain / Descending Chain condition for Artinian / Noetherian rings?

• – Watson Aug 24 '16 at 21:50

Let $J_n$ be the ideal of entire functions vanishing on the first $n$ positive integers, and let $I_n$ be the ideal of entire functions vanishing on positive integers greater than $n$. Then $$J_1\supset J_2\supset J_3\supset\cdots,$$ $$I_1\subset I_2\subset I_3\subset\cdots,$$ and all of these containments are proper. One way to see that the containments are proper is to use the fact that given any sequence of complex numbers $a_1,a_2,\ldots$, there is an entire function $f$ such that $f(n)=a_n$ for each positive integer $n$. For more on this, see these MathOverflow questions.
• I feel compelled to add that that result I mentioned is overkill for this. To see that the containments of the $J_k$s are proper you could use the polynomials $(z-1)(z-2)\cdots(z-k)$. To see that the containments of the $I_k$s are proper you could divide $\sin(\pi x)$ by those polynomials. – Jonas Meyer Sep 7 '10 at 1:44