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Question: 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 it is not a factorisation domain.

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

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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.

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Jonas Meyer
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    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 07 '10 at 01:44
  • To prove the strict inclusions of the $I_n$, wouldn't it be sufficient to argue that there exists an entire function whose only zeros, counted with their multiplicities are $n,n+1,\dots$? This result is a simple application of Weierstrass factorisation theorem, c.f. Walter Rudin's Real and Complex Analysis, 3rd ed., theorem 15.11. – Akerbeltz Apr 01 '21 at 14:07
  • @Akerbeltz: Existence of $\sin(\pi x)$ mentioned in my comment is even simpler, but my original justification was overkill definitely. – Jonas Meyer Mar 27 '23 at 07:00