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I´m trying to figure out some algebraic properties of the ring of the entire functions, so I came across with the following problem that I cannot solve.

For any $n\geq 1$, consider the entire function defined as: $$g_{n}(z)=\prod_{k=n}^{\infty}\bigg(1-\frac{z^{2}}{k^{2}}\bigg), z\in \mathbb{C}$$

How can I show that the ideal generated by this functions in the ring of entire functions $\{fg_{n}: f\in H(\mathbb{C}), n\geq 1 \}$ and that moreover this ideal is not principal?

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  • $\begingroup$ For the first part: every element of the ideal is a sum of the form $$f_1g_{n_1} + ... + f_kg_{n_k}$$ with $n_i$ an increasing sequence and $f_1,...,f_k$ entire. Do you see why this function is of the form $fg_{n_k}$? $\endgroup$ – idok Mar 7 '18 at 18:41
  • $\begingroup$ @idok. Is this something to do with the form of $g_{n}$? Or is it just commutative algebra? $\endgroup$ – Tsk Mar 7 '18 at 21:06
  • $\begingroup$ @Tsk this is because $n<m$ implies that $g_n$ is a multiple of $g_m$ by an entire function, from the definition of $g_n$. $\endgroup$ – idok Mar 8 '18 at 5:11
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This ideal is not even finitely generated. If it were, then it would be generated by $g_1,\ldots,g_N$ for some $N$, But $g_{N+1}$ is not in the ideal generated by $g_1,\ldots,g_N$, since $g_{N+1}(N)\ne0$ but $g_j(N)=0$ for $j\le N$.

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  • $\begingroup$ This does not prove that the ideal is not finitely generated, because it could be generated by a different set of functions, not necessairly all of the form $g_n$ $\endgroup$ – idok Mar 8 '18 at 5:27
  • $\begingroup$ @idok Each element of the ideal is a finite linear combination of the $g_j$ with coefficients in $H(\Bbb C)$. Any finite subset of the ideal is contained in the ideal generated by $g_1,\ldots,g_N$ for some $N$. Therefore if the ideal is finitely generated, it is generated by $g_1,\ldots,g_N$ for some $N$. $\endgroup$ – Lord Shark the Unknown Mar 8 '18 at 7:19
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In my comment, I showed that $I=\{fg_n | f \in H(\mathbb{C}), n \ge 1\}$ was indeed the required ideal. Let us assume it is finitely generated by some $h_1,...,h_n$.

Note that for all $f \in I$ there exists an integer $N$ such that all integers greater than $N$ are roots of $f$.

Since $h_1,...,h_n$ are in $I$, there must be an integer $N$ such that all integers greater than it are roots of all the $h_i$.

Since the $h_i$ are generators, all integers greater than $N$ are roots of all the functions in $I$. But $N+1$ is not a root of $g_{N+2} \in I$. Contradiction!

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