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what are some natural examples of integral domains $A$ together with sequences of elements satisfying $... | f_2 | f_1 | f_0$, but so that $f_i$ divides no power of $f_{i+1}$? So in the spectrum there will be an ascending chain of basic-open subsets.

Of course we could just take the union of the $k[y^{-n} x,y]$ in $k(x,y)$ or the union of the $\mathbb{Z}[2^{-n} x]$ in $\mathbb{Q}(x)$, or a valuation domain whose value group has arbitrary small positive elements, but I don't call these examples natural. What about rings of holomorphic functions on a connected open subset of $\mathbb{C}$?

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up vote 5 down vote accepted

In rings of analytic functions (real or complex), you could take any function with an infinite product expansion, and remove one factor at a time. E.g., you could take $f_0(z)=\sin(\pi z)$, and $f_k(z)=\frac{\sin(\pi z)}{(z-1)(z-2)\cdots(z-k)}$. As indicated in this example, you don't really need to think about infinite products. You can take any function with infinitely many zeros, and divide out by factors corresponding to those zeros (including the multiplicity of each zero, so that taking powers won't add them back in).

(It might be nicer to take $f_0(z)=\sin(\pi z)$ and $f_k(z)=\frac{\sin(\pi z)}{\left(1-\frac{z^2}{1^2}\right)\left(1-\frac{z^2}{2^2}\right)\cdots\left(1-\frac{z^2}{k^2}\right)}$, so that you're actually removing factors from the product expansion.)

You can do this on any connected open subset of $\mathbb{C}$ or $\mathbb{R}$ (connected because you want an integral domain), but another "natural" class of examples that comes to mind is the set of Blaschke products with infinitely many zeros on the open unit disk. In this case, you could take $$f_k(z)=\frac{f_0(z)}{(\phi_{a_1}(z))^{m_1}(\phi_{a_2}(z))^{m_2}\cdots(\phi_{a_k}(z))^{m_k}},$$ where $a_1,a_2,\ldots$ are the distinct zeros of the infinite Blaschke product $f_0$, $m_k$ is the multiplicity of $a_k$, and $\phi_{a_k}$ is a holomorphic automorphism of the disk that sends $a_k$ to $0$ (unique up to a constant multiple of modulus $1$). Thus each $f_k$ is a Blaschke product and has one less zero than $f_{k-1}$.

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Thanks, this is a good example! – Martin Brandenburg Nov 8 '10 at 9:46

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