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What are some "useful" sets of functions (rings under pointwise multiplication) $\mathbb{R} \rightarrow \mathbb{R}$ that are unique factorization domains, other than the polynomial ring?

I will quantify usefulness in my context: I'm only interested in functions that form an algebra over $\mathbb{R}$ (or $\mathbb{C}$) in the space of continuous functions, and preferably also forms a dense algebra in $C([a,b])$ with respect to the usual supremum norm.

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What do you mean by "useful"? There are a LOT of useful sets of functions, the list would go on and on. Are you just curious or do you have something in mind? – Patrick Da Silva Nov 16 '12 at 20:33
I've added some clarification. I'm mostly curious, I suppose; however, I cannot really think examples other than polynomials, or perhaps just taking the polynomials and tacking something onto it. – Christopher A. Wong Nov 16 '12 at 20:40
up vote 7 down vote accepted

This is a very nice question.
I'll boldly assert that outside of algebraic geometry there are probably no examples of spaces of functions that are unique factorization domains!
The reason is that most spaces of functions, be they continuous, differentiable or smooth (=$\mathcal C^\infty$) do not even constitute a domain.
Holomorphic functions on a connected open subset $U\subset \mathbb C$ (or on a connected holomorphic manifold of arbitrary dimension) do constitute a domain $\mathcal O(U)$ , but (unless I'm mistaken) are never unique factorization domains.
This is easy to see for the entire functions $\mathcal O(\mathbb C)$ : irreducible entire functions are of the form $az+b$ and $\sin(z)$ is certainly not a finite product of such affine functions.

In algebraic geometry however there do exist affine varieties $X$ for which the regular functions $\mathcal O(X)$ form a unique factorization domain : for example $\mathbb C$ or $\mathbb C^*$.
And affine varieties for which this is not the case, like the quadric in $\mathbb C^3$ given by the equation $z^2-xy=0$.
Deciding which is which for an affine variety is a very delicate problem.

And to finish on a slightly humorous note, let me remark that the domain of trigonometric polynomials $\mathbb R[\cos(\theta), \sin(\theta)]\subset \mathcal C(\mathbb R)$ is not a unique factorization domain because of the reasonably well known formula $$\cos\theta \cdot \cos\theta=(1-\sin \theta)\cdot (1+\sin \theta) $$

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thanks for your comments. I'd hoped that maybe there some other nice examples, but I guess it makes sense why there wouldn't be any common examples outside the realm of algebra. I'm not really an algebraist so I'm actually not familiar with any of the examples from algebraic geometry, either. – Christopher A. Wong Nov 17 '12 at 9:27
Why irreducible entire functions are of the form $az+b$? This may be easy but I didn't get clarification. (I was trying to post a question related to this; but the answer to my question is hidden in your statement, whose clarification I want.) – Groups Dec 11 '14 at 6:36
@groups: entire functions without zeros are invertible and thus not irreducible. On the other hand an entire function $f$ which does have a zero $c\in \mathbb C$ is divisible by $z-c$ and thus can be written $f=(z-c)g$. Hence $f$ is irreducible iff $g$ is a constant $a\in \mathbb C^*$ i.e. iff $f=(z-c)a=az+b$ . – Georges Elencwajg Dec 11 '14 at 7:35
@Georges: Thanks for clarification. – Groups Dec 11 '14 at 7:37
You are welcome, dear @Groups. – Georges Elencwajg Dec 11 '14 at 7:39

The ring of germs of functions holomorphic at the origin?

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This is an interesting ring but it is not a ring of functions. – Georges Elencwajg Nov 16 '12 at 21:31
@GeorgesElencwajg I thought it was close :) – rschwieb Nov 17 '12 at 0:44
Dear rschwieb: yes, I agree, it is close. – Georges Elencwajg Nov 17 '12 at 5:46

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