# Abstract Nonsense

## Rings of Functions (Pt. II)

Point of Post: This is a continuation of this post.

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Differentiable Like Subrings

Of course my statement that (most basic) analysis could in some sense be described as studying subrings of $C(X)$ for certain spaces $X$ is hinting at one particular group of subrings in particular, subrings of differentiable ‘type’ functions. Here we mention only two particular examples. Namely, we mention the ring $C^k(U)$ of $k$-times differentiable functions for an open subset $U\subseteq\mathbb{R}^n$ and the set $\text{Hol}(U)$ of holomorphic functions on some open subset $U\subseteq\mathbb{C}$.

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While the purpose of this section is only to introduce the notations for these structures I can’t help but note the following theorem from basic complex analysis. In particular, have you ever wondered why an open connected subset of $\mathbb{C}$ is called a ‘domain’? Well, perhaps the following might elucidate things:

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Theorem: Let $U\subseteq\mathbb{C}$ be non-empty and open. Then, $U$ is connected if and only if $\text{Hol}(U)$ is an integral domain.

Proof: Recall from basic complex analysis that if $f\in\text{Hol}(U)$ is non-zero where $U$ is open and connected then $f^{-1}(\{0\})$ has no accumulation points and thus it’s countable (this follows from the fact that every uncountable subset of a Lindelof space has an accumulation point). Thus, if $f,g\in\text{Hol}(U)$ are non-zero then $(fg)^{-1}(\{0\})=f^{-1}(\{0\})\cup g^{-1}(\{0\})$ is countable and thus not all of $U$.

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Conversely, if $U$ is not disconnected then we can write $U=V\sqcup W$ where $W,V$ are non-empty open subsets of $U$. It’s easy to see then that $\bold{1}_W,\bold{1}_V$ (the characteristic functions) are holomorphic and non-zero yet $\bold{1}_W\bold{1}_V=0$. $\blacksquare$

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Polynomials as Functions

As was mentioned in our motivation for polynomial rings there is a very real distinction between a polynomial $p\in R[x]$ and the obvious function $p\in R^R$ it represents. The example we gave was that as polynomials we have that $1\ne 1+x+x^2\in \mathbb{Z}_2[x]$ yet as functions $1=1+x+x^2\in \mathbb{Z}_2^{\mathbb{Z}_2}$ as is easily checked. That said, there is a natural map $\mathcal{P}:R[x]\to R^R$ which takes each polynomial $p$ to its associated function. It’s easy to check that the definition of $R[x]$ is precisely such that $\mathcal{P}$ is a homomorphism. The question then is, when is it an injection so that we may identify $R[x]$ with $\text{im }\mathcal{P}\subseteq R^R$? Well, as shown above we clearly ‘always’ can’t. In fact, it’s fairly clear that if $R$ is finite then $\mathcal{P}$ can never be injective since $R[x]$ is infinite and $R^R$ is not.

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So, when can we legitimately think of $R[x]$ as being a subring of $R^R$ via the above identification?  I mean, obviously this is a desirable trait. In fact, it is so natural and done so freely in non-rigorous mathematics that most of us would be caught off guard (in our dilletantish days of course) by the subtle need for the distinction. Luckily, the identification is harmless in most of the cases one is most apt to make it, most notably for $\mathbb{R}$ or $\mathbb{C}$. So, what is a sufficient condition on a ring $R$ for which this identification is ‘ok’? Rewording this, we really want to know when $\mathcal{P}$ is injective, or when $\ker\mathcal{P}=\{0\}$. Taking it one step further we come to the simplest interpretation of this problem: under what conditions on a ring $R$ does $a_nr^n+\cdots+a_0=0$ for all $r\in R$ implies $a_0=\cdots=a_n=0$? Taking our ‘known’ (more correctly stated ‘presumed’) cases of $\mathbb{R}$ and $\mathbb{C}$ one may be inclined to guess that a sufficient condition is that $R$ is an (infinite) field. Well, while this is correct it leaves out a lot of very natural rings such as $\mathbb{Z}[x]$ and, thinking ahead, $k[x,y]=(k[x])[y]$ (for some field (not necessarily infinite) $k$). Luckily, both of these are also rings for which the identification is a truthful one. It’s probably obvious at this point from the wishlist of rings that the sufficient (although not necessary) condition I seek is that $R$ is an infinite integral domain. Indeed:

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Theorem: Let $R$ be an infinite integral domain. Then, the map $\mathcal{P}:R[x]\to R^R$ which takes a polynomial to its associated function is an embedding.

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Although the proof isn’t too hard (for those acquainted with rings of fractions and the consequences of the Euclidean algorithm on $k[x]$ the proof should follow fairly quickly) we shall postpone it for now.

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

3. Cartan, Henri. Elementary Theory of Analytic Functions of One or Several Complex Variables. Paris: Éditions Scientifiques Hermann., 1963. Print.

4. Ahlfors, Lars V. Complex Analysis; an Introduction to the Theory of Analytic Functions of One Complex Variable. New York: McGraw-Hill, 1966. Print.