# Abstract Nonsense

## Rings of Functions (Pt. I)

Point of Post: In this post we discuss multiple types of rings of functions including the ring $R^X$ of functions from a set $X$ into a ring $R$ and the subring of $X^R$ of continuous functions when $X$is a topological space  and $R$ is a topological ring.

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Motivation

We now discuss a class of rings which comes up very often in mathematics, rings of functions.  Why are these rings interesting? Well, besides as serving as counterexamples for some, otherwise plausible theorems, they serve as an ambient ring for many other important rings. To be less cryptic, we shall see that such important rings as polynomial rings and the rings $C(\mathbb{R})$ and $C(\mathbb{C})$ sit naturally inside rings of functions.

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Rings of Functions

Let $R$ be a ring and $X$ a non-empty set.  We turn the set $R^X=\left\{f:f:X\to R\right\}$ into a ring by defining the sum and multiplication of functions pointwise. Namely, if $f,g\in R^X$ we define $f+g$ and $fg$ by the rules $(f+g)(x)=f(x)+g(x)$ and $(fg)(x)=f(x)g(x)$. Evidently this turns $R^X$ into a ring with additive identity the zero function $0(x)$. If $R$ is unital then $R^X$ is unital with one function $1(x)$.

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In fact, it’s fairly easy to see that as rings $R^X$ is isomorphic to the product ring $\displaystyle \prod_{x\in X}R$ (i.e. ‘$\#(X)$-copies’ of $R$). Thus, the distinction between the two is purely connotative,  especially when considering it as the ambient ring to very important subrings.

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Similar to polynomial rings we have, for each $x\in X$, the evaluation moprhism $\text{ev}_x:R^X\to R$ given by $\text{ev}_x(f)=f(x)$. This is evidently a homomorphism since $\text{ev}_x(f+g)=f(x)+g(x)=\text{ev}_x(f)+\text{ev}_x(g)$ and $\text{ev}_x(fg)=f(x)g(x)=\text{ev}_x(f)\text{ev}_x(g)$. Moreover, if $S$ is any subring of $R^X$ which contains the constant functions $c_r(x)=r$ then $\text{ev}_x:S\to R$ is surjective since for any $r\in R$ one has $\text{ev}_x(c_r)=c_r(x)=r$. From this and the first isomorphism theorem we have just proved:

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Theorem: Let $R$ be a ring, $X$ a set, and $S$ a subring of $R^X$ which contains the constant functions $c_r$ for every $r\in R$. Then, $S/\mathfrak{m}_x\cong R$ for every $x\in R$ where $\mathfrak{m}_x=\left\{f\in S:f(x)=0\right\}$.

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There is a nice way to picture this geometrically, but it may be hard to describe. To make things extra simple let’s deal with $X=R=\mathbb{R}$ and identify each function $f\in\mathbb{R}^\mathbb{R}$ with its graph $\Gamma_f\subseteq\mathbb{R}^2$. We then picture $\mathbb{R}^\mathbb{R}$ as sitting inside the infinite rectangular prism $\left(\mathbb{R}^2\right)^\mathbb{R}$ in the sense that each ‘sheet’  contains the graph of a function $f$. We can then see that if we fix some point $x\in\mathbb{R}$ and draw the vertical line $\{x\}\times\mathbb{R}$ through $x$ in each sheet that the graph $\Gamma_f$ will intersect this line precisely once. In particular, let’s fix some $\Gamma_{f_0}$  and think about the sheet, call it $\mathcal{S}_{f_0}$, that $\Gamma_{f_0}$ lives in. As we noticed we have that $\Gamma_{f_0}$ intersects the line $\{x\}\times\mathbb{R}$ in $\mathcal{S}_{f_0}$ precisely once, in particular at $f_0(x)$. So, imagine standing at $(x,f_0(x))$ in $\mathcal{S}_{f_0}$ and looking up or down at the sheets above and below you. You can see that in some sheets there is no intersection directly above or below you (i.e. that the graph of the function in that sheet doesn’t intersect $\{x\}\times\mathbb{R}$ at $f_0(x)$, perhaps it intersects at $f_0(x)+1$) and in some sheets there is. We can imagine then that modding out $\mathbb{R}^{\mathbb{R}}$ by $\mathfrak{m}_x$ has the effect of ‘crushing’ all the functions whose points of intersection do lie directly above or below you down onto yourself, so that under the identification you treated as a single point. We see then that after ‘crushing’ all the sheets down in this way for each point on the line $\{x\}\times\mathbb{R}$ we are really left with a single sheet $\mathbb{R}^2$ and all the functions have been crushed to lie only on the line $\{x\}\times\mathbb{R}$ which we evidently identify with $\mathbb{R}^{\mathbb{R}}$. And there we have it, ta-da! A convoluted geometric reasoning to why $\mathbb{R}^\mathbb{R}/\mathfrak{m}_x\ \cong\mathbb{R}$, and with a slight stretch why $R^R/\mathfrak{m}_x\cong R$.

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Rings of Continuous Functions

While $R^X$ serves as an interesting example of a ring from a pure ring theoretic standpoint there are much more ‘natural’ places that rings of functions occur in mathematics. Probably the first place one encounters this ring was in advanced calculus (a kind of belittling term for basic analysis) when the teacher said “the sum and product of continuous functions is continuous. Since obviously the constant one function is continuous this is precisely the assertion that the ring $C(\mathbb{R})$ of continuous functions $\mathbb{R}\to\mathbb{R}$ is a unital subring of $\mathbb{R}^\mathbb{R}$.

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For the more mathematically experienced reader this probably isn’t the extent of one’s exposure to rings of continuous functions. Indeed, given a compact Hausdorff space $X$ one considers the ring $C(X)$  of continuous functions $X\to\mathbb{R}$ as a Banach algebra in the general Stone-Weierstrass theorem. In fact, a lot of more advanced analysis could be described as studying various subrings of $C(X)$ for particular spaces.

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Thus, it’s clear that while the distinction between the ring of functions and product of rings is shaky at best there is a real use to considering rings of continuous functions from a topological space $X$ to $\mathbb{R}$. More generally though we can consider the ring of continuous functions $C(X,R)$ where $X$ is a topological space and $R$ is a topological ring ,i.e. a ring $(R,+,\cdot)$ which is also a topological space such that the maps $(a,b)\mapsto a+b$ and $(a,b)\mapsto a+b$ are continuous as maps $R\times R$ when $R\times R$ is given the product topology and the map $R\to R:a\mapsto -a$ is continuous. In other words, a topological ring is a ring with a topology such that the underlying group structure forms a topological group and the product map is continuous. The first thing we note is that:

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Theorem: Let $X$ be a topological space and $R$ a topological ring, then $C(X,R)$ is a subring of $R^X$ which is unital if $R$ is.

Proof: If $f,g\in C(X,R)$ then $f-g\in C(X,R)$ since $f-g$ is the composition of the maps $R\to R\times R:x\mapsto (f(x),-g(x))$ and $+:R\times R\to R$ both of which are continuous by assumption. Similarly, $fg\in C(X,R)$ since $fg$ is the composition of the maps $R\to R\times R:x\mapsto (f(x),g(x))$ and $\cdot:R\times R\to R$. Evidently if $R$ is unital then $1(x)$ is the identity for $R^X$ and since constant functions are continuous we have that $1(x)\in C(X,R)$. The conclusion follows. $\blacksquare$

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Since the constant functions are in $C(X,R)$ we also know that $C(X,R)/\mathfrak{m}_x\cong R$ by previous theorem.

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We shall see that rings of continuous functions shall serve as prime examples for many of the concepts that shall come up in the future.

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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.

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August 1, 2011 -

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1. […] Let be an infinite integral domain and the ring of all functions with the usual operations. Then the obvious map is an […]

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