Consider the ring $C = C(X) = C(X; \mathbb{R})$ of continuous functions $f:X\to \mathbb{R}$ where $X$ is a Tychonoff space. This is naturally a lattice ordered ring by setting $f\geq 0$ iff $f(x)\geq 0$ for all $x\in X$ when $f\in C$. In this ring we have naturally defined exponentiation by real numbers, given $f\in C$, $f\geq 0$ and $r \in \mathbb{R}_{>0}$ define $f^r(x) := (f(x))^r$ for all $x\in X$. Then $f^r\in C$.
For $f\in C$ let $Z(f) = \{x\in X : f(x) = 0\}$. We call $I\subseteq C$ a $z$-ideal if $I$ is an ideal and $Z(f) = Z(g)$ with $g\in I$ implies $f\in I$.
If $I$ is a prime or $z$-ideal, then $I$ is absolutely convex in the sense that if $f\in C, g\in I$ and $|f|\leq |g|$, then $f\in I$. This implies that if for $a\in C/I$ we define $a \geq 0$ iff there exists $f\in C$ with $a = f+I$ and $f\geq 0$, then $C/I$ is a lattice ordered ring and the canonical map $C\to C/I$ is a lattice homomorphism.
Case of $z$-ideals.
If $I$ is a $z$-ideal and $a\in C/I, a\geq 0, r\in \mathbb{R}_{>0}$ then define $a^r = (f+I)^r = f^r+I$ where $a = f+I$ and $f\geq 0$. This is well-defined since if $a = f+I = g+I$ and $f, g \geq 0$, then $Z(f-g) = Z(f^r-g^r)$ so that $f^r+I = g^r+I$. This ordering has the following properties:
Let $I$ be a $z$-ideal and $u, v\in C/I$ be such that $u, v\geq 0$, $n\in\mathbb{N}_{>0}$ and $s, t\in\mathbb{R}_{>0}$. Then
- $u^tu^s = u^{t+s}$
- $(u^t)^s = u^{ts}$
- $u < v \iff u^t < v^t$
- $0^s = 0$, $1^s = 1$
- $u > 1, s > t \implies u^s > u^t$
- $0 < u < 1, s > t \implies u^s < u^t$
- $u^n = \prod_{i=1}^n u$
- $u^{1/n}$ is the unique non-negative $n$th root of $u$
Proof of 3: If $u^r = v^r$ then $u = v$ so it suffices to prove $u^r\leq v^r$. Then we use a technical lemma:
Lemma. If $I$ is a $z$-ideal, then $f+I\geq 0$ iff there exists a zero set $Z$ of $I$ with $f(x)\geq 0$ for $x\in Z$.
Let $u = f+I, v = g+I$, $f, g\geq 0$. Since $u\leq v$, there exists a zero set $Z$ of $I$ such that $f(x)\leq g(x)$ for $x\in Z$. Then $f^r(x)\leq g^r(x)$ for $x\in Z$, hence $u^r\leq v^r$. $\square$
The proof of the rest is left to the reader since it can be proved elementary.
Case of prime ideals.
If $I$ is a prime ideal, then $C/I$ is totally ordered. If $a = f+I$, $f\geq 0$, $n\in\mathbb{N}_{>0}$, then $f^{1/n}+I$ is the unique non-negative $n$th root of $a$. So $a^{1/n} = f^{1/n}+I$ is well-defined.
In fact, for the following discussion let $A$ be any totally ordered ring with $n$th roots of non-negative elements.
If $a\geq 0$, $n, m\in\mathbb{N}_{>0}$ then $(a^{1/m})^n = (a^n)^{1/m}$ and $(a^{1/m})^{1/n} = a^{1/mn}$.
Proof: $((a^{1/m})^n)^m = ((a^{1/m})^m)^n = a^n$ so by definition $(a^{1/m})^n = (a^n)^{1/m}$. Similarly, $((a^{1/m})^{1/n})^{nm} = a$ so $(a^{1/m})^{1/n} = a^{1/mn}$. $\square$
If $a\geq 0$, $r = \frac{p}{q}\in\mathbb{Q}_{>0}, p, q\in\mathbb{N}_{>0}$ let $a^r := (a^{1/q})^p$. Then $a^r$ is well-defined.
Proof: If $r = \frac{p}{q} = \frac{s}{t}$, then $$(a^{1/t})^s = (((a^{1/t})^{1/q})^q)^s = (a^{1/tq})^{sq} = (a^{1/tq})^{pt} = (((a^{1/q})^{1/t})^t)^p = (a^{1/q})^p.\ \square$$
Let $A$ be a totally ordered ring with $n$th roots, $u, v\in A$ be such that $u, v\geq 0$, $n\in\mathbb{N}_{>0}$ and $s, t\in\mathbb{Q}_{>0}$. Then
- $u^tu^s = u^{t+s}$
- $(u^t)^s = u^{ts}$
- $u < v \iff u^t < v^t$
- $0^s = 0$, $1^s = 1$
- $u > 1, s > t \implies u^s > u^t$
- $0 < u < 1, s > t \implies u^s < u^t$
- $u^n = \prod_{i=1}^n u$
- $u^{1/n}$ is the unique non-negative $n$th root of $u$
Proof of 3: Let $t = \frac{p}{q}$. If $v^{1/q}\leq u^{1/q}$, then $v\leq u$, so $u < v$ implies $u^{1/q} < v^{1/q}$, hence $u^t < v^t$. $\square$
Note that this proof is different in that we used properties of total order. Again, proof of the rest is elementary and left to the reader.
In particular, $A = C/I$ has well-defined exponentiation by rational numbers. One can ask the following, suppose that $a > 0$, $a\neq 1$ and $r\in\mathbb{R}_{>0}$. Then $S = \{a^t : t < r, t\in\mathbb{Q}_{>0}\}$ and $T = \{a^t : t > r, t\in\mathbb{Q}_{>0}\}$ are sets of positive elements such that either $S < T$ or $T < S$ depending on if $a > 1$ or $a < 1$ (here I use notation that $S < T$ if $x < y$ for all $x\in S, y\in T$).
Proposition. Suppose $I$ is prime, $S, T\subseteq C/I$ are countable and $S < T$. Then there exists $c\in C/I$ with $S \leq c \leq T$ (again, interpret this notation as above).
In particular there exists an element $c$ between sets $S, T$ above. But we can't have $c\in S$ nor $c\in T$, so we in fact have $S < c < T$ or $T < c < S$. Of course this choice doesn't have to be unique. We could then naively define $a^r = c$.
This then defines $a^r$ for all $r \in\mathbb{R}_{>0}$. It satisfies 4-8 above.
- If $f\in C, f\geq 0$ and $r\in\mathbb{Q}_{>0}$ then $(f+I)^r = f^r+I$.
Proof: Let $r = p/q$, then its clear that $(f+I)^{1/q} = f^{1/q}+I$, so $(f+I)^r := ((f+I)^{1/q})^p = (f^{1/q}+I)^p = f^r+I$. $\square$
Question.
Can we define $a^r$ for $a\in C/I$ and $r\in\mathbb{R}_{>0}$, $I$ a prime ideal, so that it satisfies 1-9? If not, then which properties may fail?
