Are there logarithm functions for arbitrary rings?

Question

The logarithm function for $\mathbb{R}$ satisfies $\log xy = \log x + \log y$ whenever both $\log x$ and $\log y$ are defined.

Are their conditions for a ring $R$ which guarantee the existence of a reasonably large $D \subseteq R$ together with a function $L \colon D \rightarrow D$ that satisfies: if $d_{0}, d_{1} \in D$ then $d_{0}d_{1} \in D$ and $L(d_{0}d_{1}) = L(d_{0}) + L(d_{1})$?

If $0 \in D$ then $L$ is the zero function. I am looking for something not quite so trivial. Since the elements of $D$ are closed under multiplication this might have something to do with rings of fractions. Since ring addition is abelian the elements of $D$ will commute.

In this paper Baez, Fritz, and Leinster use convexity and logarithm functions. If $R$ is an arbitrary ring we can replace $[0, 1] \subseteq \mathbb{R}$ with a subset $C \subseteq R$ that satisfies

1. We have $0, 1 \in C$.
2. For all $c, c_{0}, c_{1} \in C$ we have $c c_{0} + (1 - c) c_{1} \in C$.

The set $C$ will work for the convexity parts of the proofs. There are parts, see corollary 4, where the nonnegative reals are used. We can use $C^{*}$ where $C^{*}$ is the smallest set satisfying

1. We have $C \subseteq C^{*}$.
2. If $c \in C$ and $c^{*}_{0}, c^{*}_{1} \in C^{*}$ then $cc^{*}_{0} +(1 - c)c^{*}_{1} \in C^{*}$.
3. For all $r \in R$; if there exists a $c \in C \smallsetminus \{ 0 \} $ with $cr \in C^{*}$ then $r \in C^{*}$.

With a logarithm function we could create a notion of entropy for a ring where the value of the entropy is an element of the ring.

Answer 1

You can define the logarithm on Banach algebras using the Taylor series whenever it converges, and more generally in any topological ring where the Taylor series makes sense and converges. This is already useful for e.g. finite-dimensional real matrices. But IIRC the correct statement is that $\log(ab) = \log(a) + \log(b)$ if everything is defined and $a, b$ commute.

There are some obstructions for more general rings, e.g. if $R$ is a commutative ring then a logarithm function needs to give a homomorphism from some subgroup of $R^{\times}$ to some subgroup of $R$ and there may not be a nonzero such homomorphism (consider $R = \mathbb{F}_p$ for $p$ prime).

It's far from clear to me that it's meaningful to generalize entropy much beyond $\mathbb{R}$. All of the motivations I know for defining entropy are pretty specific to the real numbers, and in particular order structure is very important.

Answer 2

Here is an elaboration of what Qiaochu has indicated in his answer. Let $R$ be a complete topological commutative ring, which is also a $\mathbb{Q}$-algebra, and whose topology is induced by some ideal $I$ (i.e. the powers $I^k$ are a fundamental system of open neighborhoods of $0$). Then, for every $a \in I$, the sequence defined by $s_n = \sum_{k=1}^{n} (-1)^{k+1} \frac{a^k}{k}$ is Cauchy, since for $m>n$ we have that $s_m - s_n \in I^{n+1}$, hence converges to some element of $R$, which we might call $\ln(1+a)$. We get a map $\ln : 1+I \to I$.

We can also define an exponential map in the other direction, namely $\exp: I \to 1+I$ mapping $a \mapsto \sum_{k=0}^{\infty} \frac{a^k}{k!}$. The Binomial Theorem and the Cauchy product imply $\exp(a+b)=\exp(a) \exp(b)$, so that $\exp$ is a homomorphism of groups $(I,+) \to (1+I,*)$. One can prove that $\exp$ and $\ln$ are inverse to each other, so that in particular $\ln(ab)=\ln(a)+\ln(b)$ (which seems to be quite messy to prove directly, but of course possible) and $(I,+) \cong (1+I,*)$.

In fact, it suffices to check the equations $\exp(\ln(1+a))=1+a$ and $\ln(\exp(a))=a$ in the universal example, the ring of formal power series $R=\mathbb{Q}[[x]]$ (using the continuous homomorphism $\mathbb{Q}[[x]] \to R$ defined by $x \mapsto a$). So we only have to check that the coefficients match. But these are the same as in the corresponding well-known equations of analytic power series over $\mathbb{C}$. Direct proofs are also possible, but probably more messy.