What are the operations $\oplus$ on $\mathbb{R}$ that can replace the addition, while keeping the ring structure with multiplication ?

Or, to generalize a bit: what are the operations $\oplus$ on positive reals ($\mathbb{R}_{>0}$) such that ($\mathbb{R}_{>0}, \oplus)$ is a cancellative commutative semigroup, and $\times$ distributes over $\oplus$?
Then we can probably extend with $0$ as neutral for $\oplus$, add inverse elements for $\oplus$, and define multiplication on them by $(\ominus a)\times b \stackrel {\text{def}} = \ominus (a \times b)$.

Some operations that work are $a \oplus b \stackrel {\text{def}} = (a^p + b^p)^{1/p}$, for $p \in \mathbb{R}$, $p \neq 0$. Are they the only ones?

(If they are the only ones, maybe it could be proved using Cauchy's functional equation, where the regularity to avoid pathological solutions would come from distributivity?)

Notes on these $(a^p + b^p)^{1/p}$ operations:

  • $\oplus$ derives from $+$ by the $f(x) = x^p$ bijection from $\mathbb{R}_{>0}$ to itself. Equivalently, one can use a $\log$ to go from $+$ to $\times$, then a $\log$ with a different base and on the opposite direction: this gives one of those $\oplus$ operations.
  • When $p$ is an odd integer, $\oplus$ with $\times$ is a ring on $\mathbb{R}$.
  • When $p \rightarrow +\infty$, $a \oplus b$ converges pointwise to $\max(a,b)$, and when $p \rightarrow -\infty$, $a \oplus b$ converges pointwise to $\min(a,b)$. $\max$ and $\min$ lose the cancellation property, but multiplication is still distributive upon them.
  • $(a^p + b^p)^{1/p}$ reminds of the $p$-norm (for $p > 1$): this is not a coincidence, but I won't extend this already too long question any further :-).
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    $\begingroup$ Even on $\mathbb{Z}$, there are other ring structures with the same multiplication. Indeed, the multiplicative monoids of $\mathbb{Z}$ and of $\mathbb{Z}[x]$ (polynomials) are isomorphic, because both rings are UFDs with countably many primes and only $\pm{1}$ as units. So, the ring structure on $\mathbb{Z}[x]$ transports to one on $\mathbb{Z}$ with the usual multiplication but a different addition. $\endgroup$ Commented Aug 4, 2022 at 16:15
  • $\begingroup$ @GeoffreyTrang Thanks! I am not much familiar with polynomial rings. So if I want an example of a different addition on $\mathbb{Z}$, the recipe is: take the list of irreductible polynomials of $\mathbb{Z}[x]$, map them to the primes in $\mathbb{Z}$, then look at what the addition in $\mathbb{Z}[x]$ gives in $\mathbb{Z}$? $\endgroup$ Commented Aug 4, 2022 at 22:59
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    $\begingroup$ The operation $a \oplus b=(a^p+b^p)^{1/p}$ is the conjugate of $+$ by the multiplicative automorphism $f(x)=x^p$. There are $2^{\aleph_0}$ multiplicative automorphisms of the form $f(x)=x^p$. But there are $2^{2^{\aleph_0}}$ multiplicative automorphisms of $(\mathbb R_{>0},\cdot)$ altogether, so there are a lot more distinct conjugates of $+$ than you have indicated. $\endgroup$ Commented Aug 5, 2022 at 5:28
  • $\begingroup$ @KeithKearnes Thanks. If I understand well, multiplicative automorphisms of $\mathbb{R}_{>0}$ are given by permutations of a basis of $\mathbb{R}$ as a $\mathbb{Q}$ vector space. The answer here is also handy: math.stackexchange.com/questions/3147683/… $\endgroup$ Commented Aug 5, 2022 at 11:25
  • $\begingroup$ This is an interesting question. One thing I noticed is that the map $(-)\otimes 0:\mathbb \to\mathbb R$ satisfies Cauchy functional equation, but I guess you already figured that out $\endgroup$ Commented Oct 6, 2022 at 21:50


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