TITLE: Can $\mathbb{R}^{+}$ be divided into two disjoint sets so that each set is closed under both addition and multiplication?
QUESTION [13 upvotes]: Can $\mathbb{R}^{+}$ be divided into two disjoint nonempty sets so that each set is closed under both addition and multiplication?
I know if we only require both sets to be closed under addition then this can be done. For example, this post gives an answer.
Thank you very much!

REPLY [4 votes]: This is an adaptation of the construction in the paper by Daniel Kane (linked in Jorge's answer).
Claim: There exists a non-trivial $\mathbb{Q}$-linear derivation on $\mathbb{R}$.
Before giving the proof, a non-trivial derivation $D$ lets us define a partition of $\mathbb{R}^+$ by
$$
A=\big\{x\in \mathbb{R}^+:D(x) \geq 0\big\},\,\,\,\, B=\big\{x\in \mathbb{R}^+:D(x) < 0\big\}.
$$
The sets $A$ and $B$ are closed under addition because $D$ is $\mathbb{Q}$-linear, and are closed under multiplication by the Leibniz rule. The partition is non-trivial because $D$ is non-trivial: if $x\in\mathbb{R}^+$ satisfies $D(x)\neq 0$, then $D(x)$ and $D(x^{-1})=-x^{-2}D(x)$ have opposite sign.
Proof of the claim: We construct a derivation with $D(\pi)=1$. Consider the set of pairs $(A,D)$, where $A$ is a subring of $\mathbb{R}$ containing $\pi$ and $D:A\to\mathbb{R}$ is a derivation satisfying $D(\pi)=1$. This set is non-empty because it contains $(\mathbb{Q}[\pi],\frac{d}{d\pi})$, and is partially ordered by extension. The set satisfies the hypotheses of Zorn's Lemma, so contains a maximal element $(A,D)$. It must be the case that $A=\mathbb{R}$, because if $x\in\mathbb{R}\backslash A$ we can extend $D$ to $A[x]$: if $x$ is transcendental over $A$ we set $D(x)=0$, and if $x$ is algebraic over $A$ with minimal polynomial $f$ we set $D(x)=-D(f)(x)/f'(x)$, where $D(f)$ is the polynomial obtained by applying $D$ to the coefficients of $f$.