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Do we know everything about addition and multiplication of cardinalities in choiceless set theory?

For example, let $M$ be a model of $\textsf{ZF}+\textsf{AD}+V=L(\mathbb{R})$, consider the sets $\mathbb{R}$, $\omega_1$, and $\mathbb{R}/\mathbb{Q}$ as interpreted in $M$ (feel free to add more in case this turns out to be trivial), and then consider the "semi-rng" generated by these sets under disjoint union and Cartesian product, modulo equinumerosity in $M$. Can we compute the structure of this semi-rng? Is there any $M\models\textsf{ZF}$ that produces a "freest" semi-rng?

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    $\begingroup$ Woodin's paper The cardinals below $|[\omega_1]^{<\omega_1}|$ seems relevant to your question. $\endgroup$
    – Hanul Jeon
    Commented Apr 27, 2023 at 2:10
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    $\begingroup$ You can have models of ZF in which every partial order emebds into the cardinals, and so there always large sets of pairwise incomparable cardinals. It seems to me that those will have a very crazy structure of arithmetic. But as a whole I'm not sure what is your actual question. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 27, 2023 at 3:50
  • $\begingroup$ @AsafKaragila If we only consider, e.g., finitely many parameter-free definable sets (instead of arbitrary sets), can the semi rng they generate be simply described? At first I was hoping that the structure generated by $\mathbb{R}$ and $\omega_1$ is "the symmetric algebra of the monoid $\{\emptyset,\mathbb{R},\omega_1,\mathbb{R}+\omega_1\}$", but that's far from true. But it still seems interesting if given two sets $A,B$ there is an algorithm to figure out whether they generate a symmetric algebra. And I'm more interested in determinacy model than arbitrary $\mathsf{ZF}$ model $\endgroup$
    – new account
    Commented Apr 27, 2023 at 4:41
  • $\begingroup$ @AsafKaragila But my last question does concern arbitrary $\mathsf{ZF}$ models. An attempt to formalize it: fix a bunch of definitions of cardinalities, modulo provable equinumerosity in $\mathsf{ZF}$; for each model $M$ there is an homomorphism from those definitions to actual cardinalities in $M$; is there an $M$ that is universal, namely all other homomorphisms factor through it $\endgroup$
    – new account
    Commented Apr 27, 2023 at 4:49
  • $\begingroup$ @newaccount You almost never get universality in such a sense when you're working with definitions in a logic as strong as first-order logic; consider definitions like "$0$ if $\mathsf{AC}$ and $1$ if $\neg \mathsf{AC}$." No model of $\mathsf{ZF}$ can simultaneously separate this from both $0$ and $1$. $\endgroup$
    – Noah Schweber
    Commented Apr 28, 2023 at 1:59

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