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I read years ago in the great French book "l'aventure des nombres" (Gilles Godefroy, Odile Jacob, 1997) that the axiom of determinacy was incompatible with the axiom of choice. The latter seems to be needed to allow the existence of discontinuous field automorphisms of $C$. As some of you may know, I have been working for some time on an approach of RH based on automorphisms of rigs (rings without negative) whose maximal one I conjecture to correspond to "the" algebraic closure of $Q$.

In other words, denoting by $\mathcal{M}$ the maximal "L-rig", i.e. rig whose elements are L-functions, I expect $\operatorname{Aut}(\mathcal{M})$ to be isomorphic to $\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. The key idea is to define the symmetry group $\operatorname {Sym}(F)$ of a given element $F$ of $\mathcal{M}$ as made of elements of $\operatorname {Aut}(\mathcal{M})$ preserving $F$. Then we want to establish the isomorphy of $\operatorname{Sym}(F)$ and the isometry group of the multiset of non trivial zeroes of $F$. The latter group is to be seen as the automorphism group of the Euclidean domain where this multiset lies, i.e. the critical strip, preserving this multiset. This group consists of isometries of the complex plane, where we identify two such isometries $f$ and $g$ if any non trivial zero has the same image under the action of $f$ or of $g$. Such a group is finite. RH is equivalent to this group being trivial iff $F$ is non self-dual, and of order $2$ otherwise.

That way, one has to restrict the potential candidates in $\operatorname{Aut}(\mathcal{M})$ to elements of finite order or to continuous elements of this absolute Galois group. The elements of finite order are the identity and the conjugates of the complex conjugation. The continuous elements are the identity and the complex conjugation. Such a restriction seems rather artificial, and is not quite satisfying. So, my question is:

Would working with the axiom of determinacy instead of the axiom of choice provide a more natural framework for RH eliminating discontinuous automorphisms of $C$? Has this conjecture been proven to be provable in ZF without the requirement of the axiom of choice?

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    $\begingroup$ Almost whole second paragraph is composed of a single sentence which is so convoluted I'm having hard time understanding what it is trying to say. In particular what's unclear to me is what axiom of determinacy has to do with everything else. One question I have is, without axiom of choice, can you prove a maximal L-rig even exists? I am not aware of a construction, and Zorn's lemma is not available without choice. $\endgroup$
    – Wojowu
    Commented Jan 17, 2021 at 22:16
  • $\begingroup$ Determinacy, indeed. A friend of mine also told me I tended to use too much natural language in my questions, and I have a hard time improving this. I'll try to though. $\endgroup$
    – Sylvain JULIEN
    Commented Jan 17, 2021 at 22:21
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    $\begingroup$ @Rahman.M Although ZF+AC cannot prove any arithmetic statements which ZF can't, the same is not true of ZF+AD (so in principle it could be the case ZF+AD proves RH but ZF does not). It is not clear to me how Sylvain proposes to use AD though. If he only wishes to use that C has no discontinuous automorphisms, we can use a weaker hypothesis (all sets of reals have property of Baire), which I believe does not add any arithmetic consequences to ZF. $\endgroup$
    – Wojowu
    Commented Jan 17, 2021 at 22:29
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    $\begingroup$ I might title this as: "Would the axiom of determinacy provide a good setting for studying rigs with $\operatorname{Aut}(\mathcal{M})\simeq\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$? The current title seems to promise a closer relationship with RH than is justified or explained by the body. $\endgroup$
    – user44143
    Commented Jan 17, 2021 at 22:48
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    $\begingroup$ @SylvainJULIEN Keep in mind that all concrete results gotten will only depend on the large cardinal hypotheses, not on determinacy per se. Specifically, by Shoenfield absoluteness all $\Pi^1_2$ sentences are absolute between $V$ and $L(\mathbb{R})$; if there are sufficient large cardinals in $V$, then $L(\mathbb{R})\models\mathsf{ZF+DC+AD}$, so if we can decide a $\Pi^1_2$ sentence from $\mathsf{ZF+DC+AD}$ then we can decide it the same way from $\mathsf{ZFC+}$ [large cardinals]. And for example the Riemann hypothesis is ($\mathsf{ZF}$-provably equivalent to a sentence which is) $\Pi^0_1$. $\endgroup$
    – Noah Schweber
    Commented Jan 18, 2021 at 0:39

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