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The idea of a theory of algebraic geometry over the "field with one element" $\mathbb{F}_{1}$ is to give a fully faithfully embedding of categories $$\mathsf{Sch}_{\mathbb{Z}}\hookrightarrow\mathsf{Sch}_{\mathbb{F}_{1}}$$ of the category of schemes into a larger category of "$\mathbb{F}_{1}$-schemes". The latter would have a zero-dimensional terminal object, $\mathrm{Spec}(\mathbb{F}_{1})$, and also satisfy a number of expected properties, such as having $\mathbb{F}_{1}$-vector spaces correspond to pointed sets.

In view of this last point, one of the simplest approaches to such a theory is to start directly by defining $\mathbb{F}_1$-vector spaces to be pointed sets, and then define $\mathbb{F}_{1}$-algebras, the analogue of rings in this theory, to be monoids in the category of $\mathbb{F}_{1}$-modules. These turn out to be "monoids with zero", and the corresponding notion of scheme is given by "$\mathcal{M}_{0}$-schemes". All this is developed by Connes–Consanni in Schemes over $\mathbb{F}_1$ and zeta functions, arXiv:0903.2024, and also by Berkovich in two preprints.

Now, there's an "obvious" forgetful functor from (semi)rings to $\mathbb{F}_{1}$-algebras, given by sending a (semi)ring $(R,+,\cdot,1,0)$ to $((R,\cdot,1),0)$. However, this functor is very far from being fully faithful, as there are lots of (semi)rings with the same underlying multiplicative structure.

On the other hand, one could in principle imagine somehow devising a crazy functor $\mathsf{Rings}\hookrightarrow\mathsf{Alg}_{\mathbb{F}_{1}}$ which would be fully faithful (e.g. one could wonder whether maybe first passing (via some appropriate construction) to $\lambda$-rings, thought sometimes to be "rings equipped with descent data to $\mathbb{F}_{1}$", and then from there to $\mathsf{Alg}_{\mathbb{F}_{1}}$ would solve the problem). This naturally leads to the following question:

  • Can one prove (be it constructively or not) whether there's a fully faithful embedding of categories $\mathsf{Rings}\hookrightarrow\mathsf{Alg}_{\mathbb{F}_{1}}$?

(Also, what about a fully faithful embedding $\mathsf{Semirings}\hookrightarrow\mathsf{Alg}_{\mathbb{F}_{1}}$?)

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    $\begingroup$ I don't agree with the first sentence. This would mean that $\mathbb{F}_1 \to \mathbb{Z}$ is an epimorphism. We just want to have a forgetful (left adjoint) functor. $\endgroup$
    – Martin Brandenburg
    Commented Aug 13, 2021 at 4:46
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    $\begingroup$ @MartinBrandenburg Ah, that's a very good point! I understand that the functor $\mathsf{Semirings}\to\mathsf{Mon}$ sending a semiring to its underlying multiplicative monoid has a left adjoint, the monoid semiring functor $\mathbb{N}[-]$. Do you know if the forgetful functor $\mathsf{Semirings}\to\mathsf{Alg}_{\mathbb{F}_{1}}$ also admits a left adjoint? (I'm pretty sure it does, but I haven't worked it out; I ask because I think there's a good chance you might have already figured this out before) $\endgroup$
    – Emily
    Commented Aug 13, 2021 at 5:00
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    $\begingroup$ The forgetful functor $\mathrm{Alg}(T) \to \mathrm{Alg}(T')$ associated to a morphism of monads $T' \to T$ always has a left adjoint when $\mathrm{Alg}(T')$ has coequalizers. You write it down using presentations. A reference is Durov's thesis, Prop 3.3.19. $\endgroup$
    – Martin Brandenburg
    Commented Aug 13, 2021 at 11:54
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    $\begingroup$ The contracted monoid algebra is the left adjoint of the forgetful functor from semirings to monoid. No fancy category theory is needed. You take the free $\mathbb N$-module (or some would say semimodule) with basis the non zero elements of the monoid M and define the product on the basis to agree with the monoid product where you interpret the zero products as the zero in the module. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 13, 2021 at 12:18
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    $\begingroup$ If you Google contracted semigroup algebra you will find the ring theoretical version but it is the same for semirings $\endgroup$
    – Benjamin Steinberg
    Commented Aug 13, 2021 at 12:20

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