TITLE: What does $\mathrm{Spec}(\mathbb{Z})\times_{\mathrm{Spec}(\mathbb{F}_{1})}\mathrm{Spec}(\mathbb{Z})$ look like?
QUESTION [9 upvotes]: See also: What does $\mathrm{Spét}(H\mathbb{Z})\times_{\mathrm{Spét}(\mathbb{S})}\mathrm{Spét}(H\mathbb{Z})$ look like?.

$\newcommand{\F}{\mathbb{F}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}$One of the most mysterious objects in mathematics is the elusive "field with one element", and coming with it is the arithmetic curve $\mathrm{Spec}(\mathbb{Z})\times_{\mathrm{Spec}(\mathbb{F}_{1})}\mathrm{Spec}(\mathbb{Z})\cong\mathrm{Spec}(\mathbb{Z}\otimes_{\mathbb{F}_{1}}\mathbb{Z})$. I want to know what such a thing would look like, and hence am trying to work it out in one particular model for geometry over $\mathbb{F}_1$, that of binoids. Here are some definitions (for the question, it suffices to know 1–3 only).

A binoid is a commutative monoid $M$ together with an absorbing element $0$.
An ideal of $M$ is a subset $I$ such that

$0\in I$.
If $a\in I$ and $r\in M$, then $ra\in I$.


An ideal $I$ of $M$ is prime if it is proper and whenever $ab\in I$ then $a\in I$ or $b\in I$.
The spectrum $\mathrm{Spec}(M)$ of a binoid $M$ is the set of all prime ideals of $M$.
The Zariski topology on $\mathrm{Spec}(M)$ is the topology generated by the collection $\{D(I)\}$ with $D(I)=\mathrm{Spec}(M)\setminus V(I)$, where
$$V(I)=\{\mathfrak{p}\in\mathrm{Spec}(M):I\subset\mathfrak{p}\}.$$
A distinguished open of $\mathrm{Spec}(M)$ is a set of the form $D_f=D(\{f\})$ for some $f\in A$. These form a basis for the Zariski topology on $\mathrm{Spec}(M)$.
A binoidal space is a pair $(X,\mathcal{O}_X)$ with $X$ a topological space and $\mathcal{O}_X$ a sheaf of binoids on $X$.
An affine binoid scheme is a binoidal space of the form $(\mathrm{Spec}(M),\mathcal{O}_{M})$, where $\mathcal{O}_{M}$ is defined on the distinguished opens by
$$\mathcal{O}_{M}(D_f)=M_f.$$

For example, every ring $R$ has an associated binoid, given by forgetting the addition of $R$. We have also a tensor product of binoids, and the tensor product $\mathbb{N}\otimes_{\mathbb{F}_{1}}\mathbb{N}$ is isomorphic to a countable direct sum of the multiplicative monoid of positive natural numbers, $(\mathbb{N}\setminus\{0\},\cdot,1)$, adjoined with an absorbing element $\{0\}$. It looks like this:

The binoid $\Z\otimes_{\F_{1}}\Z\cong(\Z\setminus\{0\},\cdot)^{\oplus{\N}}\sqcup\{0\}$ is pictured in the same way, we just add negative numbers.
What I'd like to ask is: What are the $\mathrm{Spec}$'s of the main objects involved here, including $\mathrm{Spec}(\mathbb{N})$ and $\mathrm{Spec}(\mathbb{Z})$ (where $\mathbb{N}=(\mathbb{N},\cdot,1)$ and similarly for $\mathbb{Z}$), and, above all,
\begin{align*}
\mathrm{Spec}(\mathbb{N})\times_{\mathrm{Spec}(\mathbb{F}_{1})}\mathrm{Spec}(\mathbb{N}) &\cong \mathrm{Spec}(\mathbb{N}\otimes_{\mathbb{F}_{1}}\mathbb{N}),\\
\mathrm{Spec}(\mathbb{Z})\times_{\mathrm{Spec}(\mathbb{F}_{1})}\mathrm{Spec}(\mathbb{Z}) &\cong \mathrm{Spec}(\mathbb{Z}\otimes_{\mathbb{F}_{1}}\mathbb{Z}),
\end{align*}
the sets of prime ideals of the binoids $\N\otimes_{\F_{1}}\N\cong(\N\setminus\{0\},\cdot)^{\oplus{\N}}\sqcup\{0\}$ and $\Z\otimes_{\F_{1}}\Z\cong(\Z\setminus\{0\},\cdot)^{\oplus{\N}}\sqcup\{0\}$?

REPLY [2 votes]: $\newcommand{\Z}{\mathbf{Z}}$Marc Hoyois pointed on a related question of mine on MathOverflow that I conflated the tensor products involved in this case. What I have been calling $\mathbf{Z}\otimes_{\mathbf{F}_{1}}\mathbf{Z}$ should have been called $\mathbf{Z}\otimes_{\mathbf{N}_{+}}\mathbf{Z}$, while what I have called $\mathbf{Z}\otimes_{\mathbf{F}_{1}}\mathbf{Z}$ is given by
\begin{align*}
    \mathbf{Z}\otimes_{\mathbf{F}_{1}}\mathbf{Z} &\cong (\mathbf{Z}\setminus\{0\})^+\otimes_{\mathbf{F}_{1}}(\mathbf{Z}\setminus\{0\})^+\\
    &\cong \left[\left(\Z\setminus\{0\}\right)\oplus\left(\Z\setminus\{0\}\right)\right]^+,
\end{align*}
where $X^+=X\sqcup\{0\}$. For $\mathbf{N}\otimes_{\mathbf{F}_{1}}\mathbf{N}$, writing $\mathbf{N}_*:=\mathbf{N}\setminus\{0\}$, we have a non-canonical isomorphism
\begin{align*}
    \mathbf{N}\otimes_{\mathbf{F}_{1}}\mathbf{N} &\cong (\mathbf{N}_*\oplus\mathbf{N}_*)^+\\
    &\cong \left((\mathbf{N},+,0)^{\oplus\mathbf{P}}\oplus(\mathbf{N},+,0)^{\oplus\mathbf{P}}\right)^+\\
&\overset{\text{non-canonically}}{\cong}((\mathbf{N},+,0)^{\oplus\mathbf{P}})^+\\
&\cong\left(\mathbf{N}_*\right)\sqcup\{0\}\\
&\cong\mathbf{N},
\end{align*}
corresponding to a bijection $\mathbf{P}\times\mathbf{P}\cong\mathbf{P}$ of the square of the set of prime numbers with itself.
However, this argument fails
for $\mathbf{Z}$, as we have $\mathbf{Z}\setminus\{0\}\cong(\mathbf{N},+,0)^{\oplus\mathbf{P}}\oplus\mathbf{Z}_2$, and thus there will be an extra factor of $\mathbf{Z}_{2}$ if we try to repeat the same argument, i.e.:
$$\mathbf{Z}\otimes_{\mathbf{F}_{1}}\mathbf{Z}\cong((\mathbf{Z}\setminus\{0\})\oplus\mathbf{Z}_2)^+.$$
Therefore, while we have a non-canonical bijection
\begin{align*}
    \mathrm{Spec}(\mathbf{Z}\otimes_{\mathbf{F}_{1}}\mathbf{Z})&\cong\mathrm{Spec}(\mathbf{Z})
\end{align*}
of sets, the associated monoidal spaces aren't isomorphic, even non-canonically.