What does $\mathrm{Spec}(\mathbb{Z})\times_{\mathrm{Spec}(\mathbb{F}_{1})}\mathrm{Spec}(\mathbb{Z})$ look like? 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 $\Z\otimes_{\F_{1}}\Z\cong(\Z\setminus\{0\},\cdot)^{\oplus{\N}}\sqcup\{0\}$?
 A: The prime ideals of $(\mathbb{N},\cdot,1,0)$
Let $\mathbb{P}$ be the set of prime numbers. There is a bijection$^1$
$$\mathcal{P}(\mathbb{P}) \to \mathrm{Spec}_{\mathbf{CMon}_0}((\mathbb{N},\cdot,1,0)),~ E \mapsto \langle E \rangle,$$
which maps a set of prime numbers to the ideal generated by it. Explicitly, we have $\langle \emptyset \rangle = \{0\}$ and $\langle E \rangle = \bigcup_{p \in E} p\mathbb{N}$ for $E \neq \emptyset$.
Proof. With the explicit description of $\langle E \rangle$ it is easy to see that $\langle E \rangle$ is indeed a prime ideal with $E = \langle E \rangle \cap \mathbb{P}$. Let $I$ be a prime ideal and $E := I \cap \mathbb{P}$. Clearly, $\langle E \rangle \subseteq I$. Conversely, if $n \in I$, w.l.o.g. $n \neq 0$, we may factor $n$ as a product of prime numbers. Since $I$ is prime, one of the prime numbers, say $p$, must be contained in $I$. Then $p \in E$ and $n \in p\mathbb{N}$, so $n \in \langle E \rangle$. $\square$
The homomorphism $(\mathbb{N},\cdot,1,0) \hookrightarrow (\mathbb{Z},\cdot,1,0)$ induces a bijection on the spectra (you only need to check that a prime ideal $I \subseteq \mathbb{Z}$ satisfies $I = (I \cap \mathbb{N}) \cup -(I \cap \mathbb{N}$), which is easy). So we also have a bijection $\mathcal{P}(\mathbb{P}) \to \mathrm{Spec}_{\mathbf{CMon}_0}((\mathbb{Z},\cdot,1,0))$, $E \mapsto \langle E \rangle$.
Direct sums
Here is a more conceptual explanation and generalization.
Let $(M_i)_{i \in \mathbb{N}}$ be a family of commutative monoids (not with zero at this point), written multiplicatively. Their coproduct $\coprod_{i \in I} M_i$ is the "direct sum" $\bigoplus_{i \in I} M_i \subseteq \prod_{i \in I} M_i$ consisting of those tuples which are $1$ almost everywhere. The inclusions $\iota_i : M_i \hookrightarrow \bigoplus_{i \in I} M_i$ induce maps $\mathrm{Spec}_{\mathbf{CMon}}(\bigoplus_{i \in I} M_i) \to \mathrm{Spec}_{\mathbf{CMon}}(M_i)$ and hence a map
$$\alpha : \mathrm{Spec}_{\mathbf{CMon}}(\bigoplus_{i \in I} M_i) \to \prod_{i \in I} \mathrm{Spec}_{\mathbf{CMon}}(M_i).$$
Conversely, given a family of prime ideals $\mathfrak{p}_i \subseteq M_i$, the ideal $\langle \bigcup_{i \in I} \mathfrak{p}_i \rangle = \bigcup_{i \in I} \langle \mathfrak{p}_i \rangle \subseteq \bigoplus_{i \in I} M_i$ is a prime ideal which restricts to the $\mathfrak{p}_i$ along $\iota_i$. If $\mathfrak{p} \subseteq \bigoplus_{i \in I} M_i$ is any prime ideal, consider some element $m = \prod_{i \in I} m_i \in \mathfrak{p}$. Since $\mathfrak{p}$ is prime, we have $m_i \in \mathfrak{p}$ for some $i$, so $m_i \in \mathfrak{p} \cap M_i$, and since $m$ is a multiple of $m_i$, we see $m \in \langle \mathfrak{p} \cap M_i \rangle$. This shows that $\alpha$ is bijective. (Actually, $\alpha$ is an isomorphism of ringed spaces.)
Adjoining a zero
Notice that for every $M \in \mathbf{CMon}$ there is a canonical bijection
$$\mathrm{Spec}_{\mathbf{CMon}_0}(M \cup \{0\}) \to \mathrm{Spec}_{\mathbf{CMon}}(M),~ \mathfrak{p} \mapsto \mathfrak{p} \cap M.$$
The prime ideals of $(\mathbb{N},\cdot,1,0)$ again
The monoid $(\mathbb{N},+,0)$ has exactly two prime ideals, namely $\emptyset$ and $\mathbb{N}^+$. Prime factorization yields an isomorphism
$(\mathbb{N},\cdot,1,0) \cong (\mathbb{N}^+,\cdot,1) \cup \{0\} \cong \bigl(\bigoplus_{p \in \mathbb{P}} (\mathbb{N},+,0) \bigr) \cup \{0\}$. Thus, the previous results show that
$$\mathrm{Spec}_{\mathbf{CMon}_0}((\mathbb{N},\cdot,1,0)) \cong \prod_{p \in \mathbb{P}} \mathrm{Spec}_{\mathbf{CMon}}((\mathbb{N},+,0)) = \prod_{p \in \mathbb{P}} \{\emptyset,\mathbb{N}^+\} \cong \mathcal{P}(\mathbb{P}).$$
Finally, the tensor product
Now we can also compute the prime ideals of $(\mathbb{N},\cdot,1,0) \otimes_{\mathbb{F}_1} (\mathbb{N},\cdot,1,0)$. It is $(\mathbb{N}^+,\cdot,1) \otimes_{\mathbb{F}_0} (\mathbb{N}^+,\cdot,1)$ with an adjoined zero. The tensor product refers to commutative $\mathbb{F}_0$-algebras aka commutative monoids; I will just write $\otimes$ since the object themselves show which category we are in. We have
$$(\mathbb{N}^+,\cdot,1) \otimes (\mathbb{N}^+,\cdot,1) \cong \bigoplus_{p,q \in \mathbb{P}} (\mathbb{N},+,0) \otimes (\mathbb{N},+,0) \cong \bigoplus_{p,q \in \mathbb{P}} (\mathbb{N},+,0)$$
and therefore
$$\mathrm{Spec}_{\mathbf{CMon}}((\mathbb{N},\cdot,1,0) \otimes_{\mathbb{F}_1} (\mathbb{N},\cdot,1,0)) \cong \mathrm{Spec}_{\mathbf{CMon}}((\mathbb{N}^+,\cdot,1) \otimes (\mathbb{N}^+,\cdot,1)) \cong \mathcal{P}(\mathbb{P} \times \mathbb{P}).$$
Explicitly, the prime ideal associated to a subset $E \subseteq \mathbb{P} \times \mathbb{P}$ is $\bigcup_{(p,q) \in E} \langle p \otimes 1, 1 \otimes q \rangle$ (I think).
$^1$It is a really good idea to not forget forgetful functors in general, especially here when we need to emphasize the multiplicative structure and the zero. This is why I don't just write $\mathbb{N}$, which is merely the underlying set of the monoid with zero $(\mathbb{N},\cdot,1,0)$.
$^2$The term "binoid" is a bad choice here, I won't use it.
A: $\newcommand{\Z}{\mathbf{Z}}$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.
