# What does $\mathrm{Spec}(\mathbb{Z})\times_{\mathrm{Spec}(\mathbb{F}_{1})}\mathrm{Spec}(\mathbb{Z})$ 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).

1. A binoid is a commutative monoid $$M$$ together with an absorbing element $$0$$.
2. An ideal of $$M$$ is a subset $$I$$ such that
• $$0\in I$$.
• If $$a\in I$$ and $$r\in M$$, then $$ra\in I$$.
3. An ideal $$I$$ of $$M$$ is prime if it is proper and whenever $$ab\in I$$ then $$a\in I$$ or $$b\in I$$.
4. The spectrum $$\mathrm{Spec}(M)$$ of a binoid $$M$$ is the set of all prime ideals of $$M$$.
5. 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 $$V(I)=\{\mathfrak{p}\in\mathrm{Spec}(M):I\subset\mathfrak{p}\}.$$
6. 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 $$\mathrm{Spec}(M)$$.
7. A binoidal space is a pair $$(X,\mathcal{O}_X)$$ with $$X$$ a topological space and $$\mathcal{O}_X$$ a sheaf of binoids on $$X$$.
8. 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 $$\Z\otimes_{\F_{1}}\Z\cong(\Z\setminus\{0\},\cdot)^{\oplus{\N}}\sqcup\{0\}$$?

• I wonder why statements like "One of the most mysterious objects in mathematics is the elusive "field with one element"" are still made. Here, $\mathbb{F}_1$ is just the monoid with zero $(\{0,1\},\cdot,1,0)$, nothing mysterious about that. Algebraic geometry over $\mathbb{F}_1$ is much simpler than algebraic geometry over $\mathbb{Z}$ - at least when we take the approach here with comm. monoids with zero. Aug 14, 2021 at 16:20
• @MartinBrandenburg Well, it is not clear at all if $(\{0,1\},\cdot,1,0)$ is the correct model for $\mathbb{F}_{1}$, so in this sense the latter is still elusive. Besides, these kind of statements make for great paragraph openings :P Aug 14, 2021 at 19:04

### 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.)

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.

• Marc Hoyois pointed out that I made a mistake in my question, which carried over to your answer (I conflated $\otimes_{\mathbf{N}_+}$ with $\otimes_{\mathbf{F}_{1}}$, so when I wrote $\mathbf{Z}\otimes_{\mathbf{F}_{1}}\mathbf{Z}$, I should really have written $\mathbf{Z}\otimes_{\mathbf{N}_{+}}\mathbf{Z}$). I wrote an answer to try to patch this, but maybe I got things wrong again... What do you think? Aug 15, 2021 at 10:33
• Alright, I see if I find some time to check it. Aug 15, 2021 at 19:54

$$\newcommand{\Z}{\mathbf{Z}}$$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.