Explicit description of $\mathbb{Z}\otimes_{\mathbb{N}}\mathbb{Z}$

Question

$\newcommand{\Q}{\mathbb{Q}}\newcommand{\N}{\mathbb{N}}\newcommand{\Z}{\mathbb{Z}}$Recall the definition of the tensor product $\otimes_{\mathbb{N}}$ of commutative monoids (see also this note by Harold Simmons).

Question. Is there an explicit/"nice" description of the tensor product $\Z\otimes_{\N}\Z$, where $\Z=(\Z,\cdot,1)$ is the multiplicative monoid of integers?

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A partial description. Eric Wofsey mentioned a really nice way to partly describe $\Z\otimes_\N\Z$ here: since $\Z \setminus \{0\} \cong\N^{\oplus\N}\oplus\Z_2$ via the map sending a non-zero integer $k= \pm 2^{n_1}3^{n_2}5^{n_3}\cdots$ to $((n_1,n_2,n_3,\ldots),\mathrm{sgn}(k))$, we may use distributivity of tensor products along direct sums to write \begin{align*} (\Z\setminus\{0\})\otimes_{\N}(\Z\setminus\{0\}) &\cong (\N^{\oplus\N}\oplus\Z_2)\otimes_\N(\N^{\oplus\N}\oplus\Z_2)\\ &\cong (\N^{\oplus\N}\otimes_\N\N^{\oplus\N})\oplus(\N^{\oplus\N}\otimes_\N\Z_2)\oplus(\Z_2\otimes_\N\N^{\oplus\N})\oplus(\Z_2\otimes_\N\Z_2)\\ &\cong \N^{\oplus\N}\oplus\Z_2^{\oplus\N}\\ &\cong (\Z\setminus\{0\},\cdot)^{\oplus\N}. \end{align*} One could then add back $0$ to each factor of $\Z\setminus\{0\}$ above, adding new elements and relations to $(\Z\setminus\{0\},\cdot)^{\oplus\N}$. What is the result of this?

Answer 1

The computation of the tensor product of commutative monoids $\mathbb{Z} \otimes \mathbb{Z}$ can be reduced to the computation of $(\mathbb{Z} \setminus \{0\}) \otimes (\mathbb{Z} \setminus \{0\})$ as follows.

More generally, let $M$ be any commutative monoid (for example, $\mathbb{Z} \setminus \{0\}$ under multiplication). Then $M \sqcup \{0\}$ is a commutative monoid with zero which contains $M$ as a submonoid. This construction is left adjoint to the forgetful functor from commutative monoids with zero to commutative monoids. But since we only work with commutative monoids, here the universal property is a bit more complicated: $$\hom(M \sqcup \{0\},N) \cong \{(u,f) \in M \times \hom(M,N) : u \text{ absorbs } f(M)\}.$$ The condition means that $u \cdot f(m) = u$ for all $m \in M$. Since the tensor product can also be defined by the hom-tensor-adjunction, this easily implies that (and a direct proof is also possible) $$(M \sqcup \{0\}) \otimes N \cong (N \oplus (M \otimes N)) / (n \cdot (m \otimes n) = n)_{m \in M, n \in N}.$$ The quotient means that we mod out the smallest congruence relation which makes $n \cdot (m \otimes n)$ and $n$ equivalent, for all $m \in M$, $n \in N$. We can use this twice to deduce $$(M \sqcup \{0\}) \otimes (N \sqcup \{0\}) \cong (M \oplus N \oplus (M \otimes N)) \sqcup \{0\}/\\ (n \cdot (m \otimes n) = n,\, m \cdot (m \otimes n) = m)_{m \in M, n \in N}.$$ As already said in my comment, notice that the tensor product is much easier if we work in the category of commutative monoids with zero, since in that case we just have $(M \sqcup \{0\}) \otimes_0 (N \sqcup \{0\}) \cong (M \otimes N) \sqcup \{0\}$ by general nonsense.

Answer 2

$\newcommand{\Z}{\mathbb{Z}}\newcommand{\F}{\mathbb{F}}\newcommand{\N}{\mathbb{N}}$Here's a remark on Martin's answer, which is unfortunately too long to post as a comment.

The point I want to mention is that the simplification noted in Martin's answer, where one considers tensor products not in $\mathsf{CMon}$, but rather in the category of commutative monoids with zero, comes from considering tensor products not over $\N$, but over the field with one element $\F_1$.

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Firstly, note that we may build $\mathsf{CMon}$ by first considering $\N$-modules, which are just monoids, and then considering commutative monoids in that category, giving us commutative monoids again, i.e. $\mathsf{CMon}\cong\mathsf{CMon}(\mathsf{Mod}_{\N})$.

We can, however, start by considering not $\N$-modules, but $\F_1$-modules. These are usually thought to be pointed sets (i.e. we define $\mathsf{Mod}_{\F_1}\cong\mathsf{Sets}_*$). Now, the category $\mathsf{Sets}_*$ admits a monoidal structure $\otimes_{\F_1}$ obtained by defining $$ (X,x_0)\otimes_{\F_1}(Y,y_0)=(X\times Y/{\sim},[(x_0,y_0)]), $$ where ${\sim}$ is the equivalence relation on $X\times Y$ obtained by declaring \begin{align*} (x,y_0) &\sim (x_0,y_0),\\ (x_0,y) &\sim (x_0,y_0), \end{align*} for each $x\in X$ and each $y\in Y$. The monoidal unit of this tensor product is then $\mathbb{F}_1=\{0,1\}$.

Now, a commutative monoid in $(\mathsf{Mod}_{\F_1},\otimes_{\F_1},\F_1)$ is precisely a commutative monoid with zero! Indeed, it will be a triple $((A,0_A),\mu,\eta)$ consisting of

• A pointed set $(A,0_A)$;
• A morphism of pointed sets $$\mu\colon(A,0_A)\otimes_{\F_1}(A,0_A)\to(A,0_A),$$ being equivalently given by a map of sets $\mu\colon A\times A\to A$, the multiplication of $A$, which is $\F_1$-bilinear in that we have \begin{align*} 0_Aa &= 0_A,\\ a0_A &= 0_A; \end{align*}
• A morphism of pointed sets $$\eta\colon(\F_1,0)\to(A,0_A),$$ picking an element $1_A$ of $A$, its unit.

So, in a sense, commutative monoids with zero are $\F_1$-algebras, and considering the tensor product of the $\F_1$-algebra $\Z$ with itself as an $\N$-algebra will give the wrong result, the complicated tensor product in Martin's answer.

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Finally, here's a picture I drew of the tensor product $\N\otimes_{\F_{1}}\N\cong(\N\setminus\{0\},\cdot)^{\oplus{\N}}\coprod\{0\}$, where $\mathbb{N}$ is the multiplicative monoid of natural numbers:

Here the blue circles depict the element $(2$, $1$, $4$, $1$, $2$, $3$, $4$, $2$, $2$, $1$, $2$, $3$, $2$, $1$, $4$, $4$, $1$, $2$, $2$, $2$, $2$, $4$, $3$, $1$, $3$, $2$, $1$, $3$, $4$, $2$, $2$, $2$, $3$, $3$, $4$, $2$, $1$, $2$, $1$, $2$, $2$, $2$, $\ldots)$ of $\N\otimes_{\F_{1}}\N\cong(\N\setminus\{0\},\cdot)^{\oplus{\N}}\coprod\{0\}$, represented as a counterclockwise spiral.

(Note that the factors of $\mathbb{N}$ in the above direct sum carry the multiplicative monoid structure, not the additive one!)

We can then picture $\Z\otimes_{\F_{1}}\Z\cong(\Z\setminus\{0\},\cdot)^{\oplus{\N}}\coprod\{0\}$ in a similar way, adding negative numbers to the above image.