The tropical integers

Question

Let \begin{align} \oplus_\mathbb{N} &= + \\ 0_\mathbb{N} &= 0 \\ \odot_\mathbb{N} &= \cdot \\ 1_\mathbb{N} &= 1 \end{align} Then $(\mathbb{N}, \oplus_\mathbb{N}, 0_\mathbb{N}, \odot_\mathbb{N}, 1_\mathbb{N})$ is the ordinary rig of natural numbers. Let $\mathbb{Z} = \mathbb{N}^2 / \sim$ where \begin{align} (a_1, a_2) \sim (b_1, b_2) &\iff a_1 \oplus_\mathbb{N} b_2 =a_2 \oplus_\mathbb{N} b_1 \\ (a_1, a_2) \oplus_\mathbb{Z} (b_1, b_2) &= (a_1 \oplus_\mathbb{N} b_1, a_2 \oplus_\mathbb{N} b_2) \\ 0_\mathbb{Z} &= [(0_\mathbb{N}, 0_\mathbb{N})]_\sim \\ \ominus_\mathbb{Z} (a_1, a_2) &= (a_2, a_1) \\ (a_1, a_2) \odot_\mathbb{Z} (b_1, b_2) &= ((a_1 \odot_\mathbb{N} b_1) \oplus_\mathbb{N} (a_2 \odot_\mathbb{N} b_2), (a_1 \odot_\mathbb{N} b_2) \oplus_\mathbb{N} (a_2 \odot_\mathbb{N} b_1)) \\ 1_\mathbb{Z} &= [(1_\mathbb{N}, 0_\mathbb{N})]_\sim \end{align} Then $(\mathbb{Z}, \oplus_\mathbb{Z}, 0_\mathbb{Z}, \ominus_\mathbb{Z}, \odot_\mathbb{Z}, 1_\mathbb{Z})$ is the ordinary ring of integers. Suppose we let \begin{align} \oplus_\mathbb{N} &= \max \\ 0_\mathbb{N} &= 0 \\ \odot_\mathbb{N} &= + \\ 1_\mathbb{N} &= 0 \end{align} instead. This makes $(\mathbb{N}, \oplus_\mathbb{N}, 0_\mathbb{N}, \odot_\mathbb{N}, 1_\mathbb{N})$ the tropical rig of natural numbers. Call $(\mathbb{Z}, \oplus_\mathbb{Z}, 0_\mathbb{Z}, \ominus_\mathbb{Z}, \odot_\mathbb{Z}, 1_\mathbb{Z})$ the "tropical integers". Which papers, if any, have studied this structure? Does it have a geometric or easily-visualizable interpretation?

Answer 1

Unfortunately, nothing like this works.

The starting point is the observation that the "max-version" of $\sim$ doesn't behave well at all:

• It's not an equivalence relation since it's non-transitive: we have e.g. $(1,0)\sim(2,2)$ and $(2,2)\sim (0,1)$ but $(1,0)\not\sim(0,1)$.

• More generally, whenever $c>a,b$ we have $(a,b)\sim(c,c)$. So the transitive closure of $\sim$ is everything.

Motivated by this specific observation we can in fact prove a general impossibility result:

Every rig homomorphism from $T=(\mathbb{N};\max,+)$ to a rig $R$ which happens to be a ring has trivial image (= sends everything to the additive identity of $R$).

The point is this: for each $m\in T$ there is some $n\in T$ such that $m+n=n$. But then if $f:T\rightarrow R$ is a rig homomorphism we have $f(m)+f(n)=f(n)$, so since $R$ happens to be a ring we have $f(m)=0$. (In fact, all we need about $R$ is that it have the cancellation property for addition.)