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One of the generalizations of algebraic geometry is provided by the theory of semiring schemes, viz. Lorscheid 2012. The theory follows the same set up of scheme theory, but we use semirings instead of rings, also known as rings without additive inverses; examples of which are the natural numbers $\mathbb{N}$ or the tropical semiring $\mathbb{T}$.

An ideal of a semiring $R$ is a set $I\subset R$ which 1) is closed under addition 2) contains $0$, and 3) is such that $IA=I$. Moreover, $I$ is prime if whenever $ab\in I$, then $a\in I$ or $b\in I$, and we write $\mathrm{Spec}(R)$ for the set of prime ideals of $R$.

What are $\mathrm{Spec}(\mathbb{N})$ and $\mathrm{Spec}(\mathbb{T})$? Do we have other nice examples of prime spectra of semirings?

Emily
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    $\mathbb{T}$ doesn't even have many ideals in the first place – it's like a field. There is a natural identification of the prime ideals of $\mathbb{Z}$ with the prime ideals of $\mathbb{N}$, but it is an interesting exercise to find a prime ideal of $\mathbb{N}$ that is not just the intersection of a prime ideal of $\mathbb{Z}$ with $\mathbb{N}$. (There is at least one, and if I recall correctly it is the only one.) – Zhen Lin Aug 14 '21 at 05:55
  • @ZhenLin Oh, that's true. Is $\mathrm{Spec}(\mathbb{N})=\mathrm{Spec}(\mathbb{Z})\cup(2,3)$? – Emily Aug 14 '21 at 07:03
  • The ideal generated by ${ 2, 3 }$, i.e. $\mathbb{N} \setminus { 1 }$, is indeed prime. As far as I know that is the only extra prime ideal but I don't remember why. – Zhen Lin Aug 14 '21 at 07:41

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First let's compute $\operatorname{Spec}\mathbb{N}$. Certainly $p\mathbb{N}$ is a prime ideal for every positive prime $p\in\mathbb{Z}$, and you can check that both $\{0\}$ and $\mathbb{N}\setminus\{1\}$ are prime ideals. As @Zhen Lin points out in the comments, these are in fact the only prime ideals of $\mathbb{N}$; to see why, suppose $P$ is a non-zero prime ideal of $\mathbb{N}$. Then there exists some non-zero natural number in $P$; if $P$ contains $1$, then $P=\mathbb{N}$, a contradiction. So $P$ contains a natural number greater than one, and hence must contain one of its prime factors $p$. If every element of $P$ is divisible by $p$, then $P=p\mathbb{N}$. Otherwise $P$ contains an element outside of $p\mathbb{N}$, and hence must contain one of its prime divisors $q$, which is necessarily distinct from $p$. By the "coin theorem", $P$ now contains every number greater than $pq-p-q$. In particular, for any $n>1$, there exists some $k$ such that $n^k>pq-p-q$, whence $n^k\in P$ and so $n\in P$. So $P=\mathbb{N}\setminus\{1\}$, as desired.

Now let's compute $\operatorname{Spec}\mathbb{T}$. By convention, I take $\mathbb{T}=\mathbb{R}\cup\{\infty\}$, with $\oplus$ as $\min$ and $\otimes$ as $+$. Note that $\{\infty\}$ is a prime ideal of $\mathbb{T}$; I claim it is the only one, and that it is in fact the only proper ideal of $\mathbb{T}$. To see this, suppose that $I$ is a proper ideal of $\mathbb{T}$ containing some $x\in\mathbb{R}$. Then $I\ni\lambda\otimes x$ for every $\lambda\in\mathbb{T}$. In particular, for any $y\in\mathbb{R}$, $I$ contains $(y-x)\otimes x=y$, and furthermore $I$ contains $\infty\otimes x=\infty$. So $I=\mathbb{T}$, a contradiction. Thus indeed $\{\infty\}$ is the only prime ideal of $\mathbb{T}$.

Atticus Stonestrom
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