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Given a (pointed or unpointed) monoid $M$, one defines an ideal of $M$ to be a subset $I$ of $M$ such that if $a\in I$ and $r\in M$, then $ra\in I$ (and such that $0\in I$ if $M$ is pointed). A prime ideal $\frak p$ of $M$ is then a proper ideal such that if $ab\in\frak p$, then $a\in\frak p$ or $b\in\frak p$. The set of all prime ideals of $M$ can is written $\mathrm{MSpec}(M)$, and it can be made into a topological space equipped with a sheaf of monoids, giving a variant of algebraic geometry where monoids play the role of rings. Good references for this are Sections 1--2 of this paper and Ogus's book.

Lately I've been trying to compile some examples of such prime spectra, and wondering about how these relate to the usual $\mathrm{Spec}$ of a ring. So far, I've found or read about the following examples (the first five examples below come from Martin Brandenburg's answer here):

  • $\mathrm{MSpec}(\mathbf{N},+)=\{0,\mathbf{N}_{>0}\}$.
  • $\mathrm{MSpec}(\mathbf{Z},+)=\{0\}$.
  • $\mathrm{MSpec}(\mathbf{N},\cdot)=\mathrm{MSpec}(\mathbf{Z},\cdot)=\mathcal{P}(\mathbf{P})$, the powerset of the set $\mathbf{P}$ of all prime numbers.
  • $\mathrm{MSpec}(\mathbf{N}\otimes_{\mathbf{N}_+}\mathbf{N})\cong\mathcal{P}(\mathbf{P}\times\mathbf{P})$.
  • $\mathrm{MSpec}(\mathbf{Z}\otimes_{\mathbf{F}_1}\mathbf{Z})\cong\mathcal{P}(\mathbf{P})\times\mathcal{P}(\mathbf{P})$.
  • $\mathrm{MSpec}(K)=\{(0)\}$ whenever $K^\times=K\setminus\{0\}$. In particular this applies to $\mathbf{F}_1=\{0,1\}$.
  • $\mathrm{MSpec}(\mathbf{F}_1[x])=\{(0),(x)\}$.
  • $\mathrm{MSpec}(\mathbf{F}_1[x,y])=\{(0),(x),(y),(x,y)\}$.
  • $\mathrm{MSpec}(\mathbf{F}_1[x_1,...,x_n])=\mathcal{P}(\{x_1,...x_n\})$.
  • $\mathrm{MSpec}(\mathbf{F}_1[t,t^{-1}])=\{(0),(t),(t^{-1})\}$.

Now, any ring $R$ has an associated monoid, given by keeping only multiplication. I've noticed that there seems to be some relation between $\mathrm{Spec}(R)$ and $\mathrm{MSpec}(R)$ when applicable in the above examples; e.g. $\mathrm{Spec}(\mathbf{Z})=\{(0)\}\cup\mathbf{P}$ naturally injects into $\mathrm{MSpec}(\mathbf{Z},\cdot)=\mathcal{P}(\mathbf{P})$, at least as a set.

Are there any results relating the monoidal spaces $\mathrm{Spec}(R)$ and $\mathrm{MSpec}(R)$?

Are there for $\mathrm{MSpec}(M)$ and $\mathrm{Spec}(\mathbf{N}[M])$ or $\mathrm{Spec}(\mathbf{N}_+[M])$?

Finally, do we always have a natural morphism of monoidal spaces $\mathrm{Spec}(R)\to\mathrm{MSpec}(R)$?

Martin Brandenburg
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Emily
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  • I had forgotten to add a pointer to Martin Brandenburg's answer in the list of examples above. My apologies. – Emily Aug 25 '21 at 11:54
  • Note: this question shares part of its body with Examples of prime spectra of monoids, which asks about more examples of $\mathrm{MSpec}$'s. – Emily Aug 25 '21 at 13:27
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    I think the map in your last question is kind of immediate, no? Every prime of a ring is a prime of its multiplication monoid, this map is continuous. For the map of sheaves, every section is locally a quotient of elements of the ring in both sides so we can think of the map as "identity" – sss89 Aug 31 '21 at 07:53
  • @sss89 Sorry for the delay in replying. I had this map in mind, but wanted to confirm that it works with someone else. Thanks! – Emily Sep 03 '21 at 08:46

1 Answers1

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If $P$ is a (monoid) prime ideal of a commutative ring $R$ then $R\setminus P$ is saturated (if $xy \in R\setminus P$ then $x, y \in R \setminus P$) and closed under multiplication. Therefore, $R \setminus P$ is a complement of a union of (ring) prime ideals. Conversely, any union of (ring) prime ideals of $R$ is a (monoid) prime ideal. Therefore, the points of the monoid spectrum are the unions of prime ideals of the ring $R$.

Topologically, the space of unions of prime ideals of $R$ can be constructed from $\mathrm{Spec}(R)$ by taking the power set of $\mathrm{Spec}(R)$ and then identifying any two points which belong to the same basic open sets. This is discussed, for example, in Example 2.5 of this paper. But this is not a purely topological construction because it requires a choice of base of $\mathrm{Spec}(R)$ (it's really just another way of saying "take the unions of prime ideals"). But it can be viewed as a kind of "completion" of $\mathrm{Spec}(R)$ by adding a bunch of points, in the sense that if you repeat the process, you don't get anything new.

Flounderer
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    Wow, this is really neat, I didn't know there was such a clean relation between both Spec's. Thank you very much! – Emily Nov 09 '24 at 13:27