0

I have heard of the spectrum of a ring before, but I don't really know what it's intended to be used for. I have also heard that, historically, spectra using maximal ideals appeared before spectra using prime ideals, but that in modern times prime ideals are used.

I'm interested in whether there are other options besides the prime ideals and the maximal ideals as a point set for the topology corresponding to the ideal lattice (which is also the congruence lattice) of a ring.

I don't know what a sheaf is, so I am focusing on the spectrum only as a topological space.

So, are there other options for the underlying point set besides the maximal and prime ideals? And what makes some choices more interesting than others?

Let $R$ be a commutative, unital ring.

If $F$ is a family of ideals, then $\cap F$ is also an ideal.

Additionally if $I$ and $J$ are ideals, then $(I, J)$ is also an ideal, and the join of $I$ and $J$ in the lattice of ideals.

Our lattice of ideals is closed under arbitrary meets and finite joins, so that makes me wonder whether we can pick a suitable point-set to turn it into a topology.

We already know that every proper ideal is contained in a maximal ideal.

Define a map $f(I)$ to be the set of maximal ideals $J \supset I$.

It is not hard to show that:

$$ f \left( \land X \right) = \cup \{ f(z) : z \in X \} $$

and

$$ f(x \lor y) = f(x) \land f(y) $$

As a quick smoke check, consider the ring $\mathbb{Z}$ and the ideals $(2)$ and $(3)$. They are both maximal ideals, thus $f((2)) \cap f((3))$ would be empty. The only ideal that is not contained in any maximal ideal is $(1)$, which checks out.

At the moment, I don't have a proof myself that for any two proper ideals $I$ and $J$, there exists a maximal ideal $K$ such that $I \subset K$ but $J \not\subset K$.

So, taking the maximal ideals as our point set might lose information. I don't think it does, but I don't have a proof that it doesn't.

Anyway, I can come up with one point set that is neither precisely the maximal ideals nor precisely the prime ideals and that is:

  1. The maximal ideals and $(0)$ when $(0)$ is a prime ideal.

The only ideal that is a subset of $(0)$ is $(0)$, so, relative to the maximal spectrum, we've added a new point that is only in the universal open set and no other open subset. And we can use the presence or absence of this point to detect whether the original ring was an integral domain, which is nice.

We can also come up with other weird examples like:

  1. The maximal ideals and the principal prime ideals.
Greg Nisbet
  • 11,657
  • I think this is an interesting question though I am unsure of your motivations. For primes and maximal ideals, there were clear geometric reasons to do so. On the other hand, I cannot imagine any geometric reason why (1) and (2) should be interesting and detecting integral domains does not seem too important because you could do it directly from the prime spectrum anyway. I upvoted since it appears well thought and I’m interested in what others have to say. – Shrugs Dec 24 '23 at 01:28
  • 1
    Regarding your comment "At the moment, I don't have a proof myself that for any two proper ideals $I$ and $J$, there exists a maximal ideal $K$ such that $I\subset K$ but $J \not\subset K$", this is not generally the case. There are rings (the local rings; https://en.wikipedia.org/wiki/Local_ring) where there is a unique maximal ideal; in those rings every proper ideal is still contained in a maximal ideal, but those maximal ideals are all the same. There are many (very interesting) examples of local rings. – Chris Eagle Dec 24 '23 at 03:43
  • 1
    Continuing the previous comment, this property you asked for isn't even true of $\mathbb{Z}$ (which is not local). For any prime number $p$, $(p^2)$ and $(p^3)$ are distinct proper ideals but the only maximal ideal containing either of them is $(p)$. More generally, if $n$ and $m$ are integers that are divisible by the same set of primes, then $(n)$ and $(m)$ are contained in the same maximal ideals. – Chris Eagle Dec 24 '23 at 03:47
  • @ChrisEagle Cool. Seems like the map from ideals to open sets of the spectrum is frequently non-injective regardless of which kind of spectrum you're working over. That's good to know. Originally when I was thinking about this I was just trying to build myself a nice point set that I could use to turn the ideal lattice into an actual topology. – Greg Nisbet Dec 24 '23 at 03:54

0 Answers0