# open generic points of affine scheme?

Let $$S=\text{Spec} A$$ be an affine scheme, we assume $$A$$ is not a field, then we know irreducible components of $$S$$ correspond to all minimal prime ideals of $$A$$, in fact, these prime ideals are generic points of irreducible components of $$S$$.

Then I want to know if we assume all these generic points are open points in $$S$$(or at least one generic point is open), what property of $$A$$ can we deduce?

And what I mean by an open point is that this point is open in the subspace topology.

Thanks!

For simplicitly let us assume that $$\mathrm{Spec}(A)$$ is integral (I'll leave you to think about the non-integral case) with unique generic point $$\eta$$. If $$\{\eta\}$$ is open then we know that there exists some neighborhood $$D(f)$$ of $$\eta$$ contained in $$\{\eta\}$$ and thus, of course, $$D(f)=\{\eta\}$$. From this, we see that

$$A[f^{-1}]=\mathcal{O}(D(f))=\mathcal{O}_{\mathrm{Spec}(A),\eta}=\mathrm{Frac}(A)$$

where the middle equality holds since there are no　neighborhoodsof $$\eta$$ properly contained in $$D(f)$$.

Conversely, we see that if there exists some $$f$$ in $$A$$ such that $$\displaystyle A[f^{-1}]=\mathrm{Frac}(A)$$ then we see that, in particular, $$A[f^{-1}]$$ is a field and so $$\mathrm{Spec}(A[f^{-1}])$$ only consists of one point. But, the map $$\mathrm{Spec}(A[f^{-1}])\to \mathrm{Spec}(A)$$ is an open embedding with image $$D(f)$$ and so, in particular, its image contains $$\eta$$. But, since $$D(f)$$ consists of only one point we must have that $$D(f)=\{\eta\}$$ and thus $$\{\eta\}$$ is open.

Thus, from the above we deduce the following:

Proposition: Let $$A$$ be a domain. Then the (unique) generic point $$\eta$$ of $$\mathrm{Spec}(A)$$ is open if and only if there exists some $$f$$ in $$A$$ such that $$A[f^{-1}]=\mathrm{Frac}(A)$$.

Let us give some simple examples/non-examples:

Example 1: Let $$\mathcal{O}$$ be a DVR with uniformizer $$\pi$$. Then, $$\mathrm{Frac}(\mathcal{O})=\mathcal{O}[\pi^{-1}]$$ and so you see that the generic point of $$\mathrm{Spec}(\mathcal{O})$$ is open. In fact, $$\mathrm{Spec}(\mathcal{O})$$ consists, as is used very often, of an open generic point $$\eta$$ and a closed point $$(\pi)$$.

Remark 1: More generally, if $$K$$ is a field and $$\mathcal{O}$$ is a so-called microbial valuation ring in $$K$$ (e.g. see [1, §I.1.5]) then $$K=\mathcal{O}[\varpi^{-1}]$$ for any pseudo-uniformizer $$\varpi$$ (e.g. see [1, Lemma I.1.5.9]) and so the generic point of $$\mathrm{Spec}(\mathcal{O})$$ is open. Such valuation rings play a pivotal role in Huber's theory of adic spaces. As an example, one can consider the the valuation induced on $$\mathrm{Frac}(\mathbb{C}_p\langle t\rangle)$$ by the valuation

$$\left|\sum_{n=0}^\infty a_n t^n\right|=\sup |a_n|$$

on

$$\mathbb{C}_p\langle t\rangle:=\left\{\sum_{n=0}^\infty a_n t^n:\lim a_n=0\right\}$$

Then, the valuation ring in $$\mathrm{Frac}(\mathbb{C}_p\langle t\rangle)$$ is an example of a microbial valuation ring. You can make even more exotic (non-rank $$1$$) examples. See [1, §I.1.5] again).

Non-example 2: Certainly $$\mathbb{Z}$$ doesn't have open generic point since there is no element $$f$$ in $$\mathbb{Z}$$ such that $$\mathbb{Z}[f^{-1}]=\mathbb{Q}$$. Indeed, this is clear by thinking about the fact that $$v_p(f)\ne 0$$ for only finitely many $$p$$ (where $$v_p$$ is the $$p$$-adic valuation).

Example/Non-example 3: If $$A$$ is finite type over a field $$k$$ (and a domain) then the generic point of $$\mathrm{Spec}(A)$$ is open if and only if $$A$$ is a finite extension of $$k$$. Indeed, one easy way to see this is that this implies that there exists some $$f$$ in $$A$$ such that $$A[f^{-1}]=\mathrm{Frac}(A)$$. Since $$\mathrm{Spec}(A[f^{-1}])\to \mathrm{Spec}(A)$$ is an open embedding this implies (e.g. see [2, Theorem 5.22(3)]) that $$\mathrm{Spec}(A)$$ has dimension zero from where the conclusion follows (e.g. see [2, Corollary 5.21]).

Remark 2: Combining Example 1 and Example/Non-example 3 we are able to observe an interesting subtelty. Namely, as we used in the latter (and is well-known) if $$X$$ is an irreducible variety then $$\dim(U)=\dim(X)$$ for any open subset $$U$$ of $$X$$. This is false for general rings as Example 1 shows since $$\dim\{\eta\}=0$$ but $$\dim \mathcal{O}=1$$ (in the DVR case).

References:

• Thanks for your answer! I have thought about my question for a few hours in the case that $A$ is irreducible, this is to say the nilpotent radical is prime, and I think this is a little complicated. I have seen people think about generic points and closed points in a scheme, but why don't people think about open points? – user832207 Oct 17 '20 at 16:54
• @Sate Well, if you want to assume just irreducible then note that your property doesn't change under passage from $A$ to $A_\text{red}$ (the reduced quotient), in which case your back in the integral situation. Does that help clarify? The answer to your second question is then, as you might expect, opens points are just exceedingly rare. For example, in the classical theory of varieties one sees that points can never be open! Indeed, the same argument as in Example/Non-example 3 shows that if $X$ is an irreducible variety with an open point $x$ then $\dim(X)=0$. – Alex Youcis Oct 17 '20 at 17:01