Are the nonmaximal point in $\operatorname{Spec} A$ unimportant in some situations?

Question

(I don't know the best way to set up the context of this question, so I am just going to say whatever I heard people keep saying, according to my poor memory.)

Let $k$ be an algebraically closed field. An affine $k$-variety is the space $\operatorname{Spec} A$ for some finitely generated integral domain $A$ over $k$. (Different people have different conventions for affine variety, so let's not pay too much attention to that.) In this context, we often ignore points in $\operatorname{Spec} A$ that are not maximal ideals. For example, this means we can identify $\operatorname{Spec} k[x_1, \cdots, x_n]$ with points in $k^n$ by the Hilbert Nullstellensatz and $\operatorname{Spec} k[x_1^{\pm 1}, \cdots, x_n^{\pm 1}]$ with the algebraic torus $(k^\times)^n$.

Honestly, I am not confident that we can always do that, but yet from my experience of hearing other people's talks, this seems to be quite a common thing to do/assume. Intuitively, every ideal is included in some maximal ideal, so this ignorance kinda makes sense, but I never recall Vakil doing that in his FOAG (maybe he does but I just forgot). I guess my (rather broad) question for today is that: can we always do that? If not, then why do people keep ignoring nonmaximal ideals in certain settings (and in which settings are this ignorance permissible)? When we write proofs, how do we even make this precise, because setwise you really can't identify $\operatorname{Spec} k[x_1, \cdots, x_n]$ with $k^n$, as $k[x_1, \cdots, x_n]$ do contains nonmaximal ideals?

Answer 1

Over an algebraically closed field $k$, the closed points of an algebraic variety uniquely determine the corresponding scheme. This is certainly true if we use the word "variety" to mean a zero set of some polynomial equations in an affine space $\mathbb{A}^n_k$, or a projective space $\mathbb{A}^n_k$, or, more generally, an open subset of such a set of zeroes.

The precise statement (but likely not the strongest possible one) is Proposition II.2.6 of Hartshorne:

Let $k$ be an algebraically closed field. There is a natural fully faithful functor $t: \mathcal{Var}(k)\to\mathcal{Sch}(k)$ from the category of varieties over $k$ to the category of schemes over $k$. For any variety $V$, its topological space is homeomorphic to the set of closed points of $t(V)$, and its sheaf of regular functions is obtained by restricting the structure sheaf of $t(V)$ via this homeomorphism.

The idea is that nonclosed points correspond to irreducible subvarieties, and over an algebraically closed field every such subvariety is uniquely determined by its closed point.

Answer 2

Question: "Honestly, I am not confident that we can always do that, but yet from my experience of hearing other people's talks, this seems to be quite a common thing to do/assume. Intuitively, every ideal is included in some maximal ideal, so this ignorance kinda makes sense, but I never recall Vakil doing that in his FOAG (maybe he does but I just forgot). I guess my (rather broad) question for today is that: can we always do that? If not, then why do people keep ignoring nonmaximal ideals in certain settings (and in which settings are this ignorance permissible)? When we write proofs, how do we even make this precise, because setwise you really can't identify Speck[x1,⋯,xn] with kn, as k[x1,⋯,xn] do contains nonmaximal ideals?"

At the following link you find a construction of a locally ringed space $(X^m, \mathcal{O}_{X^m})$ from any commutative ring $A$ that is finitely generated over a Dedekind domain $k$:

Grothendieck point of view of algebraic geometry

The above construction may be done for any scheme $X$ that is of finite type over an affine base scheme $S:=Spec(k)$ where $k$ is a Dedekind domain. The reason for the inclusion of the prime (non maximal) ideals in $A$ as points in the topological space $Spec(A)$ is because one wants to speak of multiplicities. Then one needs a functor from the category of commutative rings to locally ringed spaces. If you apply the above functor to a ring $A$ that is non reduced but finitely generated over $k$, you get a topological space $X^m$ (not containing non maximal ideals) and a sheaf of rings $\mathcal{O}_{X^m}$ where the local sections $\mathcal{O}_{X^m}(U)$ have nilpotent elements. If $A$ is an integral domain it follows the local sections $\mathcal{O}_X(U)$ are integral domains hence reduced rings.

You let $X:=Spec(A)$ and let $X^m \subseteq X$ be the topological subspace of maximal ideals. Let $i: X^m \rightarrow X$ be the canonical inclusion map and define $\mathcal{O}_{X^m}:=i^{-1}(\mathcal{O}_X)$. Here $i^{-1}$ denotes "the topological pull back" of the sheaf of rings $\mathcal{O}_X$. It follows the pair $(X^m, \mathcal{O}_{X^m})$ is a locally ringed space and there is an isomorphism

$$ \Gamma(X^m, \mathcal{O}_{X^m}) \cong A.$$

You may do this for any finitely generated ring over any Dedekind domain (or any scheme of finite type over $S:=Spec(A)$ with $A$ a Dedekind domain). The non-maximal ideals are not points in $X^m$, but the sheaf of rings $\mathcal{O}_{X^m}$ may locally have nilpotent elements.

Question: "If not, then why do people keep ignoring nonmaximal ideals in certain settings (and in which settings are this ignorance permissible)? When we write proofs, how do we even make this precise, because setwise you really can't identify Speck[x1,⋯,xn] with kn, as k[x1,⋯,xn] do contains nonmaximal ideals?"

If they work over a field or a Dedekind domain, they ignore the non maximal ideals for the above reason.