# Grothendieck point of view of algebraic geometry

Given a ring $$R$$ and $$I\subseteq R[x_1,\dots ,x_n]$$ an ideal, define the functor $$V_R(I):\operatorname{Alg}_R\to \operatorname{Sets}$$, that sends a $$R$$-algebra $$A$$ in the subset of points $$\mathbf a \in A^n$$ such that $$f(\mathbf a)=0\$$for all $$f\in I$$. Recall that in the case $$R=k$$ an algebraically closed field, $$V_k(I)(k)$$ is in bijective correspondence with the maximal ideals of $$k[x_1,\dots ,x_n]$$.

Now, in the course, we talked about the geometric points of a ring $$A$$, defining them as the equivalence classes of ring homomorphisms $$A\to K$$, where $$K$$ is a field (not fixed). The motivation that we were given is that this is a generalization of the construction above, since in this way we can obtain a correspondence between homomorphisms and prime ideals, not only maximal ideals; moreover, one can consider any ring, not necessarily a $$k$$-algebra.

However, I don't understand in what sense this is a generalization: it seems to me that they are just different constructions. For example, even if I take $$k[x_1,\dots ,x_n]$$, $$k$$ an algebraically closed field, in order to capture all its spectrum I need to consider not only the homomorphisms in $$k$$, but also in some trascendental extension $$k\subset K$$; so we are not generalizing, we are just adding something in my opinion, because these two constructions (i.e. homomorphisms in $$k$$ fixed and homomorphisms in any field) don't reduce to the same when considering the case of finitely generated $$k$$-algebras. Am I right? Moreover, I have not clear the sense of arriving to the prime ideals, but not to any ideal (i.e. why geometric points of a ring $$A$$ are not defined as the classes of homomorphisms from $$A$$ to any ring). Thanks for any clarify, I know that these question may sound stupid but I'm quite new to this things.

• I don't understand how you can say "thinking about homomorphisms to any field" is not a generalization of "thinking about homomorphisms to $k$". $k$ is a field! – KReiser Nov 14 '20 at 11:11
• @KReiser Because these two constructions (i.e. homomorphisms in $k$ fixed and homomorphisms in any field) don't reduce to the same when considering the case of finitely generated k-algebras. – Dorian Nov 14 '20 at 11:17
• I think my professor presented the second construction as a construction that, considered on a finitely generated $k$-algebra, give the same result of the first one. However it seems to me that it's not true – Dorian Nov 14 '20 at 11:20
• I mean, it would be a generalization if one could capture every prime ideal of $k[x_1,\dots, x_n]$ just considering the homomorphisms in $k$, but still need to consider homomorphisms in any field in order to capture every prime ideal of a generic ring; however as I said this is not hte case. – Dorian Nov 14 '20 at 11:25

I want to talk about the meaning of spectrum in the classical sense. let $$A$$ be a $$k$$ algebra as you said for every Ideal $$I$$ you have a closed subset of $$A(k)$$ defined by $$V(I)$$. this subsets could be reducible meaning that it may be possible to write it non-trivially as the union of two closed subsets of $$A(k)$$.the irreducible closed subsets of $$A(K)$$ are in bijection with prime ideals of $$A$$.for example if $$A$$ is an integral domain $$A(K)$$ itself corresponds to the ideal $$(0)$$.

now that we know the meaning of prime ideals of $$A$$. why do we like to see them as point? in the classical language of algebraic geometry (and of course in the usual Geometry) there is a lot of situations you want to talk about things like that when a general point of a variety has some property. this is only make sense for irreducible varieties because reducible varieties consist of several parts and general points on different parts might behave differently.(just consider closed subset of a plane consisting of two different pairs.)

now the idea of Grothendieck was to add a generic point to each irreducible closed subset that controls the behavior of a general point on that set. so he has a point for every irreducible closed subset so equivalently for each prime ideal.

now this kills two birds with one stone: when you want to work with arbitrary rings $$A$$ it does not make sense to only considering $$k$$ points of $$A$$ for a fixed field $$k$$. you have to consider all maps $$A\to K$$ for every field $$K$$ and by taking kernels you recover prime ideals

Question: "I mean, it would be a generalization if one could capture every prime ideal of $$k[x_1,…,x_n]$$ just by considering the homomorphisms with values in $$k$$, but still we need to consider homomorphisms in any field in order to capture every prime ideal of a generic ring; however as I said this is not the case."

In general the ring $$A$$ is not a $$k$$-algebra for $$k$$ a field.

Lemma: If $$A$$ is a commutative unital ring and $$\mathfrak{p}\subseteq A$$ is a prime ideal, there is a field $$K$$ and a map of unital rings $$\phi: A \rightarrow K$$ with $$\mathfrak{p}=ker(\phi)$$.

Proof. Let $$u:A\rightarrow A_{\mathfrak{p}}$$ be the canonical map into the localization of $$A$$. It follows prime ideals $$\mathfrak{q}* \subseteq A_{\mathfrak{p}}$$ corresponds 1-1 to a prime ideal $$\mathfrak{q}\subseteq \mathfrak{p}$$ in $$A$$. Hence the zero ideal $$(o) \subseteq \kappa(\mathfrak{p})$$ in the residue field $$\kappa(\mathfrak{p})$$ has inverse image $$\mathfrak{p}$$ under the canonical map $$\phi_{\mathfrak{p}}: A \rightarrow \kappa(\mathfrak{p})$$. Hence you may choose $$\phi:=\phi_{\mathfrak{p}}$$.

Hence any prime ideal in $$A$$ is the kernel of some map into a field $$K$$. The problem is that the "set of all fields $$K$$" is a "very large set" - it is not even a set! The set of prime ideals in $$A$$ is "smaller" and this is one reason to work with $$Spec(A)$$: Since $$A$$ is a set it follows $$Spec(A)$$ is a set - it is a subset of the set of all subsets of $$A$$.

Example. If $$A$$ is a $$k$$-algebra of finite type over a field $$k$$, a non-maximal prime ideal $$\mathfrak{p}$$ will in general have a residue field $$\kappa(\mathfrak{p})$$ that is no longer a finite extension of $$k$$. If $$\mathfrak{p}$$ is maximal it follows $$k \subseteq \kappa(\mathfrak{p})$$ is a finite extension.

Here you find a discussion on the problem of defining algebraic "varieties" avoiding prime ideals and instead using maps to fields or the space of maximal ideals:

https://mathoverflow.net/questions/377922/building-algebraic-geometry-without-prime-ideals/378961#378961

Note that if $$A,B$$ are commutative unital rings that are finitely generated over a field $$k$$ or the integers $$\mathbb{Z}$$ the following holds:

Lemma.Let $$\phi:A \rightarrow B$$ be a map of rings and let $$\mathfrak{m}\subseteq B$$ be a maximal ideal. it follows $$\phi^{-1}(\mathfrak{m})\subseteq A$$ is a maximal ideal.

Proof. Assume $$k$$ is a field and let $$\mathfrak{n}:=\phi^{-1}(\mathfrak{m})$$. We get an inclusion of rings

$$k \subseteq \kappa(\mathfrak{n}) \subseteq \kappa(\mathfrak{m})$$

and $$\kappa(\mathfrak{n})$$ is an integral domain. Since $$\kappa(\mathfrak{m})$$ is a finite extension of $$k$$ the following holds: Let $$0\neq x\in \kappa(\mathfrak{n})$$ be an element. There is a canonical surjective map

$$\psi: k[t]\rightarrow k(x)$$

defined by $$\psi(t):=x$$, with $$ker(\psi):=(p(t))$$ for a non-zero ideal $$p(t)\in k[t]$$. Since $$k(x)$$ is an integral domain it follows $$(p(t))$$ is a maximal ideal, hence $$x$$ is a unit in $$\kappa(\mathfrak{n})$$, and it follows $$\mathfrak{n}$$ is a maximal ideal. A similar proof holds when $$k$$ is the ring of integers. QED.

If we let $$Specm(A)$$ denote the set of maximal ideals in $$A$$ we get by the Lemma an induced map

$$\phi_m: Specm(B) \rightarrow Specm(A)$$

defined by $$\phi_m(\mathfrak{m}):=\phi^{-1}(\mathfrak{m})$$, and the map $$\phi_m$$ is continuous in the topology induced by the Zariski topology. Hence the classical map defined for the affine spectra

$$\phi: Spec(B) \rightarrow Spec(A)$$

induce a map at the level of maximal ideals.

There is a class of rings - Hilbert-Jacobson rings - where the following result holds:

Theorem. Let $$A$$ be a Hilbert-Jacobson ring and let $$\phi: A\rightarrow B$$ be a map of commutative unital rings where $$B$$ is a finitely generated $$A$$-algebra. It follows $$B$$ is a Hilbert-Jacobson ring, and for any maximal ideal $$I\subseteq B$$ it follows $$J:=\phi^{-1}(I)\subseteq A$$ is a maximal ideal and the canonical map $$A/J \rightarrow B/I$$ is a finite extension of fields.

A ring $$A$$ is a Hilbert-Jacobson ring iff every prime ideal in $$A$$ is the intersection of the maximal ideals containing it.

For this class of rings you get a well defined continuous map at the level of max-spectra and you can from such a ring $$A$$ construct a locally ringed space where the underlying topological space $$X^m$$ is the set of maximal ideals in $$A$$ with the Zariski topology, and where the structure sheaf $$\mathcal{O}_{X^m}$$ has nilpotent elements. Hence the ringed space $$(X^m, \mathcal{O}_{X^m})$$ is in a sense a "compromise" between a "classical algebraic variety" and a "scheme". If the base-ring is finitely generated over a field $$k$$ it follows the points in the topological space corresponds to solutions of systems of polynomial equations with values in a finite extension of $$k$$, hence the definition is in a sense "more intuitive" and "pedagogical" and it may be interesting for people teaching algebraic geometry.