$\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\ev{ev}$Teaching algebraic geometry, in particular schemes, I am struggling to provide intuitive proofs. In particular, I find it counter-intuitive that points are prime ideals. I discovered a trick which I suspect is not new. Basically, you build the functor of points into the definition. I want to modify the definition of $\Spec(R)$ as follows:

As a set, $\Spec(R)$ is simply all pairs $x=(k_x, \ev_x)$ where $k_x$ is a field and $\ev_x:R\to k_x$ is a homomorphism. Then as usual, elements of $R$ are called functions and the value of a function $f\in R$ at a point $x$ is $f(x)\mathrel{:=}\ev_x(f)\in k_x$. Then it continues as usual: closed set is where some collection of functions vanishes. Basic open set is where some function is invertible.

Of course, there are some problems with this approach:

  1. The class of all fields is not a set. Technically, we can limit ourselves to some very large set of "test fields". So this can be swept under the rug.

  2. $\Spec(R)$ with this definition is not $T_0$. But after getting used to spaces being not Hausdorff it should be easy to take it to the next level with spaces being not $T_0$. Of course, to every non-$T_0$ space there is a canonically associated $T_0$ space where you identify topologically indistinguishable points, so you recover the usual construction of $\Spec(R)$ this way.

Nevertheless, I find this approach much more intuitive, because it seems like a natural question to solve some system of equations in some unknown field, rather then studying prime ideals (which is of course basically the same thing, language aside).

Is this not new? Are there any lecture notes following this approach? Of course, the full "functor of points" approach sort of contains this one, but notice that to do what I want I do not need Yoneda lemma, I do not ask for functoriality, so I do not need to sweep under the rug all the tedious checks of naturality. So I find it more basic than functor of points.

Here is an example. When we construct the localization of a ring $R$ with respect to a multiplicative set $S$ we prove that prime ideals of $S^{-1}R$ are in bijection with a subset of ideals of $R$. With this approach the corresponding statement is a simple consequence of the universal property of the localization, there is nothing more to prove.

Another example. Prove that the map $\mathbb{A}^1\to \mathbb{A}^3$ given by $t\to (t^3, t^4, t^5)$ has image $Z(xz-y^2, x^3-yz, x^2 y -z^2)$. This becomes simply high school algebra.

  • 2
    $\begingroup$ How do you deal with field extensions? If we have a point $R \to \mathbb{R}$, then we get another point by inclusion into $\mathbb{C}$. Are these just two different points now? $\endgroup$ – Brian Shin Dec 2 '20 at 0:50
  • 3
    $\begingroup$ @BrianShin: isn't that exactly the $T_0$ issue mentioned in 2.? $\endgroup$ – Sam Hopkins Dec 2 '20 at 1:04
  • 2
    $\begingroup$ Why take points over fields when you can take points over arbitrary commutative $k$-algebras? (Aren't you losing nilpotents with your approach, too?) Marc Nieper-Wißkirchen's Algebraische Geometrie lecture notes (used to be on his website; now I can only find them on libgen) take a functor-of-points approach; from what I recall at least one version of EGA tried the same -- but I guess both are above the level at which you're trying to teach. $\endgroup$ – darij grinberg Dec 2 '20 at 1:07
  • 3
    $\begingroup$ @darijgrinberg the spectrum doesn’t see the nilpotents. You need the structure sheaf for that $\endgroup$ – Benjamin Steinberg Dec 2 '20 at 2:42
  • 1
    $\begingroup$ I thought of this a while back. Interestingly, this generalizes to a notion of spectrum for non commutative rings (you replace fields by matrix algebras, more generally, what we really want to consider is the "set" of all possible linear representations of your algebra). I don't know if this is a fruitful generalization. $\endgroup$ – Asvin Dec 2 '20 at 8:23

Actually, you have rediscovered a nice motivation of using prime ideals as points. Indeed, your collection of points are triples $(R, k_x, \mathrm{ev}_x)$ where , $\mathrm{ev}_x \colon R \to k_x$ is a homomorphism. The collection of all such triples is a class rather a set. In any case, you should not change the universe to get the underlying topological space whose functions are given by $R$, much as you won't change the universe when you reconstruct a differential manifold from the algebra of differentiable functions.

A nice solution is to impose an equivalence relation on the set of points. Define $$(R, k_x, \mathrm{ev}_x) \sim (R, k_y, \mathrm{ev}_y)$$ whenever there are field extensions $i_1 \colon k_x \hookrightarrow K$ and $i_2 \colon k_y \hookrightarrow K$ such that $i_1 \circ \mathrm{ev}_x = i_2 \circ \mathrm{ev}_y$. After all, a point with coordinates in a field remains the same if we consider the coordinates in a bigger field. Now take the quotient set of the equivalence relation. It is clear that the triples $(R, k_x, \mathrm{ev}_x)$ are classified by $\mathrm{Im}(\mathrm{ev}_x)$, equivalently by $\mathrm{Ker}(\mathrm{ev}_x)$, that turns out to be a prime ideal. Thus, every equivalence class has a canonical representative $(R, \kappa(\mathfrak{p}), \mathrm{ev}_\mathfrak{p})$ where $\mathfrak{p}$ is a prime ideal in $R$, $\kappa(\mathfrak{p}) = R_\mathfrak{p}/\mathfrak{p}R_\mathfrak{p}$, the residue field of $\mathfrak{p}$ and $\mathrm{ev}_\mathfrak{p} \colon R \to \kappa(\mathfrak{p})$ the canonical map. So in fact points as maps to fields are classified by primes, have a canonical field where elements of $R$ may be evaluated and the collection that form the equivalence classes is clearly a set.

Of course, the next step is to define a sheaf of rings, that in some sense, might be interpreted as a sheaf of functions on $\mathrm{Spec}(R)$. This is exactly the motivation I use for the philosophy points are primes in algebraic geometry in my graduate courses under the name "the sermon of points". Of course, this point of view is well-known though it is rarely displayed in print.

  • $\begingroup$ Is there a reason you include $R$ in $(R, k_x, \ev_x)$? It would be nice to have a reference. Do you have notes? Sometimes when I prove something I read a proof in a standard reference and maybe I don't find time to optimize it by translating into this language, some other times I get a neater proof. It would be nice to have it applied systematically. $\endgroup$ – Anton Mellit Dec 2 '20 at 8:57
  • 2
    $\begingroup$ I put the $R$ in the notation to emphasize that $\mathrm{Spec}(R) $ is a functor over rings. Notice that points are (contravariantly) functorial wrt to homomorphism of rings. As a byproduct you obtain that a prime ideal contracts to a prime. As for a written account, I would send you if I ever get it to typeset my notes, but I don't know if I would do it soon, unfortunately. $\endgroup$ – Leo Alonso Dec 2 '20 at 9:02

This nice approach to points on schemes in fact becomes crucial once one leaves the world of schemes and travels to the galaxy of stacks.

For an algebraic stack $X$, one defines a point of $X$ to be a morphism $\mathrm{Spec}(k) \to X$, modulo the natural equivalence relation discussed above.

See in particular:



  • $\begingroup$ Oh, nice to see that this is how it's done for stacks. So it's good to put stress on this approach from the start. There is one difference though: your equivalence relation (two morphisms are equivalent if there is a third one covering both) is not exactly my equivalence relation (topologically indistinguishable). But for decent spaces these should be the same (stacks.math.columbia.edu/tag/03K3). Or do you think the two equivalence relations are equivalent more generally? $\endgroup$ – Anton Mellit Dec 2 '20 at 10:45
  • $\begingroup$ I don't fully understand the difference between what you are suggesting and what the stacks project is doing. For schemes the two definitions seem to coincide and there are only potential differences once you start looking at algebraic spaces. In any case, I believe that stacks project definition should be the right one to use. $\endgroup$ – Daniel Loughran Dec 2 '20 at 12:30
  • $\begingroup$ @AntonMellit In your definition two points are topologically indistinguishable exactly when their kernel has the same elements (since those are exactly the basis open sets that do not contain your point), so this definition coincides with the definition in the stacks project $\endgroup$ – Denis Nardin Dec 2 '20 at 19:02
  • $\begingroup$ @DenisNardin for schemes yes, I was just wondering if we obtain a different notion for stacks. $\endgroup$ – Anton Mellit Dec 3 '20 at 16:09
  • $\begingroup$ This sort of perspective is similarly related to what happens when you "beef up" your prime ideals into valuations e.g. in rigid analytic geometry (the kernel of a valuation is a prime ideal, but the valuation contains more informaion). The valuation comes with a codomain, but you end up identifying valuations with different value groups in a similar way to here. $\endgroup$ – Tim Campion Feb 4 at 12:34

In this thread

Why must nilpotent elements be allowed in modern algebraic geometry?

you find the following construction: Let $k$ be a field or the ring $\mathbb{Z}$ of integers and let $X$ be a scheme of finite type over $k$. By this we mean $X$ has a finite open affine cover $X=\cup_{i=1}^n Spec(A_i)$ where $A_i$ is a finitely generated $k$-algebra for all $i$.

Definition 1. Let $X^m$ be the set of closed points in $X$ with $u:X^m \rightarrow X$ the canonical inclusion map and with the induced topology. Let $\mathcal{O}_{X^m}:=u^{-1}(\mathcal{O}_{X})$ be the topological inverse of $\mathcal{O}_X$.

There is the following Lemma:

Lemma 1. Let $\phi: A\rightarrow B$ be a map of finitely generated $k$-algebras where $k$ is a field or the ring of integers. If $\mathfrak{m}\subseteq B$ is a maximal ideal it follows $\mathfrak{n}:=\phi^{-1}(\mathfrak{m})\subseteq A$ is a maximal ideal.

Lemma 1 remains true if $A,B$ are essentially of finite type over $k$, since the maximal ideals in a localization $S^{-1}B$ are in 1-1 correspondence with maximal ideals in $B$ that do not meet $S$. Hence the definition and Lemma 1 makes sense for any local ring $B_{\mathfrak{p}}\cong S^{-1}B$ where $S:=B-\mathfrak{p}$.

Hence the classical map $\phi^*: Spec(B)\rightarrow Spec(A)$ induced by the map of rings $\phi: A\rightarrow B$, induce a map

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

which is continuous in the topology induced from the Zariski topology on $S:=Spec(A)$ and $X:=Spec(B)$. The map of structure sheaves

$$\phi^{\#}: \mathcal{O}_S \rightarrow \phi_*\mathcal{O}_X$$

induce a map

$$\phi(m)^{\#}: \mathcal{O}_{S^m}\rightarrow \phi(m)_*(\mathcal{O}_{X^m}).$$

Hence we get from the map $\phi: A\rightarrow B$ a map of locally ringed spaces

$$(\phi(m),\phi(m)^{\#}):(X^m, \mathcal{O}_{X^m})\rightarrow (S^m, \mathcal{O}_{S^m}).$$

It follows we may associate to a finitely generated $k$-algebra $B$ the pair $(X^m, \mathcal{O}_{X^m})$ which is a locally ringed space. This construction is functorial and we get a functor

$$F: \underline{Alg^f(k)} \rightarrow \underline{RSpace}$$

where $\underline{Alg^f(k)}$ is the category of commutative unital $k$-algebras that are fintely generated over $k$ and $\underline{RSpace}$ is the category of locally ringed spaces. When we pass to global sections we recover the original map $\phi$. If we require a map of locally ringed space to be "local" it seems the following holds:

$$Hom_{Rings}(A,B) \cong Hom_{RSpace}((X^m,\mathcal{O}_{X^m}), (S^m, \mathcal{O}_{S^m}))$$

Hence using the functor $F$ we get an embedding of the category $\underline{Alg^f(k)}$ as a sub category of the category of locally ringed spaces - we may view $F$ as a "geometrization functor". If $\phi$ is a morphism in $\underline{Alg^f(k)}$, it follows $F(\phi)$ is an isomorphism iff $\phi$ is an isomorphism.

Example. The ringed space $(X^m, \mathcal{O}_{X^m})$ is a "compromise" between a "classical algebraic variety" and a "scheme" - it is more intuitive and you can speak about nilpotent elements in the structure sheaf. If $I \subseteq A$ is an ideal and $X:=Spec(A/I), X(n):=Spec(A/I^{n+1})$, it follows $X^m$ and $X(n)^m$ are isomorphic as locally ringed spaces iff $n=0$. The underlying topoplogical spaces are isomorphic since a maximal ideal $\mathfrak{m}\subseteq A$ contains $I$ iff it contains $I^{n+1}$.

Lemma 2. If $X^m:=Spec(B)^m$ and $\mathfrak{m}\in X^m$ is a point it follows

$$ \mathcal{O}_{X^m,\mathfrak{m}}\cong \mathcal{O}_{X,\mathfrak{m}}$$

Proof: There is the embedding $u: X^m \rightarrow X$ and by definition

$$\mathcal{O}_{X^m,\mathfrak{m}}:=lim_{\mathfrak{m}\in U^m}u^{-1}(\mathcal{O}_X)(U^m) \cong $$

$$ lim_{\mathfrak{m}\in U}\mathcal{O}_X(U) \cong \mathcal{O}_{X,\mathfrak{m}}.$$ QED

Hence the sheaves $\mathcal{O}_{X^m}$ and $\mathcal{O}_X$ have the same stalks at maximal ideals.

Example. If $(A,\mathfrak{m})$ is a local ring and $X^m:=Spec(A)^m$, you get

$$\mathcal{O}_{X^m}(X^m) \cong A \cong A_{\mathfrak{m}}\cong \mathcal{O}_{X^m,\mathfrak{m}}$$

since $A$ is local and the multiplicative set $S:=A-\mathfrak{m}$ consists of units. Hence when you localize $A$ at $S$ you recover $A$. The topological space $X^m$ is the one point space, but the information on the "subvarieties" through the point $\mathfrak{m}$ is encoded in the structure sheaf: It is the set of prime ideals in the ring of global sections of the structure sheaf. Hence the information on these subvarieties is not lost when passing to the ringed space $X^m$.

In the above thread I explain why $X^m$ is a locally ringed space that is "similar" to a "classical algebraic variety" as defined in Hartshornes book in chapter I as the set of zeros of an ideal in a polynomial ring in a finite set of variables. The difference is that Hartshorne starts with a fixed algebraically closed field $k$ and an ideal $I:=\{f_1,..,f_n\} \subseteq B:=k[x_1,..,x_n]$ in a polynomial ring $B$ in a finite set of variables.

Example: Hartshorne defines $V(I)$ as the "set of $n$-tuples" $\{t:=(t_1,..,t_n)\in k^n\}$ such that $f(t_1,..,t_n)=0$ for all poynomials in $I$. Hence Hartshornes $n$-tuples $t$ have their coefficients in the base field $k$. Since the field $k$ is algebraically closed it follows the set of maximal ideals $\mathfrak{m}$ in $X^m$ all have residue field $k$. Hence for an algebraically closed field $k$ it follows $X^m=V(I)$ if $X:=Spec(A)$ and $A:=B/I$. The above Definition 1 makes sense for any Hilbert-Jacobson ring $k$. You need the property that any prime ideal $\mathfrak{p}$ in $k$ is the intersection of maximal ideals.

Example. Let $A:=\mathbb{R}[x]$ and let $C:=Spec(A)$. It follows $C^m$ is the set of irreducible polynomials in $A$. A polynomial $p(x)$ in $A$ is irreducible iff $p(x)=x-r$ with $r\in \mathbb{R}$ or $p(x)=(x-z)(x-\overline{z})=x^2-4ax+a^2+b^2$ with $a,b\in \mathbb{R}$ and $b \ne 0$. In the above thread I explain how you may use nilpotent elements to Taylor expand sections of $\mathcal{O}_{C^m}(U)$. To explain Taylor expansion for sections of sheaves to students using real curves is "easier to understand" and "more intuitive". Sometimes students have problems understanding the field of complex numbers.

The structure sheaf $\mathcal{O}_{X^m}$ has the property

P1. $\Gamma(X^m, \mathcal{O}_{X^m})=i^{-1}(\mathcal{O}_X)(X^m):=lim_{X^m \subseteq U}\mathcal{O}_X(U) \cong \Gamma(X, \mathcal{O}_X)$,

hence $\mathcal{O}_{X^m}$ and $\mathcal{O}_X$ have the same global sections. Hence if $X:=Spec(A)$ it follows $\Gamma(X^m, \mathcal{O}_{X^m})=A$ and you recover the ring $A$ from the locally ringed space $(X^m, \mathcal{O}_{X^m})$.

Example: If $S:=A[x_1,..,x_n]/I$ where $I$ is a homogeneous ideal and $A$ is a Hilbert-Jacobson ring we may define $X^m \subseteq X:=Proj(S)$. If $\mathcal{E}$ is any finite rank locally trivial $\mathcal{O}_X$-module, we may define $\mathbb{P}(\mathcal{E}^*)^m \subseteq \mathbb{P}(\mathcal{E}^*)$. If $I \subseteq A\otimes A$ is the ideal of the diagonal and if

$\mathcal{P}(l):=Spec(A\otimes A/I^{l+1})$

we may define $\mathcal{P}(l)^m$. It has the property that

$\Gamma(\mathcal{P}(l), \mathcal{O}_{\mathcal{P}(l)})=A\otimes A/I^{l+1}$.

The canonical surjective map $m: A\otimes A/I^{l+1} \rightarrow A$ gives a one-to-one correspondence between $\mathcal{P}(l)^m$ and $Spec(A)^m$. Hence $\mathcal{P}(l)^m$ has the same points as $Spec(A)^m$ but it has non-trivial nilpotent elements in the structure sheaf - it is a "classical algebraic variety" with a non-reduced structure sheaf.

Comment: "Of course, there are some problems with this approach: The class of all fields is not a set. Technically, we can limit ourselves to some very large set of "test fields". So this can be swept under the rug."

Remark: The ring $A$ can be an arbitrary finitely generated $k$-algebra and it can be non-reduced. Hence this gives a way of introducing nilpotent elements for "classical algebraic varieties" without speaking about "classes of fields". Since $X^m$ is defined as the "set of closed points" in the "set" $X$, there are no "set theoretic difficulties" and if you want to teach algebraic geometry to students you want to avoid this. Moreover you don't use prime ideals, hence the construction gives an answer to the original question: Yes you may do this if the base ring is a finitely generated ring over a field or the integers $\mathbb{Z}$. Some people "complain" about the fact that a "scheme" is unintuitive, but if the base field is not algebraically closed and if you construct the space $X^m$ above, it follows a section $s\in \mathcal{O}_{X^m}(U^m)$ gives rise to a function $s^m: U^m(k) \rightarrow k$ at the level of $k$-rational points. It does not give a function on $U^m$ in general.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.