$\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\ev{ev}$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.

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    $\begingroup$ 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, 2020 at 0:50
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    $\begingroup$ @BrianShin: isn't that exactly the $T_0$ issue mentioned in 2.? $\endgroup$ Dec 2, 2020 at 1:04
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    $\begingroup$ 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. $\endgroup$ Dec 2, 2020 at 1:07
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    $\begingroup$ @darijgrinberg the spectrum doesn’t see the nilpotents. You need the structure sheaf for that $\endgroup$ Dec 2, 2020 at 2:42
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    $\begingroup$ 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, 2020 at 8:23

2 Answers 2


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. $\endgroup$ Dec 2, 2020 at 8:57
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    $\begingroup$ 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, 2020 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? $\endgroup$ Dec 2, 2020 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. $\endgroup$ Dec 2, 2020 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 $\endgroup$ Dec 2, 2020 at 19:02
  • $\begingroup$ @DenisNardin for schemes yes, I was just wondering if we obtain a different notion for stacks. $\endgroup$ Dec 3, 2020 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, 2021 at 12:34

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