Building algebraic geometry without prime ideals

Question

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

Answer 1

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.

Answer 2

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:

https://stacks.math.columbia.edu/tag/01J5

https://stacks.math.columbia.edu/tag/04XE