$\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\ev{ev}$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:

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.

$\Spec(R)$ 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.

Algebraische Geometrielecture 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 1:075more comments