I’m working through beginner scheme theory and the affine real line is weird. There’s all the usual points, and sure, the generic point is smeared all over, but then we have a ton of extra closed points too. I’m imagining it as the closed upper half plane, but that doesn’t seem quite right
How do people visualise the closed points associated to irreducible polynomials in general fields? Do they move to the algebraic closure to split everything at once or do Galois splitting fields one by one? Or does it not matter?
Oct 27, 2026, 7:05 AM@gthm I think the standard way would be to think of A^1 of the algebraic closure, modulo the action of the absolute Galois group. In cases like k = R this is reasonably explicit/geometric, in cases like k = Q not so much.
@jdw as a geometer who never even took a class on Galois theory I’m having a lot of “huh” moments
Oct 27, 2026, 8:20 AM@gthm Let's forget about the generic point for a moment, and let's assume that everything is separable (for example in char. 0). Let K be the algebraic closure of k. There is a natural map K -> A^1_k mapping an element z \in K to the ideal of polynomials vanishing on z. This is a prime ideal, so it is a point in A^1_k.
The map is surjective onto the closed points of A^1_k: Every maximal ideal in k[x] is generated by an irreducible polynomial f which has a root z \in K. This z is a preimage.
Two elements in K are identified under the map (i.e. they have the same minimal polynomial) if and only if they are conjugate under a Galois automorphism. This is a standard result from Galois theory. The non-trivial direction is that elements with the same minimal polynomial f are conjugate. In a first step it is enough to show that they are conjugate in the splitting field of f, this result is typically formulated as »the Galois group of a polynomial acts transitively on its roots« in introductions to Galois theory. The automorphism can then be extended to the algebraic closure.
Hence the map induces a bijection between points of K modulo the Galois group onto the set of closed points of A^1_k. The only additional point is the generic point.
Oct 27, 2026, 8:26 AM@gthm For example if k = R and K = C, both elements i and -i of C are mapped to the prime ideal generated by (x^2 + 1). The element 5 is mapped to the prime ideal (x - 5). In general, two elements of C have the same minimal polynomial over R if and only if they are equal or one is the complex conjugate of the other. We have A^1_R = C / (complex conjugation) + generic point.
