In adic spaces, I believe analytic can refer to points associated to valuations whose kernel is not open in the topology of the ring. But I don't know whether that is the referent in this case. I don't like the terminology though. It gives me incredible urgers to define $\mathrm{Swalnut}\mathbb{C}[[t]], \mathrm{Spumpkin}\mathbb{C}[[t]], \ldots$

Wise words from Grothendieck, "A good mathematician always soaks his nuts while trying to solve a problem."

@ArunDebray With all respect to Mumford, his great expertise in the practice of algebraic geometry did not carry over to expertise in the explanation of algebraic geometry to non-mathematicians. It seems like his idea of a non-mathematician is someone who has at the very least a college degree or graduate degree in math or physics, but may not professionally practice math.

@EugeneEisenstein +1 for seconding Chandan's appreciation for the recommendation of Borger's viewpoint.

This sounds like a great premise for a new Saw-like horror movie. A group of strangers wake up in a dark, empty space. They need to determine the dimension of the ambient space, but their time is running out!

Would someone be willing to write the gist of the Mutylin's construction? The relevant pages of the google book are unavailable to me.

Thank you, Professor Conrad. I was familiar with that previous post, and your answer here will make a good complement.

I'm sure all of you are familiar, but for reference, the finitely generated case is covered by the Cassels embedding theorem, which ensures an embedding into infinitely many $\mathbb{Q}_p$.

This is related to Weierstrass preparation, correct? Yes, I understand the argument, but I would like to have more knowledge/control of the unit $u$ that can be factored off $b$.

I've got 99 vertices, but an edge ain't one.

Some more good info is in Ottem's answer here: mathoverflow.net/questions/47682/….

For starters, there's en.wikipedia.org/wiki/Weighted_projective_space and also the fact that the Veronese embedding exhibits a number of different gradings with isomorphic scheme structures, namely that scaling and shifting the grading unaffect the result.

Using the nuanced insight of @DeaneYang, we can simplify many known results: Let $p$ be a prime. Is there an expression $p = a^2 + b^2$ as a sum of squares? It depends. QED

The coefficients of the expansion are only unique up to a choice of representatives for $\mathbb{Z}/b\mathbb{Z}$, and the base expansions themselves are a result of the fact that $\mathbb{Z}$ is a discrete valuation ring, for various valuations.