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Let $R$ be a finitely generated $\mathbb{Z}$-algebra, and $\mathfrak{m}\subset R$ a maximal ideal. We wish to show $R/\mathfrak{m}$ is a finite field. Let $i: \mathbb{Z}\to R$ be the unique ring map; …

Here's an explicit example. Let $R=\mathbb{Z}[\sqrt{2}]$, let $X=\mathbb{P}^1_R$, and let $Y$ be the smooth projective conic defined by the equation $$(2-\sqrt{2})x^2+y^2+(2-\sqrt{2})z^2+xy+yz+(3-2\sq …

This is to expand on Akhil's answer. Locally free implies flat easily, so let's look at the other direction. It suffices to assume $A$ is local with maximal ideal $\mathfrak{m}$. Pick a basis of $M …

Suppose the structure morphism $g: X\to \operatorname{Spec}(A)$ is separated and of finite type, and $f: \operatorname{Spec}(B)\to \operatorname{Spec}(A)$ is faithfully flat; furthermore, assume $A, B …

No. Consider $\mathfrak{m}:=(x,y,z)\subset k[x,y,z]_{(x,y,z)}=:R$. Then the kernel of the map $$R^3\to \mathfrak{m}$$ defined by the minimal generating set $x,y,z$ is minimally generated by $$k_1:=( …

Let's make several reductions. First, the condition implies $f$ is injective, so letting $K=\operatorname{Frac}(R)$, we have that $f: K\to S'$ is an epimorphism, letting $S'$ be the localization of $ …

I've heard the term "conservative" used to describe functors which take nonzero modules to nonzero modules; see e.g. problem 4 from this problem set from a course on schemes, which is the converse of …

$\mathbb{Q}$ is flat over $\mathbb{Z}_p$, but not projective.

Should one learn point-set topology before real analysis or before studying metric spaces a bit? There are some advantages to doing so -- a more unified approach to real analysis or the study of metr …