The Stacks project

In the proof of theorem 00ZP you write "the sheafification functor is the right adjoint of the inclusion functor", it should be the left adjoint.

Thanks and fixed here.

Dear Miles, what about $(x + y)/(x + 2y)$? This corresponds to the function which is $1$ when $y = 0$ and $2$ when $x = 0$, so it corresponds to $(1, 2)$ in $k(x) \times k(y)$. OK?

4th paragraph of proof says: "Let \overline{x_i} \in coker(\phi)$." Should say something like "Let$\overline{x_1},\dots,\overline{x_n} \in \coker(\phi)$ be a finite set of generators."

Not sure this is useful, but the lemma works for any valuation ring $R$: instead of Krull's intersection theorem, use the fact that $a$ has a content ideal (Lemma 0ASX) which is principal by Comment 8087. (I do realize that Lemma 0ASX comes a bit later).

It follows from Comment 8086 that the content ideal is principal.

It may be worth pointing out that $F'$ is of finite type. Namely, there is $F''\subset F'$ of finite type such that $x\in F''\otimes_R M$, and $F''=F'$ since $F' is minimal.

Maybe one should mention oi Lemma 26.12.7, that the factorisation is unique. This is given by Lemma 26.4.6.

https://stacks.math.columbia.edu/tag/02LV

I find your Lemma 02LX fishy. Suppose R = k[x,y]/(x*y). The zero divisors of R are the two minimal prime ideals (x) and (y). The multiplicative set S consists of anything not in (x) or (y) -- so anything with a constant term, or anything like x^n + y^m. The total ring of fractions is certainly a subring of the Cartesian product of two fields k(x) x k(y). I believe that it consists of pairs f(x), g(y) such that both are regular functions at x=0 and y=0, and take the same value f(0)=g(0). In particular the elements (1,0) and (0,1) of the Cartesian product are not in S^-1R.

Matsumura's 2 books and the Wikipedia page on total ring of fractions just say the obvious definition without discussing what it means or any serious examples. In your treatment, I think Q(R) should be a subring of the Cartesian product

A slightly more complicated example would be k[x,y]/(x^n,y^m), where the total ring of fractions is a subring of k(x)[y]/y^m x k(y)[x]/x^n with the first and second factors equal modulo smaller powers of x,y. Tricky to state correctly, although the idea is clear.

I'm trying to read Mumford, Curves on a surface, Lecture 9: Cartier divisors.

Best, Miles

Dear Anonymous, this is a rare instance where I slightly disagree with you. I think that this lemma and its proof never refers to modules in the Zariski topology. However, in translating to the case of schemes (at the very end), one just needs to know that the categories of (quasi-)coherent modules on a locally Noetherian scheme $U$ are the same as the corresponding categories of (quasi-)coherent modules on $U$ viewed as an algebraic space to be able to conclude. This was discussed for quasi-coherent modules much earlier, but was somehow missing in this section. So I've added that in this commit and I hope that this will help the future readers.

Thanks and fixed here.

Thanks and fixed here.

Please always comment on the page of the tag you are commenting on. Thanks. Fixed here.

Not going to do this. The thing with the short exact sequence is explained in Remark 13.12.4; as you can see we just use $K/\tau_{\leq a}K$ for the thing that fits into the ses.

Other people have mentioned they prefer to use superscripts, so use $\tau^{\leq n}$ and $\tau^{\geq n}$ when using truncation functors for complexes with upper numbering (so for cochain complexes). If you read this and have an opinion, please leave a comment.

Thanks and fixed here.

Thanks and fixed here.

First line of proof: $\mathcal{O}_{X_\eta}$ should be $\mathcal{O}_{X_\eta,x}$. But in fact it is not very clear whether one is proving that $X$ is reduced or that $X_\eta$ is. I propose to change the first 3 lines as follows:

Assume the special fibre is reduced. Let $x \in X$ be any point, and let us show that $\mathcal{O}_{X,x}$
is reduced. (This will prove that $X$ and $X_\eta$ are reduced). Let $x \leadsto x'$ be a specialization with $x'$ in the special fibre; such a specialization exists as a proper morphism is closed. Consider the local ring $A = \mathcal{O}_{X, x'}$. Then $\mathcal{O}_{X,x}$ is a localization of $A$, so it suffices to show that $A$ is reduced. Let $\pi \in R$ be a uniformizer. (The rest is unchanged)

THanks and fixed here.

Thanks and fixed here.

Please comment on each of the items separately (on the page of the tag itself) and tell me exactly what to fix.