The Stacks project

The assumptions on $D$ are very restrictive. A more general (and natural) statement, with essentially the same proof, would be:

Let $X$ be a locally Noetherian scheme. Let $U\subset X$ be open and dense. Assume that $X\setminus U\subset \mathrm{Reg}(X)$. (This holds in particular if $X\setminus U$ is the support of an effective Cartier divisor which is regular, by Algebra, Lemma 10.106.7.) Let $Y\to U$ be a finite étale morphism, unramified over $X$ in codimension $1$. Then (...)

In the paragraph right above Lemma 0EYD, it's written "In particular, we see that Y×USpec(Kξ) is the spectrum of a finite separable algebra Lξ/K.". I wonder if you mean Lξ/Kξ here.

In the last paragraph, of the 2 compositions that need to be equal one should have X_2 in it instead of X_1

"... that $f_\ast:CH_k(Y)\to CH_k(X)$ is an isomorphism." I think it should be "... that $f_\ast:CH_k(Y)\to CH_{k+r}(X)$ is an isomorphism."

Line 3 of the proof: should it be ``$f(x)$ is in fact in the fibre category...''?

Typo one line 2 of proof: "imlpications".

To prove 1 implies 2 wouldn't it also be correct the following argument? $N=0 \iff 0 \rightarrow N \rightarrow 0$ exact $\iff 0 \rightarrow N\otimes M\rightarrow 0$ exact $\iff N\otimes M=0$ To me it feels very slightly cleaner. Or am I missing something?

Okay, I feel this proof can be slightly upgraded. First, lemma 10.44.2 only works in characteristic $p$. Of course, the characteristic zero case is obvious, but why not just use 10.43.6 instead?

Now for the tricky part. The term universal homeomorphism is not yet defined, it only appears in definition 29.45.1 and in the context of schemes, despite its use in the title of section 10.46 (maybe that's a bit confusing, but this is not the main issue here).

It seems that throughout the Stacks Project, a great care is taken to not use the term universal homeomorphism for morphisms of rings. Instead, it is written in terms of morphisms on spectra (for instance in the proof of lemma 58.14.2), or with a sentence such as "induces a universal homeomorphism on spectra" (see for instance lemma 29.46.11).

So if we want to keep this consistency, I propose to first say that by lemma 10.45.4 (or by definition) $k^{perf}/k$ is purely inseparable, so we can apply lemma 10.46.10 to find that $k^{perf}\otimes_{k}K$ has a unique prime ideal.

If this is not nitpicking for you, then I would suggest to also edit the very few remaining cases of universal homeomorphisms of rings in the Stacks Project: tags 29.47.6, 37.24.7, 110.40 (twice) and maybe 29.46.8.

In Lemma 00ZO I think that $\varphi(id_V)$ should be $\varphi(\alpha)$.

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.