Recent comments—The Stacks project

Comments 1 to 20 out of 4063 in reverse chronological order.

On Jul 17, 2019 dpw left comment #4308 on Section 12.5 in Homological Algebra

In lemma 12.5.17, both the object $y\times_z\text{Ker}\gamma$ and the morphism $u\to\text{Coker}\alpha$ are denoted by $p$ . I think it looks a bit confusing.

On Jul 16, 2019 Rankeya left comment #4307 on Lemma 10.114.7 in Commutative Algebra

Does the proof use that S is a domain anywhere?

On Jul 15, 2019 correction_bot left comment #4306 on Lemma 28.45.7 in Morphisms of Schemes

The proof says we'll only prove (1) and (2) but then talks about $X^{sn}$ which only plays a role in (3) and (4) (should be $X^{awn}$ instead).

On Jul 14, 2019 David Roberts left comment #4305 on Lemma 14.19.13 in Simplicial Methods

The right hand side has U,V and W mixed up.

On Jul 13, 2019 bogdan left comment #4304 on Proposition 10.156.15 in Commutative Algebra

Two small typos:

1) The phrase ''Let $S'⊂K'$ be the subring...'' should read ''Let $S'⊂K''$ be the subring...''

2) The sentence ''We claim that $I$ is generated by all linear forms $aX+b$ such that $ax=b$ in $K$ '' should read ''We claim that $I$ is generated by all linear forms $aX-b$ such that $ax=b$ in $K$ ''.

On Jul 12, 2019 David Speyer left comment #4303 on Section 104.16 in Examples

It seems to me that $\mathrm{Rad}(M) = \mathfrak{m} B + \mathfrak{n} B$ should read $\mathrm{Rad}(M) = \mathfrak{m} B \cap \mathfrak{n} B$.

On Jul 12, 2019 Laurent Moret-Bailly left comment #4302 on Section 16.1 in Smoothing Ring Maps

Line 11: I guess "filtered limit" should be "filtered colimit".

On Jul 11, 2019 Rankeya left comment #4301 on Section 16.1 in Smoothing Ring Maps

Okay, may be you are assuming $A$ is henselian in para 4 as well, in which case my question is redundant since henselian $G$ rings are excellent.

On Jul 11, 2019 bogdan left comment #4300 on Lemma 10.155.16 in Commutative Algebra

The proof implicitly uses the fact that $S_{\mathfrak q}$ is noetherian (when it cites 00PD). This follows from Tag 00PG but this result is not explicitly cited now (the proof can be significantly simplified in this situation).

On Jul 10, 2019 Mark Bowron left comment #4299 on Section 5.26 in Topology

"but the converse does not holds in general" has a typo.

On Jul 10, 2019 newmeditech left comment #4298 on Lemma 10.136.7 in Commutative Algebra

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On Jul 09, 2019 Bogdan left comment #4297 on Lemma 10.136.7 in Commutative Algebra

A formula $μ(dx_λ−df_λ)+λ(dx_μ−df_μ)−dx_{λμ}−df_{λμ}=0$ should read $μ(dx_λ−df_λ)+λ(dx_μ−df_μ)−dx_{λμ}+df_{λμ}=0$

On Jul 08, 2019 Goodluckthere left comment #4296 on Section 5.8 in Topology

Possible typo: In the proof of the third part of lemma 004.W you take $A$ to be a set of irreducible sets but later you consider $A'\subset A$ to be a set of indices of elements in $A$ .

On Jul 06, 2019 comment_bot left comment #4295 on Lemma 36.54.11 in More on Morphisms

It would be useful to strengthen this statement: it suffices to assume that $X$ is a local complete intersection (with $Y$ still regular). A similar comment applies to https://stacks.math.columbia.edu/tag/0E9J

On Jul 05, 2019 Xuande Liu left comment #4294 on Lemma 28.12.2 in Morphisms of Schemes

Without loss of generality we can assume that $S=\mathit{Spec}(B)$ is affine. Let $A=\Gamma(X,\mathcal O_X)$ . And hence we can factorize $f$ as $X\to \mathit{Spec}(A)\to\mathit{Spec(B)}$ . The former morphism is an open immersion and the latter is morphism between affine schemes. Therefore $f$ is separated.

On Jul 05, 2019 Xuande Liu left comment #4293 on Lemma 28.12.2 in Morphisms of Schemes

Without loss of generality we can assume that $S=\mathit{Spec}(B)$ is affine. Let $A=\gamma(X,\mathcal O_X)$ . And hence we can factorize $f$ as $X\to \mathit{Spec}(A)\to\mathit{Spec(B)$ . The former morphism is an open immersion and the latter is morphism between affine schemes. Therefore $f$ is separated.

On Jul 05, 2019 Bogdan left comment #4292 on Lemma 15.45.5 in More on Algebra

Let K be the fraction field of $k[[x_1,…,x_d]][y_1,…,y_m]$ should read $k[[x_1,…,x_n]][y_1,…,y_m]$ .

I guess $A=\cap_{k'} A′$ (the last line) should read $A^p=\cap_{k'} A′$ .

On Jul 04, 2019 Laurent Moret-Bailly left comment #4291 on Lemma 7.47.11 in Sites and Sheaves

I think one can simplify the exposition by reducing (via 00Z7(1)) to the case of just one presheaf.

On Jul 04, 2019 Kazuki Masugi left comment #4290 on Lemma 7.47.11 in Sites and Sheaves

I can't make sure that $\varphi \mapsto s$ is right inverse to the map in the statement. Would you please describe the precise way to verify?

On Jul 04, 2019 David Hansen left comment #4289 on Lemma 35.3.7 in Derived Categories of Schemes

In the commutative square here, A and B should be swapped.