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


On 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 Rankeya left comment #4307 on Lemma 10.114.7 in Commutative Algebra

Does the proof use that S is a domain anywhere?

On 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 left comment #4305 on Lemma 14.19.13 in Simplicial Methods

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

On 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 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 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 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 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 left comment #4299 on Section 5.26 in Topology

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

On left comment #4298 on Lemma 10.136.7 in Commutative Algebra

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On Bogdan left comment #4297 on Lemma 10.136.7 in Commutative Algebra

On 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 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 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 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 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 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 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 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.