Comments 1 to 20 out of 5637 in reverse chronological order.


On left comment #6057 on Section 26.21 in Schemes

On the last sentence of the first paragraph of lemma 01KP should be $A\otimes_RB\to(A\otimes_RB)/J$ .

On Rachel Webb left comment #6056 on Lemma 37.35.3 in More on Morphisms

In the lemma statement, $Y \rightarrow S$ should be $Y \rightarrow T$.

On Chern left comment #6055 on Section 57.31 in Fundamental Groups of Schemes

Typo: "for every for every"

On Jonas Ehrhard left comment #6054 on Lemma 10.131.4 in Commutative Algebra

I think this lemma should be stated after the discussion of diagram 10.131.5.1 / 00RQ, since the diagram explains the colimit $\text{colim}_i \Omega_{S_i / R_i}$.

Also, I just spend half an hour searching for the existence of colimits of rings, until I found exercise 109.2.2 / 078J. Maybe this could be referenced?

On Jonas Ehrhard left comment #6053 on Section 10.131 in Commutative Algebra

This might be nitpicking, but shouldn't it be $\text{d} \varphi(r) = 0$ just in front of definition 10.131.2?

On Hans Schoutens left comment #6052 on Section 20.30 in Cohomology of Sheaves

In 2nd line of the statement of Lemma 0FKT, ...where F^\bullet be a bounded... should be ...where G^\bullet is a bounded...

On Ashwin Iyengar left comment #6051 on Remark 31.29.5 in Divisors

Should the first equation read $U = \mathrm{Spec}(A) \backslash \left( V(\mathfrak p_1) \cup \cdots \cup V(\mathfrak p_r)\right)$?

On left comment #6050 on Lemma 10.158.1 in Commutative Algebra

What is meant is that there is an isomorphism of fields $\varphi : K \to k'$ which induces the indentity on $k$. More precisely, if $i : k \to K$ and $j : k \to k'$ are the inclusion maps, then $\varphi \circ i = j$. I will improve the text the next time I go through all the comments.

On Mark left comment #6049 on Lemma 10.159.6 in Commutative Algebra

It might be better to add a citation for the sentence "Then clearly $R$ is a discrete valuation ring." I think it's regular+dimension 1 (Tag 00PD (3)). (At least this was not clear to me at first glance, I wondered quite a while around noetherianness.)

On Mark left comment #6048 on Lemma 10.158.1 in Commutative Algebra

In the statement of the lemma: what does it mean for a field extension $k\subset k'$ to be isomorphic to another field extension $k\subset K$?

On left comment #6047 on Section 10.25 in Commutative Algebra

@#6046. No. I suggest trying to make a counterexample.

On Shurui Liu left comment #6046 on Section 10.25 in Commutative Algebra

Lemma 00EW (3) doesn't need R to be reduced and hence lemma 02LX needn't require $\mathfrak q_1 \cup \ldots \cup \mathfrak q_ t$ to be the set of zerodivisors of R (since it is automatically true).

On Harry Gindi left comment #6045 on Lemma 16.2.8 in Smoothing Ring Maps

I found this last one a bit tricky. I wrote up a proof if you want to add it:

Observe that the ideal generated by the coefficients (f_1,...,f_r) c B is the unit ideal, so the family of maps {B→B_{f_i}}{i=1}^r is a Zariski open cover. It suffices therefore to show that each of the B{f_i} is smooth over R. But observe that B_{f_i}≅A_{f_i}?91?x1,...,x_{i-1},x_{i+1},...,x_r] (omitting the ith indeterminate) for each 1≤i≤r, which is witnessed by the map sending x_i to the polynomial

1- ((f_1/f_i) x_1 + … + (f_{i-1}/f_i) x_{i-1} + (f_{i+1}/f_i) x_{i+1} + … + (f_r/f_i) x_r)

in A?91?x_1,...,x_{i-1},x_{i+1},…,x_r].

But since f_i belongs to H_{A/R}, we know that R→A_{f_i} is smooth, and polynomial rings in finitely many indeterminates are smooth so we see A_{f_i}→B_{f_i} is smooth and therefore R→A_{f_i} → B_{f_i} is smooth, as desired.

On Harry Gindi left comment #6044 on Lemma 16.2.2 in Smoothing Ring Maps

Should the statement say there exists a in A, a not in q, c≤min(n,m), subsets of cardinality c, etc, such that if ã is a lift of a to the polynomial ring, then ã satisfies the above properties in terms of the jacobian determinant and where ãf_ℓ belongs to the above ideal?

If you don't say something like 'a lift ã of a', you're multiplying an element of the quotient by a polynomial in the polynomial ring, so it doesn't typecheck.

On Harry Gindi left comment #6043 on Section 16.8 in Smoothing Ring Maps

(It's true more generally for sifted colimits (and since we're looking at products rather than finite limits) in the category of sets, but we're using here that filtered colimits in CRing are computed as the filtered colimit of the underlying sets, which I don't think holds for sifted colimits in CRing).

On Harry Gindi left comment #6042 on Section 16.8 in Smoothing Ring Maps

Small nitpick in the hint for 07F3. This is only true for filtered colimits not all colimits (more or less by definition).

On Hans Schoutens left comment #6041 on Section 20.15 in Cohomology of Sheaves

In the second displayed formula, $h(\xi)$ should be $h(\alpha)$

On Shurui Liu left comment #6040 on Section 10.11 in Commutative Algebra

A stupid remark: In the statement of lemma 0G8N and 0G8P, it should be N instead of K.

On left comment #6038 on Definition 6.21.7 in Sheaves on Spaces

See https://stacks.math.columbia.edu/tag/008K#comment-5957

On Marin Genov left comment #6037 on Definition 6.21.7 in Sheaves on Spaces

Hi! Perhaps I am missing something, but what is the raison d'etre of this definition and the short exposition that follows it? More concretely, how is this any different from a morphism of (pre)sheaves G --> f_{*} F ?

If the idea is to give another interpretation of the adjunction b/w the pullback and pushforward of sheaves, perhaps it's reasonable to establish the equality of both sides of the adjunction with the set of collections of maps \phi_{VU}: G(V) \to F(U) compatible with restrictions (as Ravi Vakil does in The Rising Sea FAG, p. 93, Ex. 2.7.B).

What do you think?