Comments 1 to 20 out of 6958 in reverse chronological order.


On left comment #7459 on Section 10.132 in Commutative Algebra

The omitted part is something about exterior powers and has nothing to do with differentials. It just says that given a ring $R$, an $R$-module $M$, and an integer $p \geq 1$ the map $M \times \ldots \times M \to \wedge^p_R(M)$, $(m_1, \ldots, m_p) \mapsto m_1 \wedge \ldots \wedge m_p$ is the universal map which is $R$-linear in each entry and alternating. This belongs in Section 10.13.

On ? left comment #7458 on Section 10.132 in Commutative Algebra

?The most difficult part is omitted?

On nhw left comment #7457 on Lemma 4.41.1 in Categories

In the first line should "categori" be category?

On mi left comment #7456 on Lemma 10.157.6 in Commutative Algebra

In the statement of (1), shall we also say a is not a unit?

On WhatJiaranEatsTonight left comment #7455 on Lemma 20.20.3 in Cohomology of Sheaves

I think here we shall add $\cap_{i\in I}U_i$ and $gcd(n_i)$ but not $\cup_{i\in I}U_i$.

On Hao Peng left comment #7454 on Definition 8.3.1 in Stacks

I am confused by the cocycle condition. My worry is that the composition doesn't make sence: While $pr^\star_{01}\phi_{ij}$ is a morphism from $(U_{ijk}\to U_{ij})^\star(U_{ij}\to U_i)^\star X_i$ to $(U_{ijk}\to U_{ij})^\star(U_{ij}\to U_j)^\star X_j$, $pr^\star_{12}\phi_{jk}$ is a morphism from $(U_{ijk}\to U_{jk})^\star(U_{jk}\to U_j)^\star X_j$ to $(U_{ijk}\to U_{jk})^\star(U_{jk}\to U_k)^\star X_k$. Thus it is preassumed that $(U_{ijk}\to U_{ij})^\star (U_{ij}\to U_j)^\star X_j=(U_{ijk}\to U_{jk})^\star (U_{jk}\to U_j)^\star X_j$, which is not in general true. Is this a mistake or we insect a natural isomorphism between them, or we can choose the cleavage such that this is true for any fiber products?

On Hao Peng left comment #7453 on Lemma 15.22.10 in More on Algebra

Sorry, in the argument above, change "$R$ is a domain" to "$M$ is torsion-free".

On Hao Peng left comment #7452 on Lemma 15.22.10 in More on Algebra

I noticed that after minor change this lemma holds for a Bezout domain too:

Let $\sum a_ix_i=0$ where $a_i\in R^*, x_i\in M$, then consider $(a_1, \ldots, a_n)=(a)$, thus $a_i=ab_i$ for some $b_i\in R$, and $\sum c_ib_i=1$ for some $c_i\in R$. Because $R$ is a domain, $\sum_i b_ix_i=0$, so we can take $\overrightarrow{y}=A\overrightarrow{x}$, where $A=1-\overrightarrow{c}\overrightarrow{b}^t$. Then $\overrightarrow{b}^tA=\overrightarrow{b}^t-\overrightarrow{b}^t\overrightarrow{c}\overrightarrow{b}^t=0$.

On Logan Hyslop left comment #7451 on Section 14.19 in Simplicial Methods

I believe there may be a rather minor typo in Lemma 14.19.2, specifically "Taking $n=k$" should say "Taking $n=k=k^\prime$."

On Christophe Marciot left comment #7450 on Section 53.23 in Algebraic Curves

In the intro at \emph{we conclude these points are nodes and smooth points on both} $C$ and $X'$, would it a bit better for comprehension to add a \emph{respectively} after the $X'$ ?

On Laurent Moret-Bailly left comment #7449 on Lemma 10.161.11 in Commutative Algebra

@ #7448: True, but if $R$ is a noetherian domain with perfect fraction field $K$ of char. $p>0$, then $R=K$. Otherwise, by Krull-Akizuki, there is a discrete valuation ring $V$ between $R$ and $K$, and a uniformizer of $V$ cannot be a $p$-th power in $K$. (There may be simpler arguments).

On Haohao Liu left comment #7448 on Lemma 10.161.11 in Commutative Algebra

From the proof we see that the condition "fraction field has characteristic zero " can be relaxed to "fraction field is perfect".

On Torsten Wedhorn left comment #7447 on Section 38.21 in More on Flatness

One could mention, that if $S'$ is a universal flattening, then $S' \to S$ is a surjective monomorphism.

On old friend left comment #7446 on Lemma 42.9.1 in Chow Homology and Chern Classes

Thanks David Holmes for confirming. I was confused by the definition given in e.g. Fulton or 3264. I think stacks insists on using O_X,ξ rather than O_Z,ξ because they later define cycle associated to a coherent sheaf F on X. There the length of F_ξ only makes sense over O_X,ξ but when F is O_Z, we could take length over O_Z,ξ.

On Jianing Li left comment #7445 on Lemma 10.161.8 in Commutative Algebra

@7439 Thanks for pointing out this. I now realize my mistake. The normality is necessary for the assertion that, if $x$ is integral over $R$, then $Trace_{L/K}(x)\in R$.

On left comment #7444 on Section 29.37 in Morphisms of Schemes

OK, thanks to both of you. This has been added here.

On David Holmes left comment #7443 on Lemma 42.9.1 in Chow Homology and Chern Classes

Dear old friend, What you propose seems to me to work, and 00IX seems simpler to apply here than 02M0.

On old friend left comment #7442 on Lemma 42.9.1 in Chow Homology and Chern Classes

"Hence, it also has finite length over OX,ξ, see Algebra, Lemma 10.52.12." In this case, the two lengths be would even be equal right? We have OX,ξ ---> OZ,ξ is surjective. If this is the case, citing tag 00IX would be helpful too.

On left comment #7441 on Lemma 10.134.16 in Commutative Algebra

Good catch! The fix is as follows. Let $\beta' : R[x_1, \ldots, x_n, x] \to S_g$ be the representation given in Lemma 10.134.12 using $\alpha$. Then we know that $NL(\alpha) \otimes_B B_g$ is homotopy equivalent to $NL(\beta')$. We know that $NL(\beta)$ and $NL(\beta)$ are homotopy equivalent by Lemma 10.134.2. Hence we see that $NL(\alpha) \otimes_B B_g$ and $NL(\beta)$ are homotopy equivalent. Finally, we apply Lemma 10.134.14.

On left comment #7440 on Lemma 31.22.12 in Divisors

OK, good catch! The problem only happens for the "quasi-regular" case. The argument in this case can be fixed as follows. Let $f : X \to S$ be the given morphism. We may replace $Y$, $X$, $S$ by compatible arbitrarily small affine neighbourhoods of $y$, $x$, $s$ throughout the proof. Thus we can use Lemma 29.36.20 to factor $f$ as $X \to \mathbf{A}^n_S \to S$ where the first morphism $\pi : X \to \mathbf{A}^n_S$ is \'etale. This reduces the problem to the case (a) $X = \mathbf{A}^n_S$ which is handled by the proof given and (b) the case where $f$ is \'etale.

Actually, the \'etale case is kind of fun! Namely, if $f$ is \'etale and we've shrunk so that $i$ and $j$ are closed immersions, then $f$ induces an isomorphism between the formal completion of $X$ along $i(Y)$ and the formal completion of $S$ along $j(Y)$ --- more precisely, for each $n$ the $n$th infinitesimal neighbourhood of $i(Y)$ in $X$ maps isomorphically to the $n$th infinitesimal neighbourhood of $j(Y)$ in $S$. This works as long as $f$ is formally \'etale. Hence we see that $i$ is quasi-regular if and only if $j$ is quasi-regular because this notion only depends on the system of infinitesimal neighbourhoods (the same is not true for the notions of Koszul regular and $H_1$-regular immersions, but the current proof already works for those).