
Comments 1 to 20 out of 3520 in reverse chronological order.

On left comment #3646 on Lemma 10.122.2 in Commutative Algebra

Thanks for noticing this! It was a combination of Gerby and MathJax causing this. When Johan does the next update it should be fixed.

On Brian Conrad left comment #3645 on Lemma 15.12.6 in More on Algebra

I should have also mentioned that currently this Lemma is only mentioned in two proofs, neither of which is impacted by imposing the requirement $V(J)=V(IB)$ (though the second proof which claims to invoke this Lemma doesn't really clearly indicate where it is used -- that proof says this Lemma is used, but never explicitly invokes it with a cross-reference at any step).

On Brian Conrad left comment #3644 on Lemma 15.12.6 in More on Algebra

At the start of the proof, the wrong lemma is cited: should cite Lemma 09XK. In the second sentence, the invocation of the universal property doesn't make any sense unless $J$ lands in $I^h(A^h \otimes_A B) = I(A^h \otimes_A B)$, which would hold if $J=IB$. But there is the flexibility to replace $J$ with any bigger ideal of $B$ without affecting the henselian hypothesis on $(B, J)$, so you should really assume $V(J)=V(IB)$: this would ensure $(B,IB)$ is also henselian by Lemma 09XJ and would ensure that when $A$ is local with maximal ideal $I$ we can take $J$ to be the Jacobson radical of $B$ (rather than demanding $J=IB$, which would be an unpleasant requirement for such cases).

On Brian Conrad left comment #3643 on Lemma 15.12.2 in More on Algebra

In the second proof, which at the end invokes a lemma that has locality hypotheses, you need to first indicate why $A^h$ is local. One way based on general principles beyond the local setting is to recall that for any pair $(B,J)$ the ideal $J^h$ of $B^h$ is contained in the Jacobson radical (by the henselian property for the pair $(B^h, J^h)$) and by construction $J^h = JB^h$ yet always $B^h/JA^h = B/J$ (so when $B$ is local with maximal ideal $J$, the ideal $J^h$ is maximal and so is the unique maximal ideal of $J^h$, moreover with the same residue field as $B$).

Note that in this argument, we are using two facts currently recorded in Lemma 0AGU, which currently appears just after the present Lemma in section 0EM7 (no circularity, since Lemma 0AGU doesn't use the present lemma, so their order should be swapped).

Also, this shows that before swapping the order of the two lemmas, the "first proof" isn't entirely satisfactory since it isn't giving that $\mathfrak{m}^h$ is also the maximal ideal (which would however be "known" if Lemma 0AGU were put before the present lemma, as recommended above).

On Brian Conrad left comment #3642 on Lemma 17.17.2 in Sheaves of Modules

Near the start of the proof, replace "source is a coprod $\mathcal{F}_0$" with "source $\mathcal{F}_0$ is a coproduct", 2nd line after the displayed expression replace "coprod" with "coproduct", and on 3rd to last line replace $U_b$ with $V_b$.

On Brian Conrad left comment #3641 on Lemma 6.29.1 in Sheaves on Spaces

In the proof of (4), it should be mentioned that the hypothesis on $U$ forces it to be quasi-compact, so injectivity holds by (2).

On Brian Conrad left comment #3640 on Lemma 54.77.3 in Étale Cohomology

The reference to Lemma 0902 in the proof is a typo; you meant to refer to Lemma 0903 (and more specifically to part (1) of that Lemma).

On Brian Conrad left comment #3639 on Lemma 36.38.13 in More on Morphisms

In the parenthetical early in the 3rd paragraph of the proof, replace "quasi-compact etale" with "quasi-compact separated etale" (it is implicit in the Lemma invoked there, and also necessary as well), so earlier when $U$ is made affine it should have been noted (with cross-reference) that $U \to X$ is thereby separated. In the 2nd to last line of this same paragraph, don't call the neighborhood of $z$ in $Z$ by the name $V$, since the notation $V$ already has an entirely different meaning in the overall proof.

On Brian Conrad left comment #3638 on Lemma 54.77.2 in Étale Cohomology

In the final paragraph of the proof, replace "every closed subscheme $T \subset X$" with "every non-empty closed subscheme $T \subset X$".

On Brian Conrad left comment #3637 on Lemma 15.11.11 in More on Algebra

On Brian Conrad left comment #3636 on Lemma 15.11.9 in More on Algebra

The proof that (1) implies (2) seems to be incomplete as written since to do the second factorization lifting you'd need to know that the lifted monic factorization in $(A/I)[T]$ satisfies the "generate the unit ideal" condition, which is not addressed. But anyway it seems easier to just argue by the same method as in the proof of the converse: using notation as set up there, in the composition $B \to B/IB \to B/JB$ the first and second maps induce bijections on sets of idempotents bu the assumption (1), so the composite is also such a bijection and hence (2) holds (by one of the equivalent characterizations of "henselian pair").

On Brian Conrad left comment #3635 on Lemma 15.11.6 in More on Algebra

In the statement of (5), replace "radical" with "Jacobson radical" (a property used in the proof that (5) implies (2), as it must be, via the fact that $A\to A'_1$ is integral etale and an isomorphism modulo $I$ forcing its locally constant fiber degree to be constant due to $I$ being in the Jacobson radical).

You can replace $A/IA$ with $A/I$ throughout.

In the first paragraph of the proof, replace "not contained in $I$" with "which is congruent to 1 modulo $I$", and near the end of it replace "idempotent of $B$ lifts to an idempotent of $B/IB$" with "idempotent of $B/IB$ lifts to an idempotent of $B$".

In the proof of $(3) \Rightarrow (1)$, replace "$\overline{e}$ be the nontrivial idempotent" with "$\mathbf{e}$ be the idempotent" (maybe $g_0=1$ or $h_0=1$), and near the end of that paragraph the displayed expression for factoring $f$ makes no sense as written: you mean to speak in terms of characteristic polynomials of linear endomorphisms of a finite locally free $A$-module of constant rank, not determinants as written (which are elements of $A$!).

In the proof that $(1) \Rightarrow (5)$, you should mention that $I$ is contained in the Jacobson radical of $A$ by the definition of "henselian pair".

On Brian Conrad left comment #3634 on Lemma 15.10.4 in More on Algebra

On Brian Conrad left comment #3633 on Lemma 15.9.10 in More on Algebra

"disjoint union because $b_1 b_2 = 0$" should say "disjoint union because $(b_1,b_2)=B$, and covers ${\rm{Spec}}(B)$ because $b_1b_2=0$".

On Brian Conrad left comment #3632 on Lemma 10.122.12 in Commutative Algebra

The treatment of the "base" of the induction in this argument should be handled more clearly: in the case $b_m \in \mathfrak{q}$ necessarily $m>0$ because otherwise we have $b_0 = 0$ which lies in $\mathfrak{q} \cap S'$, so the case $m=0$ is vacuous and hence tautologically true. Of course, from the expository viewpoint that is bad writing and instead what is really being shown after noting that necessarily $m>0$ is that we can eventually put ourselves into the case $b_m \not\in \mathfrak{q}$ with a possibly smaller $m$. So the final case in the proof should be done first for clearer exposition.

On Brian Conrad left comment #3631 on Lemma 10.122.4 in Commutative Algebra

Why not localize at $a_k$ throughout from the start (harmless for the proof), to instantly reduce to the case $a_k=1$ handled by the Lemma invoked near the end of the argument and thereby avoid all of the futzing around here?

On Brian Conrad left comment #3630 on Lemma 10.122.2 in Commutative Algebra

The end of the proof is not processing on my screeen (maybe a MathJax bug?), though it looks fine in the .pdf output.

On Rene Schoof left comment #3629 on Lemma 10.31.4 in Commutative Algebra

$\ldots$ Then $xy=1-z$ for some $z\in I$. This means that $1\equiv z$ modulo $xR$. Since $z$ lies in the locally nilpotent ideal $I$, we have $z^N=0$ for some sufficiently large $N$. It follows that $1=1^N\equiv z^N=0$ modulo $xR$. In other words, $x$ divides $1$ and is hence a unit.

On Kestutis Cesnavicius left comment #3628 on Lemma 16.2.2 in Smoothing Ring Maps

In the statement of the lemma, one should mention that $\mathfrak{q}$ is a prime ideal of $A$.

On Dragon Bender left comment #3627 on Lemma 45.15.2 in Dualizing Complexes

Who is this "Janos Kollar".