From Kwangho CHOIY
p58, 5 lines below from Kummer theory; I think that, in the definition
of 'exponent n', "the smallest positive integer for which this is true"
can be removed.
Efthymios Sofos points out that in (5.19) on groups of order 60, s_2(A_5)=5 and s_2(A_5)=15 are contradictory. In fact, the case s_2=15 doesn't occur.
From Soli Vishkautsan
Page 4 line -1, should be the intersection of all the ideals containing S (not A).
Page 8, In proof of Thm 1.8. there seems to be some confusion, The
S_m are subsets of A[X_1,...,X_r] and not A, therefore the ideal a is
of A[X_1,...,X_r] which you didn't prove is noetherian (yet), so you
can't use lemma 1.7. Also you say f = g_1*f_1 + .... + g_s*f_s where
g_i are in A, but they should be in A[X_1,...,X_r]. Hope I am not
missing something. [Yes, the g_i should be in the polynomial ring.]
From Nousin Sabet:
I would like to make a comment
about the Exercise 2.24, page 33, which is not correct.
For example, if instead of $\Gamma(N)$, one takes $H=\Gamma_0(p)$
and $x_0=\infty$, then $[G:H]=p+1$ and $[G_\infty:H_\infty]=1$,
and the formula of that exercise gives us $p+1$ for the number
of inequivalent cusps for $\Gamma_0(p)$. However, we know
that $\Gamma_0(p)$ has only 2 cusps.
[The exercise is correct if one requires H to be normal: Let Y be the set of orbits of H in X. Then G acts transitively on Y and the stabilizer of Hx_0 is HG_0.
Hence the number of orbits is
(G:HG_0)=(G:H)/(HG_0:H)=(G:H)/(G_0:H_0).]
Tim Dokchitser points out that I prove Zarhin's trick (13.12) only
over an algebraically closed field , and then immediately apply it in (13.13) over a finite field.
This is doubly confusing because (13.10) is certainly false over nonalgebraically closed field (over such a field
an abelian variety need not be isogenous to a principally polarized abelian variety).
However, I believe everything is O.K. Specifically, the proof of Zarhin's trick requires only (13.8), and,
because this holds over an algebraically closed field, it holds over every perfect field (see my 1986
Storr's article Abelian Varieties 16.11 and 16.14).
From Sunil Chetty. Near the start of I 14 (Rosati involution): in (\alpha\beta)^\dagger = \beta\alpha there should be a dagger on each of \beta and \alpha.
From Roy Smith (on proofs of Torelli's theorem III 13)
You ask on your website for advice on conceptual proofs of Torelli. ... here goes.
There are many, and the one you give there is the least conceptual one, due I believe to Martens.
Of course you also wanted short, ....well maybe these are not all so short.
The one due to Weil is based on the fact that certain self
intersections of a jacobian theta divisor are reducible, and is
sketched in mumford's lectures on curves given at michigan. Indeed
about 4 proofs are sketched there.
The most geometric one, due to Andreotti - Mayer and Green is to
intersect at the origin of the jacobian, those quadric hypersurfaces
occurring as tangent cones to the theta divisor at double points, thus
recovering the canonical model of the curve as their base locus, with
some few exceptions.
To show this works, one can appeal to the deformation theoretic
results of Kempf. i.e. since the italians proved that a canonical
curve is cut out by quadrics most of the time, one needs to know that
the ideal of all quadrics containing the canonical curve is generated
by the ones coming as tangent cones to theta. the ones which do arise
that way cut out the directions in moduli of abelian varieties where
theta remains singular in codimension three.
But these equisingular deformations of theta embed into the
deformations of the resolution of theta by the symmetric product of
the curve, which kempf showed are equal to the deformations of the
curve itself. hence every equisingular deformation of theta(C) comes
from a deformation of C, and these are cut out by the equations in
moduli of abelian varieties defined by quadratic hypersurfaces
containing C. hence the tangent cones to theta determine C.
This version of Green's result is in a paper of smith and varley,
in compositio 1990.
Perhaps the shortest geometric proof is due to andreotti, who
computed the branch locus of the canonical map on the theta divisor,
and showed quite directly it equals the dual variety of the canonical
curve. this is explained in andreotti's paper from about 1958, and
quite nicely too, with some small errata, in arbarello, cornalba,
griffiths, and harris' book on geometry of curves.
There are other short proofs that torelli holds for general
curves, simply from the fact that the quadrics containing the
canonical curve occur as the kernel of the dual of the derivative of
the torelli map from moduli of curves to moduli of abelian varieties.
this is described in the article on prym torelli by smith and varley
in contemporary mat. vol. 312, in honor of c.h. clemens, 2002, AMS.
there is also a special argument there for genus 4, essentially using
zariski's main theorem on the map from moduli of curves to moduli of
jacobians.
There are also inductive arguments, based on the fact that the
boundary of moduli of curves of genus g contains singular curves of
genus g-1, and allowing one to use lower genus torelli results to
deduce degree torelli for later genera.
Then of course there is matsusaka's proof, derived from torelli's
original proof that given an isomorphism of polarized jacobians, the
theta divisor defines the graph of an isomorphism between their
curves.
For shortest most conceptual, I recommend the proof in Arbarello,
Cornalba, Griffiths Harris, i.e. Andreotti's, for conceptualness and
completeness in a reasonably short argument..
From Eric Moorhouse
p156 The expression appearing in the middle of the page (bounded by two powers of $q$) should be $|\alpha|$.
Herve Jacquet has pointed out that Statement 3.3 page 106 does not (appear to) follow directly from the definition of the local Artin map and formula (30), as I claim, but that it can be proved using (3.6). I don't know right now whether it can be proved without using (3.6).
From Corinne Sheridan:
there is a mistake on page 154, in the paragraph before Theorem 3.5 (Reciprocity Law).
On both lines 2 and 3, it says that a homomorphism admits a modulus if there
exists a modulus m such that S(m) is contained in S .... It
should say S(m) contains S. [Yes, if the map admits a modulus m
and m|n, then it also admits the modulus n.]
From Jonah Sinick
p223 (first page of Chapter VIII): "On may hope..." should be "One may hope..."
From Yu Zhao
Page 97, 3rd line, $v_i \in U_K^{(i)}$ should be $v_i \in U_L^{(i)}$
because one wants an element in $U_L$ to get the surjectivity.
From Kwangho CHOIY
p3, 3 lines below from section "QUADRATIC EXTENSION OF Q"; I don't think "A prime number q(not equal to p) splits in Q[square root of p*] if and only if p* is a square module q" is true.p5, line 15; I^S(m) imbedding to I^S(m') can be replaced by I^S(m')
imbedding to I^S(m) The followings are things that you wanted to add "finite", I think;
p5, Example0.6; we should assume all (p_i)* is positive and also we
should let -d=(p_1)*...(p_t)* instead of d=(p_1)*...(p_t)*, in order to
say both that "K is ramified exactly at the prime ideals (p_1),
(p_2)....,(p_t)" and that "K[sqrt (p_i)*] is unramified over K". But I
don't think that this is a matter with the result about the narrow class
group in the example.
From Victor Petrov:
From Xiandong Wang:
p100, 3 lines from the bottom; L* --> Nm(L*)
p186, the bottom line; (zeta)_L,S(s) = (zeta)_K,T(s)^[L:K] -->
(zeta)_L,T(s) = (zeta)_K,S(s)^[L:K]
p213, the first line below section "7 Application to the Brauer Group";
K^(a...) --> K^(al)*
p4, line1; An unramified abelian - > An "finite" unramified(abelian),
p4, theorem0.3; every unramified abelian - > every "finite" unramified
(abelian),
p4, just below theorem0.3; an unramified abelian - > an "finite"
unramified(abelian),
p5, theorem0.4; every abelian -> every "finite" abelian,
p153, Corollay3.4; For any abelian extension -> For any "finite"
abelian extension,
p157, line2; the abelian extensions -> the "finite" abelian extensions
p157, Theorem3.16; a Galois extension -> a "finite" Galois extension.
p148, Example1.8(c); I think that in the first short exact sequence,
the kernel {-1, 1} depends on the modulus (m). For example, if m=2, U_m,1
is equal to {-1, 1}, thus the kernel is {1}. Also, in the trivial case m=1
the kernel is equal to {1}. The case m=1 also belongs to the second exact
sequence.
p185, Example2.13(b); in the formula, you wanted (zeta)_K instead of
just (zeta).
p192, line 1 in the proof of Theorem4.9; you wanted i(K_m,1) instead of
K*.
p196, (c); you wanted u_E/L instead of u_L/K.
p206, in the proof of Corollary4.8; you wanted P|p not in S.
p228,Remark3.2; you wanted "5^2, 7^2, 10^2, ....." instead of "2".(See
Exercise5.3 pp360, Cassels-Frohlich)
p238, in the middle of the section THE POWER RESIDUE SYMBOL;
you wanted (zeta)^((q-1)/n) instead of u^((q-1)/n).
p238, 2 lines above (5.1), (5.1) itself, (5.2); you wanted p not in S(a)
Algebraic groups and arithmetic groups v1.01 (AAG)
11.34. The last sentence should read: In nonzero characteristic, the only connected algebraic groups whose
representations are all simple are the tori.
Section 23, p167. In the definition of Phi^+ and Phi^-, need to take the intersection with Phi.
In the definition of P^+, need to replace Phi with Phi^{\vee}
p110...the point-wise definition of normalizers and centralizers of algebraic
subgroups as in Proposition 13.18 (p. 110) seems to be not quite correct. At
least both Demazure, Gabriel in "Groupes algebriques" (Proposition II.1.3.5) and
Demazure, Grothendieck in SGA 3 (Expose I, Definition 2.3.3) define the value of
the normalizer of a subgroup $H$ in a group $G$ at a ring $R$ to be the set of
all elements $g\in G(R)$ such that $gH(S)g^{-1}$ is contained in $H(S)$ for any
ring extension $S$ over $R$ (and not just for $R$ itself as written in your
notes).
The criticism is correct: the definition I give in 13.18 is only valid when
H(k) is Zariski dense in H, which is true if H is connected, k is infinite, and
either H is reductive or k is perfect (Borel 1991, 18.3). In general, H(k) might
be too small, and so its centralizer and normalizer in G(k) might be too big.
p131. The claim that the map GL_2(R)->PGL_2(R) is surjective for all rings R
is incorrect. The map is surjective if all f.g. projective R-modules of rank 1 are
free, e.g., if R is local. However, let R be a Dedekind domain whose class
number is divisible by 2. Then there is a f.g. projective R-module L of rank 1 such
that L+L is free (cf. ANT 3.33), and the etale (or Amitsur) cohomology sequence
of 1->Gm->GL_2->PGL_2->1 shows that GL_2(R)->PGL_2(R) is not surjective.
The proof of LEMMA 18.3(b) (formula (68)) has some inaccuracy. This can be checked by using (65) and the definition directly.
Algebraic groups, Lie groups, and their arithmetic groups v1.01 (BAG)
From Darij Grinberg
Example 2.14. This needs to be fixed. For instance, over Z/4Z, the line (1,2,-2) has 2nd and 3rd elementary
symmetric polynomials equal to 0, but is not monomial.
In Exercise 3-4, I would write (Z/pZ)_k instead of Z/pZ in
accordance to your notations.
Example 4.1: "Let $\mathfrak{a}$ be an ideal $k[X_1,...,X_n]$" -
there is an "in" missing here.
In Example 4.3, you are speaking of a regular map V -> A^1. You are
probably abbreviating (V, O(V), alpha) by V here, but you only
introduce this abbreviation further below.