Errata
This file contains miscellaneous errata and additional remarks for my course
notes that I haven't yet incorporated into the versions on the web.
Most are taken from e-mail messages -- I thank everyone who has contributed!
Fields and Galois Theory
Group Theory
Algebraic Geometry
Algebraic Number Theory
Modular Functions and Modular Forms
Elliptic Curves
Abelian Varieties
Lectures on Etale Cohomology
Class Field Theory
Algebraic groups and arithmetic groups
Fields and Galois Theory v4.20 (FT)
p35, 3.2. As Maren Baumann pointed out, there is no need to assume that f is monic (here, and probably in other places).
p41, Example 4.7: should read "must have exactly two nonreal roots" not "to nonreal roots".
Group Theory v3.01 (GT)
No known errors.
Algebraic Geometry v5.10 (AG)
p154, 9.20 Lars Kindler points out that, in the proof
of 9.20, it is not obvious that the map \alpha\circ\pi is
given globally by a system of polynomials (rather
than just locally). It is in fact given globally, and this
is not too difficult to prove: a regular map from a variety V
to P^n corresponds to a line bundle on V and a set of global
sections, and all line bundles on A^n are trivial (see,
for example, Hartshorne II 7.1 and II 6.2). [I should include
all this in the next version.]
Algebraic Number Theory v3.00 (ANT)
Francesc Castellà points out that the proof of the uniqueness of the
factorization of ideals on p43 is not completely convincing. It would be better
to consider two factorizations of an ideal a. After adding some
factors with exponent zero (if necessary), we may suppose that
the same prime ideals occur in both factorizations. Now the argument
shows that the exponents in the two factorizations are the same.
From Roger Lipsett:
Page 61 (Norms of ideals): in the displayed equation
following "We take this as our definition:" you say Nm(P) =
P^...; ... you mean Nm(P)=p^... (small p, not large).
Page 80, last sentence of proof of 5.8. I don't think
this is correct. In fact, all elements of the kernel map to
zero in C, so one can't conclude much. I think you meant an
argument such as: an element $\alpha$ in the kernel is such
that $|\sigma_i\alpha|=1$ for all $i$, [and so the kernel is
finite by Proposition 5.5.]
From Francesc Castellà
I think that on your ANT notes p95 after the formula
defining the Bernoulli numbers should read "p is *not*
regular [i.e. irregular] iff p divides the numerator of some
$B_k$ with k=2,4,...,p-3". [Yes, it is irregular if it
divides...]
From Kwangho CHOIY
p32, 3 lines below from PROOF(a) of Proposition2.40; RHS
"(alpha)_(r+i-s) - (conjugate of(alpha))_(r+i-s)" in the equation of
'sign', "-s" can be removed.
p51, line 9; free R-modules --> free A-modules
p57, 3 lines from the bottom in REMARK 3.50; I don't think that "this
in the set of p for which (m/p)=1" is true. Because, for m=5 and p=2,
(5/2)=1, but 2 does not split in Q[(square root of 5)](just unramified)..
In order to make the statement correct, I think that we need "the set of
"odd" primes in Spl(K) of prime numbers that split in K".(see EXAMPLE
3.44(iii), p56, ANT)
p69, THEOREM4.19, 2 lines from the bottom; /If none of the points of
... is in T/ --> /If none of the points of ... is in T "other than the
origin"/.
p69, THEOREM4.19;(i)Is the statement necessary for the proof;"It will
contain only finitely many such points because (lamda) is discrete and
(1+epsilon)T is compact", where it is located at 5 lines above from
bottom? (ii) Does the theorem still hold if we replace "T is compact" with
"T is closed"?
p73, in the middle; we need "+-" in this equation,
det(A)=(-2i)^(-s)det(B)="+-"(-2i)^(-s)disc(...)^(1/2).
p73, bottom line; in the equation, you wanted (delta)_K instead of
(delta).
p78, 6 lines below from the section "Statement of the theorem";
K(tensor)C --> K(tensor)R.
p84, line 3; In particular, for any a inside "K" --> "U_K".
p112, 6 lines above from the bottom; in the equation [n/p^2], a_2*p
--> a_2.
p124, line 3 in the proof of Corollary7.59; "|sigma(Phi) - Phi| =<
|Phi|^(i+1) can be replaced by "|sigma(Phi) - Phi| < |Phi|^(i)" And
"sigma(Phi) = Phi + a*(Phi)^(i+1)" was correct. - this is what I mistook
in the last time.-
this is another proof for proposition 3.18, p45; "...(this part is
same with the notes). Because p is prime, it will contain one of the
irreducible factors, 'phi'. So we have the following chain of ideals; (0)
< (phi) < p < A.(where neither 1st inclusion nor the last one can be
equal). Since in an unique factorization domain every irreducible element
is a prime element(the converse is always true in a domain), (phi) is a
prime ideal. It follows from 3.3(c) that (phi) and p are same. Thus, p is
principal."
Moreover, let me prove that in a unique factorization domain A every
irreducible element is a prime element(the converse is always true in a
domain); "let phi be an irreducible element. suppose that phi divides ab
for some elements a, b in A. Then ab = (phi)c for some element c in A.
Since phi is an irreducible element in A, it follows from the uniqueness
of factorization that a or b has a factor 'phi'. Hence, phi is a prime
element in A."
p79, line 2&3; there are 2 parts in which you wanted "( )"
p131, in the equations in the proof of THEOREM 8.8; we need "("
Modular Forms and Modular Functions v1.10 (MF)
From Ulrich Goertz:
pdf (2 pages)
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).]
Abelian Varieties v2.00 (AV)
No known errors (but this draft is still very rough).
Lectures on Etale Cohomology v2.10 (LEC)
No known errors.
Class Field Theory v4.00 (CFT)
p2, line 4. I should have said that alpha is an algebraic integer (so f
is a monic polynomial with coefficients in O_K.
Then
disc(f)=disc(O_K[alpha]/O_K)=(O_L:O_K[alpha])^2 x disc(O_L/O_K)
(see 2.25).
Therefore, if a prime is ramified in L, then it divides disc(f), but there may be
primes dividing disc(f) that are not ramified in L. For example, let L be the
the quadratic field generated by a square root of m. If m is congruent
to 1 mod 4, then 2 divides the discriminant of f but is not ramified in L.
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.
For p*=p=5 and q=2, 5 is a square modulo 2, however 2 does not split in
Q[square root of 5]. In other words, the ideal (2) is ramified in
Z[square root of 5] as (2, 1+square root of 5)^2, whereas (2) = (2) in
Z[(1+square root of 5)/2], i.e. it is unramified(not split) in
Z[(1+square root of 5)/2]. Because the minimal polynomial of (1+square
root of 5)/2, X^2 -X^-1 is irreducible modulo 2, but but that of (square
root of 5), X^2 -5 factors into (X^2 +1)^2 modulo 2.
Nevertheless, I think the statement, "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",
holds if q is an "odd" prime.(We already assume that p is an odd prime as
well as p is distinct from q). Because, in this case, "the minimal
polynomial of (1+square root of p*)/2, X^2 -X^+(1-p*)/4 factors into 2
distinct linear factors modulo q" if and only if "the minimal polynomial
of (square root of p*), X^2 -p* does", and of course, if and only if "p*
is a square modulo q". This equivalence is based on the fact "p and q are
distinct 'odd' primes".
p4, bottom line; totally real -> totally "positive" real
p5, line12; Theorem 0.2 -> theorem 0.4
p5, line15; m|m' -> m'|m
p7, 2 lines above from the section "L-SERIES"; /The Frobenius density
theorem/ can be replaced by /The "Chebotarev" density theorem/
[True, but the Frobenius theorem is older and much more elementary than
the Chebotarev theorem.]
p7, 4 lines above from the bottom; in the equation of L(s,kai),
/kai(Na)^(-s)/ can be replaced by /kai"(a)"*N(a)^(-s)/
p7, 2 lines above from the bottom; /an ideal class group/ can be
replaced by /the "ray" class group/.
p9, 4 lines above from the bottom; /in which every prime of K/ can be
replaced by /in which every prime of K "outside S"/
p16, Theorem0.9; /L=K[alpha]/ can be replaced by /"B=A[alpha]", and we
need to rearrange sentences in this Theorem(see ANTv.3.00 theorem3.41).
p19, 2 lines above from the bottom; /sigma(alpah)=(alpha)^q for
all.../can be replaced by /sigma(alpah)=(alpha)^q "mod m_L" for all.../
p43, 4 lines below from a section "THE RAMIFICATION GROUPS..."; /i>=0/
can be replaced by /i>=1/.
p97, 6 lines below from the section "THE INVARIANT MAP"; "H^2(G,L*) ->
H^2(G,Z)" instead of 1st cohomolgies, i.e. both H^1's must be H^2's.
p105, Theorem3.1; in the front H^r(Gal(L/K),L*), L*-->Z.
p108, proposition4.1; I think that we need to give a degree of the
cyclic extension of K. So, I think that we can add "of degree dividing n"
after "every cyclic extension of K" in the sentence.(I referred to Serre's
Local Fields)
p109, 8 lines below from the section "THE GENERAL CASE"; you wanted to
have the sequence 0 -> Z -> Z -> Z/nZ -> 0 where the map Z -> Z is 'n',
instead of 0 -> Z/nZ -> Z -> Z -> 0.
p171, line2; "phi(id(a))" -> "psi(id(a))"
p173, bottom line; (idele) I_L -> (idele) I_K
p5, line 15; I^S(m) imbedding to I^S(m') can be replaced by I^S(m')
imbedding to I^S(m) m'|m, but according
to the other statements below this is correct, instead of m|m' -> m'|m.
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)*
The followings are things that you wanted to add "finite", I think;
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.
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}
From Victor Petrov:
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.