Has anyone found an error in an early version of Neukirch? I remember a friend in graduate school throwing an early edition of Jurgen Neukirch's Algebraic Number Theory book against a wall (so hard that it split the binding) after he had worked for a number of days to reconcile something he realized was an error (or typo) in the book.
For the life of me, I cannot remember what this error was. 

Has anyone come across a particularly
  annoying error in an early edition of
  Neukirch? If so, what was it?


To give context to some of the answers below, I include the original statement of this question:

What are your "favorite" annoying errors and typos in otherwise excellent sources?

Please especially try to include subtle errors that can mislead, as these are very important.
Disclaimer: The author of this quesion thinks that working to resolve an error like this is a very worthwhile exercise, at least from the standpoint of having cool war stories that make a topic memorable. I'm not sure, therefore, whether I would want to post an answer to this question myself. I posted the question because I thought it would be fun.
 A: In my edition of Neukirch, Chapter I.9, Exercise 2:
If $L|K$ is a Galois extension of algebraic number fields, and $\mathfrak{P}$ a prime ideal which is unramified over $K$ (i.e. $\mathfrak{p} = \mathfrak{P} \cap K$ is unramified in $L$), then there is one and only one automorphism $\phi_{\mathfrak{P}} \in G(L|K)$ such that
$\phi_{\mathfrak{P}}a \simeq a^q \ mod \ \mathfrak{P}$ for all $a \in \mathcal{O}$,
where $q = [\kappa(\mathfrak{P}) : \kappa(\mathfrak{p})]$. It is called the Frobenius automorphism. The decomposition group $G_{\mathfrak{P}}$ is cyclic and $\phi_{\mathfrak{p}}$ is a generator of $G_{\mathfrak{P}}$.
Typo: That should be $q = |\kappa(\mathfrak{p})|.$
A: In the last edition of Neukirch, Chap. II, section 10:
For a field with a discrete valuation, let $v$ be the valuation, $\mathcal{O}$ the ring of integers, $m$ the maximal ideal, $U^{(0)}=\mathcal{O}^\times$ and $U^{(s)}=1+m^s$ (for $s\ge 1$) the groups of $s$-th units (appropriately subscripted).
Let $L/K$ be a finite Galois extension where $K$ has a discrete valuation that has a unique extension to a valuation of $L$. Let $G=\text{Gal}(L/K)$ be the Galois group and for $s\ge -1$ let
$G_s=\{\sigma\in G,v_L(\sigma a -a)\ge s+1 \mbox{ for all }
a\in\mathcal{O}_L\}$
be the higher ramification sugroups. Then Proposition (10.2) claims that for all $s\ge 0$, the mapping
$G_s/G_{s+1}\to U_L^{(s)}/U_L^{(s+1)} \quad,\quad \sigma\mapsto \sigma\pi_L/\pi_L$
is an injective group homomorphism that does not depend on the choice of a uniformizer $\pi_L$ for $\mathcal{O}_L$.
In fact, the injectivity statement holds for example if $\pi_L$ generates $\mathcal{O}_L$ as an $\mathcal{O}_K$-algebra, or if the extension of residue fields is separable, but it fails in general. Here is a counterexample.
Consider the field of Laurent series $k=\mathbb{F}_p((t))$ in the variable $t$. Let $C$ be a Cohen ring for $k$; this is a complete discrete valuation ring of characteristic $0$ with uniformizer $p$ and residue field $k$, and it is unique up to (nonunique) isomorphism with these properties. It may be described concretely as the set of formal series $\sum_{n\in\mathbb{Z}} a_nt^n$ whose coefficients are $p$-adic integers that converge to $0$ when $n\to-\infty$. Consider $\mathcal{O}_L=\mathcal{O}_K:=C[\zeta]$ where $\zeta$ is a primitive $p$-th root of unity; this is a complete dvr with uniformizer $\zeta-1$. Finally let $L=K$ be the fraction field of $\mathcal{O}_K$. The Frobenius morphism of $k$ lifts to an endomorphism of $C$ that takes $t$ to $t^p$ and acts as the identity on coefficients. This in turn extends to a morphism $K\to L$ that makes $L$ a Galois extension of $K$ with group $\mathbb{Z}/p\mathbb{Z}$, generated by $\sigma(t)=\zeta t$. One has $G_0=G$, $G_1=\{1\}$ and the map $G=G_0/G_{1}\to U_L^{(0)}/U_L^{(1)}$ is trivial.
A: Bosch, Algebra (one of the best new textbooks in German) used to have this slip in his Witt vectors chapter:

The lemma states a congruence modulo $p$, and the proof begins by WLOG assuming that $p$ is invertible in the ground ring.
It was fixed in the 7th edition in a way I don't really like (the absurd sentence has been replaced by "we can assume WLOG that $p$ is not a zero-divisor in $R$", which is correct but not quite obvious at the point).
A: Less on the absurd side, more on the subtle one: There is a basic fact in the theory of Clifford algebra most of whose proofs in literature are either ugly or long or incorrect. 
It is the fact that the graded algebra associated to the Clifford algebra of a vector space with a symmetric bilinear form is isomorphic to the exterior algebra of that vector space.
The ugly proofs are those which use orthogonal decomposition. These proofs usually require working over a field (sometimes even algebraically closed) of characteristic $0$, and the form must be symmetric. The theorem generalizes rather straightforwardly to commutative rings, and not necessarily symmetric bilinear forms.
The long proofs either use the diamond lemma from computer science (at least it is usually considered computer science, although it's actually a basic mathematical principle) or tons of computations.
The well-known Lawson-Michelson "Spin Geometry" gives an incorrect proof (proof of Proposition 1.2 in Chapter 1 §1). It states that "the $r$-homogeneous part of $\varphi$ is then of the form $\varphi_r = \sum a_i\otimes v_i\otimes v_i\otimes b_i$ (where $\deg a_{i}+\deg b_{i}=r-2$ for each $i$)". But why that? What if our representation of $\varphi$ in the form $\varphi = \sum a_i\otimes \left(v_i\otimes v_i+q\left(v_i\right)\right)\otimes b_i$ involves some $a_{i}$ and $b_{i}$ of extremely huge degree which cancel out in the sum?
I think this error is not limited to the Spin Geometry book, and to Clifford algebras. In algebra, we often prove that "a filtered algebra $A=\bigcup\limits_{n\in\mathbb N}A_n$ is generated by some elements $x_i$", but what we later want to use is that each element of $A_n$ is generated by those $x_i$ with $i\leq n$ and not by the higher ones.
A: Here are some errors in the statements of the exercises in Neukirch:
Section I.8 Exercise 2 reads "For every integral ideal $\mathfrak{A}$ of $\mathcal{O}$, there exists a $\theta \in \mathcal{O}$ such that the conductor $\mathfrak{F} = \{\alpha \in \mathcal{O} \mid \alpha\mathcal{O} \subseteq \mathcal{O}[\theta]\}$ is prime to $\mathfrak{A}$ and such that $L = K(\theta)$." However, this is incorrect. See Keitch Conrad's notes on Factorization after Dedekind, Example 7 (last line on first paragraph on page 6): http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dedekindf.pdf
Also, Section I.3 Exercise 5 says "The quotient ring $\mathcal{O}/\mathfrak{a}$ of a Dedekind domain by an ideal $\mathfrak{a}\neq 0$ is a principle ideal domain." This is false and the quotient ring $\mathbb{Z}/(2)$ provides a counterexample. Any such quotient ring is a principal ideal ring, but not necessarily a domain.
A: Hopefully this error isn't too obvious to be considered especially annoying, but I'll post it anyways. In my edition, exercise 2 in section $I.6$ states

Show that the quadratic fields with discriminant $5,8,11,-3,-4,-7,-8,-11$ have class number 1.

The issue here is that there is no field with discriminant 11 since $11\equiv 3 \mod 4$. (Recall Stickelberger's criterion says the discriminant can only be $0$ or $1$ mod 4.) I'm fairly certain he meant to write 13 instead of 11 here since 13 is the maximal quadratic discriminant for which one can easily use the Minkowski bound to conclude the class group is trivial.  
A: There is a mistake at the end of Chapter III, §2: There is a smooth proper curve over $\mathbf{Q}$ with good reduction everywhere, namely $\mathbf{P}^1_\mathbf{Z}/\mathrm{Spec}\,\mathbf{Z}$. There are no such curves with genus $> 0$.
A: Towards the end of ch. VII, §10, Neukirch proves that for Abelian extensions, every Artin $L$-series coincides with some corresponding Hecke $L$-series, thereby proving the Artin conjecture for all Abelian extensions. He then claims (p. 527):

This also settles the Artin conjecture for every solvable extension $L\mid K$.

This is manifestly incorrect, as the Artin conjecture for all soluble extensions is an open problem. For further details, see here.
