# 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.

• Could you begin by telling us what the typo in Neukirch's book was ?  – Chandan Singh Dalawat Dec 9 '10 at 13:45
• I feel like this thread might be too general to be helpful. We have had good threads on errors in Cassels and Frolich, and Griffiths and Harris. How would a searcher who wanted to find errata for the book they were studying find this thread? – David E Speyer Dec 9 '10 at 13:57
• In Silverman's nice book "Arithmetic of elliptic curves", he develops the algebraic geometry he needs just over perfect fields (since this is said to suffice for arithmetic applications, even though function fields of varieties of positive dimension over finite fields are never perfect). But for the proof of the translation-invariance of global 1-forms on an elliptic curve $E$ over a (perfect) field $k$, he applies the previously developed results on elliptic curves to the base change of $E$ over the imperfect field $k(E)$. After one knows schemes it is easy to fix, less so for a beginner. – BCnrd Dec 9 '10 at 14:13
• Continuing the theme of incorrect proofs (in otherwise good books) concerning invariant 1-forms on algebraic groups, in Shafarevich's 1st volume on algebraic geometry, he gives a purely algebraic proof that every cotangent vector at the identity on a smooth algebraic group $G$ over an algebraically closed field extends to a left-invariant global 1-form. In his proof, at the crucial step he rather explicitly uses the fact that the Zariski topology on $G \times G$ is the product topology...oops. – BCnrd Dec 9 '10 at 14:19
• There are a great many typos in the English translation of Serre's "Local fields", so if you're stuck on something that you think could be an error, it may help to compare with a French edition. Or just read it in French, of course.... – D. Savitt Dec 9 '10 at 14:51

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 $q = |\kappa(\mathfrak{p})|.$

• This answers the modified question. Thanks, Frank. – Jon Bannon Dec 14 '10 at 1:12

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).

• Your image seems to be gone now. – Daniel Litt Jan 18 '14 at 20:59
• Changed the imagehoster; thanks for noticing. – darij grinberg Jan 19 '14 at 0:27
• I think the question was about Neukirch's books. – Ben McKay Oct 17 '14 at 19:45

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.

• Good point. The corresponding statement in Corps locaux (ch. IV §2 prop. 7) is under the standing assumption of a separable residue extension. – Torsten Schoeneberg Jan 19 '14 at 22:28

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.

• I was under the impression that Clifford algebras take quadratic forms as input rather than bilinear forms. The difference is substantial when 2 is not invertible. – S. Carnahan Dec 9 '10 at 15:20
• You're right. I was misled by my own proof, which works in almost full generality (commutative ring, etc.) but only for quadratic forms induced by bilinear forms. – darij grinberg Dec 9 '10 at 17:46
• What is / where can one find the good proof? – user39190 Dec 29 '14 at 3:08
• See web.mit.edu/~darij/www/algebra/algebra.html#chevalley , but use my PDF only as a last resort if you cannot find the references (Bourbaki's Algèbre IX, §9, no. 2-3, or Ricardo Baeza's Quadratic Forms over Semilocal Rings, Lecture Notes in Mathematics 655, Springer 1978). – darij grinberg Dec 29 '14 at 13:25

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.

• In your last paragraph $\mathbf Z/(2)$ is a principal ideal domain, as are all fields. – KConrad Oct 1 '15 at 8:15

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.

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 $> 0$.