# Issue in the first French edition of Serre's local fields

I've been reading Serre's Corps Locaux, and I believe my copy is a first edition, as there's only one copyright date listed, 1968.

I believe I found an issue on page 57, which (if you're looking at English or a different edition) should correspond to the first page of the chapter titled *Discriminant and Different." I'm not sure if this is genuinely an error (which seems unlikely, since if it is, it's horribly grievous) or if I'm just misunderstanding the French.

Given a Dedekind domain A with fraction field K, and a finite-dimensional K-vector space V, we define a réseau (I'm not sure what to call it in English - réseau translates to network in other contexts, so I'm thinking lattice seems reasonable?) of V over A to be a finitely generated A-submodule of V which spans V. Since V is torsion-free over A, it follows immediately that any réseau is projective of rank equal to the dimension of V, and moreover free in the case that A is a PID.

Previously, Serre had defined the Euler characteristic of a finite-length module over a Dedekind domain, establishing an isomorphism between the finite-length Grothendieck group and the group of ideals. He first chooses to set $\chi(A/\mathfrak p)$ equal to the prime ideal $\mathfrak p$. Since every finite-length module has a finite filtration whose factors are of the form $A/\mathfrak p$, and since the prime ideals appearing don't depend on the filtration, we get a definition for $\chi$ in general, by multiplying together the various ideals $\mathfrak p$.

Now he remarks that if $X_2\subseteq X_1$ are both réseau, then $X_1/X_2$ has finite length, so we can get a sort of pairing $\chi(X_1,X_2)=\chi(X_1/X_2)$. He wants to generalize this to the arbitrary réseau $X_1$ and $X_2$, which he does as follows:

$$\chi(X_1,X_2)=\chi(X_1/X_3)\cdot\chi(X_2/X_3)^{-1},$$

for any réseau $X_3\subseteq X_1\cap X_2$.

My concern is this: unless I'm misunderstanding this, $X_1\cap X_2$ does not typically contain any réseau! Consider for instance $A=\mathbb Z$, and $V=K=\mathbb Q$. Set $X_1=\mathbb Z$ and $X_2=2\mathbb Z$. Then $X_1\cap X_2=0$. In fact, the case where we can choose such an $X_3$ appears to be quite exceptional!

I haven't been able to find any errata about this, and in either case I really doubt that Serre wrote something as stupid as what I just described, so I can only naturally assume I've totally misunderstood something!

• No, your example is not good, since $2\mathbb Z\subset\mathbb Z$, and the intersection is the smaller set, namely $2\mathbb Z$. Jun 27 '14 at 5:34

A "réseau" is a lattice. (My high-school French is a bit rusty, but the definitions agree, and my translation of Corps Locaux uses that term.) A lattice $X$ is a finitely-generated $A$-submodule with $KX = V$. Let $X_1, X_2$ be lattices in $V$, and put $M = X_1\cap X_2$. There exists for any $x\in V$ some $\alpha, \beta\in A$ with $\alpha x\in X_1$ and $\beta x\in X_2$, and thus $\alpha \beta x\in X_1\cap X_2$. Hence $KM = V$. If $M$ is not finitely-generated, we can still take a finitely-generated submodule that spans $V$ (over $K$), since $V$ is finite-dimensional.
• Also, a satisfyingly slick proof (in my opinion) that the intersection of full lattices is a full lattices occurred to me as soon as I realized that I was wrong: the intersection $M\cap N$ is isomorphic to $\frac{M\oplus N}{M+N}$. Since $M+N$ contains a full lattice, it is one as well, and so projective of rank $d$. Therefore, $M\cap N$ is projective of rank $2d-d=d$, and hence is a full lattice. Jun 27 '14 at 6:20
• (Of course, the quotient of two projective modules isn't obviously projective, but we happen to know that this particular quotient is f.g. over a Dedekind domain and lives in a torsion-free $A$-module.) Jun 27 '14 at 6:21