Birational invariants, purity and the Gersten conjecture Lectures at ...

6
c) In the rational function field C(t) in one variable over C, any non-zero element,
all valuations of which are even, is a square. Indeed, any z ∈ C(t) ∗ after suitable
multiplication by a square in C(t) may be rewritten as a polynomial in C[t] with simple
roots z = ∏ i∈I (t − e i). If I is not empty, i.e. if z ≠ 1, then for any i ∈ I, we may consider
the valuation v i associated to the prime ideal (t − e i ), and v i (z) = 1 is not even.
d) Putting a), b), c) together, we find that the field L is not C-isomorphic to the field
C(t).
Exercise 1.1.1 : Let p and q be two coprime integers, and let e i , i = 1, . . . , q be distinct
elements of C. Let L be the function field L = C(x)(y), with y p = ∏ i=i=1,...,q (x − e i).
Arguing with elements all valuations of which are divisible by p, show that L is not purely
transcendental over C.
Let us slightly formalize the proof of Proposition 1.1. Given a field k, char(k) ≠ 2, and
a function field F/k (by function field F/k we mean a field F finitely generated over the
ground field k) we consider the group
Q(F/k) = {α ∈ F ∗ , ∀v trivial on k, v(α) ∈ 2Z}/F ∗2 .
Let Ω be the set of discrete valuation rings A (of rank one) with k ⊂ A and with
fraction field K. The group Q(F/k) may also be defined as :
Q(F/k) = {α ∈ F ∗ /F ∗2 , ∀A ∈ Ω, α ∈ Im(A ∗ /A ∗2 ).}
Our proof of the non-rationality of L = C(x)(y) fell into the following parts :
a) We showed that if F = C(t), then Q(F ) = 0. As the reader will easily check,
this result may be extended. Namely, if t 1 , . . . , t n are independent variables, the natural
map k ∗ /k ∗2 → Q(k(t 1 , . . . , t n )/k) is an isomorphism. Even more generally, if K/k is
a function field, the inclusion K ⊂ K(t 1 , . . . , t n ) induces an isomorphism Q(K/k) ≃
Q(K(t 1 , . . . , t n )/k).
b) We produce a non-trivial element in Q(L).
In the next section, we shall generalize the formalism above to functors F other than
F (A) = A ∗ /A ∗2 . In most cases, the analogue of a) will be easy to check. However the
computation of the analogue of Q(L) will often turn out to be tricky. Before getting to
this, we might comfort ourselves with the remark that in the case at hand, a few more
elementary arguments enable one to compute the exact value of Q(L).
If k is algebraically closed, the group k ∗ is divisible, hence any discrete valuation on
F is trivial on k. In that case, we simply write Q(F ) = Q(F/k). In the above proof, we
showed Q(C(t)) = 0 and Q(L) ≠ 0 for L = C(x, √ x(x − 1)(x + 1)). We shall go further
and actually compute Q(L).
First note that given any extension L/K of function fields over a field k, there is an
induced map Q(K/k) → Q(L/k) (for any discrete valuation v on L, either v is trivial
on K, or it induces a discrete valuation on K). When the extension L/K is finite and
separable, there is a map the other way round :