Goethe-Universität Frankfurt
Institut für Mathematik
Sommersemester 2017
19, April 2017
Non-archimedean geometry
Dr. Alejandro Soto
Rosemarie Martienssen
Exercise sheet 1 - Solutions
Exercise 1. Let K be a field with an absolute value | · |. Show that if 0 < t < 1
then | · |t is an absolute value. Furthermore, if | · | is non-archimedean, then | · |t is
an absolute value for every t > 0.
Solution. Clearly | · |t satisfies that |x|t = 0 if and only if x = 0 for any t and that
for every x, y ∈ K and any t we have |xy|t = |x|t |y|t because of the multiplicativity
of | · |. To show that | · |t is an absolute value, we just need to prove the triangle
inequality. As
|x + y|t ≤ (|x| + |y|)t
for x, y ∈ K, we obtain
|x + y|t ≤ (|x| + |y|)t ≤ |x|t + |y|t
so we just need to prove that (|x|+|y|)t ≤ |x|t +|y|t , for 0 < t < 1. We may assume
that x, y ∈ K × , and write the above statement as |y|t (|x|/|y| + 1)t ≤ |y|t (|x|t/|y|t + 1)
therefore it suffices to show that
(|x| + 1)t ≤ |x|t + 1,
∀x ∈ K × .
Define ϕ : R>0 → R by λ 7→ (λ + 1)t − (λt + 1). Clearly as λ + 1 > λ > 0 and
1 − t > 0, we have (λ + 1)1−t > λ1−t . This is equivalent to (λ + 1)t−1 < λt−1 which
implies that ϕ0 (λ) < 0 for every λ ∈ R>0 . This implies that ϕ is monotonically
decreasing, and as lim− ϕ(λ) = 0, then ϕ(λ) ≤ 0 for all λ > 0. This implies that
λ→0
(λ + 1)t ≤ λt + 1 and therefore that |x + y|t ≤ (|x| + |y|)t for every x, y ∈ K and
0 < t < 1.
If | · | is non-archimedean, it satisfies the strong triangle inequality, which yields
|x + y|t ≤ max{|x|, |y|}t = max{|x|t , |y|t }
for all t, so | · |t is also a non-archimedean absolute value.
Exercise 2. Let K be a field with an absolute value | · |. Show that it is nonarchimedean if and only if the set |Z| := {|n · 1| : n ∈ Z} ⊂ R is bounded.
Solution. If | · | is non-archimedean, then for every n ∈ Z we have |n| ≤ 1, which
shows that |Z| is bounded. Now assume there is a M > 0 such that |n| ≤ M , for
all n ∈ Z. Then for every x, y ∈ K
X n k n−k X n k n−k
n
x y ≤
|(x + y) | = |x| |y|
≤ M (n + 1)max{|x|, |y|}n ,
k
k
k
k
that is
|x + y| ≤ (M (n + 1))1/n max{|x|, |y|}.
Taking the limit n → ∞ gives us the result.
Exercise 3. Let | · | be a non-archimedean absolute value on a field K and let
v(·) := − log(| · |). Show that the valuation ring K ◦ is noetherian if and only if the
value group is isomorphic to Z. Show that if K is algebraically closed, the valuation
group is a divisible additive subgroup of R.
Solution. Suppose K ◦ is noetherian and let I ⊂ K ◦ be an ideal. It is finitely
generated, i.e. I = (x0 , x1 , . . . , xk ), with xi ∈ K ◦ . Suppose |x0 | ≥ |xi |, for every
i > 0. Then |xi /x0 | ≤ 1, i.e. xi /x0 ∈ K ◦ . This implies that xi = (xi /x0 )x0 .
Therefore, the ideal I = (x0 ) is principal. In particular, since K ◦◦ is an ideal in
K ◦ , we obtain that K ◦◦ = (π), with π ∈ K ◦ . Using Krulls intersection theorem for
noetherian rings we get
∞
\
(π n )K ◦ = {0}.
n=1
Now let x 6= 0 be an element of K ◦ . There exists a maximal n0 ∈ Z such that for
all n with n ≤ n0 we have x ∈ π n K ◦ and we can therefore write x as
x = π n0 u
for a unit u of the valuation ring K ◦ . (This description of x is even unique.) For
the valuation of x we get
v(x) = v(π n0 ) + v(u) = n0 v(π)
and therefore the value group is isomorphic to Z. (By multiplication with a scalar,
we can even ’normalise’ the valuation such that v(π) = 1 to make the value group
equal to Z.)
Conversely, suppose that the value group is isomorphic to Z and let I ⊂ K ◦
be an ideal. Let x0 ∈ I be such that v(x0 ) = inf{|x| : x ∈ I}. We will show, that
2
I = (x0 ). It is clear, that (x0 ) ⊂ I so we just need to show the other inclusion. For
y ∈ I, we have v(y) ∈ Z, and we can do division with remainder in Z :
v(y) = sv(x0 ) + r with s, r ∈ Z and 0 ≤ r < v(x).
Consequently, r = v(y) − sv(x0 ) = v(yxs0 ) and we can deduce that r ∈ v(I). Since
we chose the valuation of x0 to be minimal, it follows that r = 0 and y = xs0 , so
y ⊂ (x0 ).
Valuations (or absolute values) with value group isomorphic to Z are called
discrete valuations. The element π ∈ K ◦ ocurring in the first part of the proof is
called the ’uniformizer’ or the ’uniformizing element’.
Example: Consider the field Q of rational numbers, endowed with the p-adic absolute value. In this case, the uniformizer of the valuation is p.
For the last part of this exercise, we will show, that the value group Γ :=
v(K ∗ ) of an algebraically closed field K is a divisible additive subgroup of the
real numbers. Conversely, the group given by |K ∗ | is a divisible multiplicative
subgroup.
In order to show divisibility of Γ, we have to show that for all g ∈ Γ and n > 0
there exists h ∈ Γ such that g = n ∗ h. Let x be the element of K ∗ with v(x) = g.
Because K is algebraically closed, there is a solution y ∈ K fot y n = x. Since
v(y n ) = nv(y) = v(x) = g, we can set h = v(y).
Exercise 4. Let K be a field endowed with a non-archimedean absolute value | · |.
Show that K is a K ◦ -algebra of finite type.
Solution. Let u ∈ K such that |u| > 1, then K = K ◦ [u], and every element x ∈ K
can be written as x = y ∗ un for some y ∈ K ◦
Exercise 5. (Gauß norm) Let K be a field endowed with a non-archimedean
absolute value | · | and let K[T ] be the polynomial ring in one variable over K.
Show that the map
|| · || : K[T ] → R+ ,
P
given by || ak T k || := max{|ak |} defines a non-archimedean multiplicative norm
on K[T ].
Solution. Let f =
n
P
ak T k and g =
k=1
m
P
bk T k Clearly, ||f || = 0 if and only if f = 0
k=1
and
||f + g|| = max |ak + bk | ≤ max |ak |, |bk | = ||f || + ||g||.
k
k
3
What is left is to show multiplicativity. It is straightforward to show submultiplicativity:
X
||f g|| = max aj bk ≤ max{|ai |} max{|bi |} = ||f ||||g||.
i j+k=i
We can deduce, that the set OK[T ] := {f ∈ K[T ] : ||f || ≤ 1} is a subring of K[T ].
n
P
Let π : OK[T ] → K̃[T ] = K ◦ /K ◦◦ [T ] the map taking a polynomial f =
ak T k
to its reduction
n
P
k=1
k
ak T . By multiplication with scalars, we can assume, that
k=1
||f || = ||g|| = 1. Assuming ||f g|| < 1 we get that either π(f ) = 0 or π(g) = 0
because K̃[T ] is an integral domain. This is a contradiction to ||f || = ||g|| = 1
which proves the claim.
Exercise 6. Consider the following field extension Q(ξ)/Q, with ξ 5 = 1 and ξ 6= 1.
Determine the number of extensions of the 11-adic valuation of Q to Q(ξ).
Solution. We start by determining the minimal polynomial of ξ, which is the
irreducible monic polynomial f of minimal degree, such that f (ξ) = 0. The
polynomial T 5 − 1 is the first guess, but it is not irreducible since T 5 − 1 =
(T − 1)(T 4 + T 3 + T 2 + T + 1). Because of ξ 6= 1, we get f = T 4 + T 3 + T 2 + T + 1.
In order to find out about the number of extensions of | · |11 to the field Q(ξ), we
need to find out if f factors over the completion of Q with respect to | · |11 . This
completion is Q11 . In particular, f is an element of the polynomial ring over the
valuation ring Z11 , since all the coefficients are 1. Now we can use Hensels Lemma
to determine if f ∈ Z11 [T ] is irreducible by looking at the reduction in F11 . Here it
is easy to calculate, that the zeros of f are the elements 3, 4, 5, 9 ∈ F11 , so we have
f = (T − 3)(T − 4)(T − 5)(T − 9)
as an element of Q11 [T ]. Then, according to Proposition 1.5.3 there are 4 extensions
of the 11-adic valuation of Q to Q(ξ).
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