D-MATH
Prof. Gisbert Wüstholz
Diophantine Geometry
FS 2013
Exercise Sheet 1
Z
√
]. What is its residue
1. a) Show that (2)
is a maximal ideal in the ring [ 5+1
2
√
5+1
field, i.e. [ 2 ]/(2) = ?
√
√
0
b) √
Let O = [ −5], and
√ let us take the ideals p = (2, 1 + −5), p = (2, 1 −
−5), q = (3, 1 + −5). Show that these are maximal ideals of O, and
calculate the residue fields O/p, O/p0 , O/q.
√
c) Show that 2 and 1 + −5 are
√ irreducible elements of O. Furthermore, show
0
that (2) = pp and (1 + −5) = pq. This demonstrates that the ideal
generated by an irreducible element is not necessarily a prime ideal.
Z
Z
d) Show that O is not a unique factorization domain. (A unique factorization
domain is an entire ring in which every element a can be written as a product
a = εp1 . . . pn , where ε is a unit and the pi are irreducible elements, and this
factorization is unique up to reordering and multiplying by units.)
e) Show that a unique factorization domain is integrally closed in its fraction
field.
Q
√
√ √ 2
2. Show that B = (1, 3 2, 3 2 ) is an integral basis in the number field K = ( 3 2).
Calculate the discriminant of K using the definition from the lecture. Compare
the result with the discriminant of the polynomial X 3 − 2.
3. Let A be an integrally closed entire ring, and K = Frac(A) the fraction field of
A. Let L/K be a finite field extension, and let α ∈ L. The minimal polynomial
of α over K is P (X) = X n + a1 X n−1 + . . . + an ∈ K[X].
a) Show that α is integral over A if and only if P ∈ A[X].
b) If α is integral over A, then NmL/K (α), TrL/K (α) ∈ A.
c) Let B be the integral closure of A in L. Show that α is a unit in B if and
only if NmL/K (α) is a unit in A.
Bitte wenden!
4. Let K be a number field with integral basis B = {ω1 , . . . , ωn }.
a) Show that
Q, (ω, ω ) 7→ Tr Q(ωω )
is a symmetric, non-degenerate Q-bilinear map. (Non-degenerateness means
β :K ×K →
0
0
K/
that if ω ∈ K is such that β(ω, ω 0 ) = 0 for every ω 0 ∈ K, then ω = 0.)
b) Show that the discriminant of K is ∆K = det(TrK/Q (ωi ωj ))1≤i,j≤n . Using
this definition for the discriminant, deduce that ∆K ∈ , and also show that
∆K 6= 0.
Z