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NOTE TO THEOREM 2.13.6.
IAN KIMING
In the proof of Theorem 2.13.6 (Hermite’s Discriminant Theorem) it is better to
define the set M slightly differently. Define M as the set of (λ1 , . . . , λn ) ∈ Rn such
that:

p
|λ1 |
< 2r1 −1 |DK | + 1





For r1 > 1 :
for i = 2, . . . , r1
|λi |
< 12 ,




 2
λj + λ2j+r2 < 1 ,
for j = r1 + 1, . . . , r1 + r2
For r1 = 0 :











|λ1 |
<
1
2
|λ1+r2 |
<
p
λ2j + λ2j+r2
|DK |
< 1,
for j = 2, . . . , r2 .
Then,
vol(M ) =
 r p
 2 1 |DK | + 1 · π r2
for r1 > 1
p
2 |DK | · π r2 −1
for r1 = 0 ,
p
whence vol(M ) > 2r1 +r2 |DK | in all cases. As
bounded,
centrally
pM is nownopen,
p
r1 +r2
−r2
symmetric, and convex with volume > 2
|DK | = 2 · 2
|DK | we deduce
from Minkowski’s lattice point theorem the existence of a lattice point α 6= 0 with
α ∈ M.

For the end of the argument where we show that α generates K, we may clearly
assume n > 1.
In both cases, i.e., r1 > 1 and r1 = 0, we have then |gj α| < 1 for j = 2, . . . , n, so
that we deduce |g1 α| > 1.
If r1 > 1 we then see that i = 1 is the only i with |gi α| > 1.
If r1 = 0 we have |gi α| > 1 precisely for i = 1 and for i = 1+r2 . Now α can not be
a real number since we would then obtain the contradiction 1 < |g1 α| = |Re(α)| < 12
(again because α ∈ M ). Hence Im(α) 6= 0, and so we see that there is exactly one
i such that |gi α| > 1 and Im(gi α) > 0 (remember that gi+r2 α is the complex
conjugate of gi α).
References
[1] H. Koch: ‘Number Theory. Algebraic Numbers and Functions’. Graduate Studies in Mathematics 24, AMS 2000.
1
2
IAN KIMING
Department of Mathematics, University of Copenhagen, Universitetsparken 5, DK2100 Copenhagen Ø, Denmark.
E-mail address: [email protected]
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