• Nie Znaleziono Wyników

1. Suppose that k0k = 0 and that for all a, b ∈ F we have

N/A
N/A
Protected

Academic year: 2021

Share "1. Suppose that k0k = 0 and that for all a, b ∈ F we have"

Copied!
1
0
0

Pełen tekst

(1)

Berkovich spaces, Problem List 1

Let F be a field and k · k : F → R

>0

be a function.

1. Suppose that k0k = 0 and that for all a, b ∈ F we have

• kabk = kakkbk,

• k1 + ak 6 1 + kak.

Show that k · k is a semi-norm.

2. Suppose that k · k is a semi-norm on F which is equivalent to the trivial norm. Show that k · k is the trivial norm.

3. Assume that k · k is a norm. Show that the following are equivalent.

(a) The norm k · k is Archimedean.

(b) For all x ∈ F

there is n ∈ N such that kn · xk > 1.

(c) There is n ∈ N such that kn · 1k > 1.

(d) The function

N 3 n 7→ kn · 1k ∈ R is unbounded.

4. Show that if k · k is a norm which is equivalent to an Archimedean norm, then k · k is Archimedean.

5. Suppose that k · k is an Archimedean norm on F . Show that char(F ) = 0.

6. Let p, q be different prime numbers. Show that | · |

p

and | · |

q

are non- equivalent non-Archimedean norms.

7. Suppose that k · k is a non-Archimedean non-trivial norm on Q. Show that (a) there is a prime number p such that

pZ = {a ∈ Z | kak < 1};

(b) the norm k · k is equivalent to the p-adic norm | · |

p

.

8. Suppose that k · k is an Archimedean norm on Q. Show that k · k is equivalent to the absolute value | · |

.

9. Suppose that k · k is a non-trivial norm on F (X) which is trivial on F . Show that k · k is equivalent to | · |

or there is an irreducible polynomial f ∈ F [T ] such that k · k is equivalent to | · |

f

.

10. Let k·k be a norm, (F, d

k·k

) be the corresponding metric space and ( b F , b d

k·k

) be its (metric) completion. Find a structure +, b b ·, d k · k of a normed field on F such that b

d b

k·k

= d

k·kc

.

Show that the natural map F → b F is a normed field homomorphism.

11. Show that F ((T )) is complete (with the T -adic norm).

1

Cytaty

Powiązane dokumenty

Let Z, N, Q be the sets of integers, positive integers and rational numbers respectively.. In this note we prove the following

LOWER BOUNDS FOR THE SOLUTIONS IN THE SECOND CASE OF FERMAT’S EQUATION WITH PRIME POWER

For general k the proof is very long and involved (it is worth mentioning that the proof offered by Marcinkiewicz and Zygmund has a lacuna filled by Fejzic and Weil [3]).. The

The proofs of these results depend on the method of Roth and Halber- stam on difference between consecutive ν-free integers, the results of Baker [1] on the approximations of

Consequently, the bounds for hyper-Kloosterman sums of prime power moduli proved by Dąbrowski and Fisher [1] (see (19) and (20) in Section 4) can be rewritten and improved for large

In this paper, we will study class numbers of certain real abelian fields by using the polynomial g(t).. Our work is based on the observation that g(t) comes from a

I would also like to thank Professor Warren Sinnott and Professor Karl Rubin for helpful

Indeed, Proposition 1 gives a lower bound for the number of reduced ideals in an order, and Lemma 1 an upper bound for the number of reduced ideal in each class of that order.. Then