Tentamenskrivning i Algebraisk talteori 01-04-07

Lösningar

F.1 Firstly, if n is a prime, then k_{n} = n − 1, that is

(p − 1)! ≡ −1 (mod p).

Proof : The numbers {1, ..., p−1}, with the exception of p−1, are grouped in pairs x, x^{−1} (mod p).

Next, if n = 4 then k_{4} = 2, that is 3! ≡ 2 (mod 4).

Finally I claim that, if n is composite and > 4, then k_{n} = 0, i.e.: that for n > 4 and composite we have

(n − 1)! ≡ 0 (mod n).

Proof : Let p be a prime divisor of n and suppose p^{l} ∥ n. We must show that p^{l} divides (n − 1)!. If l = 1 then, since n is not prime, we have p < n and indeed p divides (n − 1)!. So suppose that l > 1. It clearly suffices to have lp < n, and hence suffices to have

lp < p^{l}.  (1)

It’s easy to prove (by induction, for example), that (1) holds unless p = l = 2, which corresponds to the exceptional case n = 4. q.e.d.

F.2 (i) Dirichlet’s approximation theorem, p.43 in Baker or p.128 in my notes.

F.3 (i) Sats 16, s.21 in my notes.

(ii) Suppose otherwise. We have

\frac{1}{ζ(s)} = ∑ \frac{μ(n)}{n^{s}} = ∑ \frac{μ(n)/ ^{4} \sqrt{n}}{n^{s−1/4}}.

If the partial sums of the numerator are bounded, then Dirichlet’s criterion implies that the series converges uniformly in any half-plane Re(s) > 1/4+δ, and hence, by Weierstraβ’ theorem, defines an analytic function in Re(s) > 1/4. That is, 1/ζ(s) is analytic in Re(s) > 1/4, hence has no poles there, which means that ζ(s) has no zeroes in the region. But this contradicts our knowledge that ζ has zeroes along the line Re(s) = 1/2.

F.4 p.28-9 in Baker.

F.5 (i) We first of all seek a solution to the congruence

h^{2} ≡ 185 (mod 4 · 17)

and find that

7^{2} = 185 − 2 · (4 · 17) .

This implies (see Sats 47, p.63 and its’ proof) that the form 17x^{2} +7xy −2y^{2} has discriminant 185 and represents 17 in (x y) = (1 0). It remains to reduce the form. Its’ matrix is

A = ( \begin{matrix}17 & 7/2 \\ 7/2 & −2 \end{matrix} ) .

If we take

M_{1} = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) , M_{2} = ( \begin{matrix}1 & −2 \\ 0 & 1 \end{matrix} ) ,

then one checks that

(M_{1}M_{2})^{T} A(M_{1}M_{2}) = ( \begin{matrix}−2 & 1/2 \\ 1/2 & 23 \end{matrix} ) ,

which is the matrix of the reduced form −2x^{2} + xy + 23y^{2}. This form represents 17 in

( \begin{matrix}x \\ y \end{matrix} ) = (M_{1}M_{2})^{−1} ( \begin{matrix}1 \\ 0 \end{matrix} ) = ( \begin{matrix}−2 \\ −1 \end{matrix} ) .

(ii) For example, the form f(x, y) = xy represents every positive integer, with f(n, 1) = n.

Remark : It is perhaps interestign to note that there is no such form of a non-square discriminant. For suppose the discriminant is d. By Prop. 48, p.68, it suffices to find an odd prime p such that (\frac{d}{p}) = −1. By quadratic reciprocity, this is equivalent to finding a prime satisfying a finite number of congruences. Such primes exist, by Dirichlet’s theorem on arithmetic progressions (See F.1 on the January exam).

F.6 Theorem 73, p.95 in my notes.

F.7 (i) Theorem 55, p.75 in my notes, and Baker p.39-40 for a complete proof.

(ii) For any natural number x, we have x^{2} ≡ 0, 1 or 4 (mod 8). It follows easily that no number of the form 8n + 7 can be written as the sum of three or fewer squares.

F.8 (i) A ring A is Noetherian if it has no infinite ascending chains of ideals or, equivalently, if every ideal is finitely-generated as an A-module.

A ring A is local if it has exactly one maximal ideal.

(ii) (a) A = Z/NZ where N = 3 · 5 · 7 · 11 · 13 · 17 · 19 · 23 · 29 · 31 · 37.

(b) A polynomial ring over C in infintely many variables is non-Noetherian (See supplementary exercise no. 17(ii)). Localising at an appropriate maximal ideal gives a non-Noetherian local ring.

(c) The polynomial ring C[x] in one variable is Noetherian (from algebraic structures you know it is a PID). But it has the infinite descending chain of ideals

(x) ⊃ (x^{2}) ⊃ (x^{3}) ⊃ ⋯