Let R=Z[−19] and L=Z[2α+1] which is Dedekind by theorem 3.19. For an ideal i we have L/i≅in/in+1 as discussed during the last lecture. Also, we have ik⊂ik−1⊂L⇒(L/ik)/(ik/ik+1)≅L/ik by the Third isomorphism thoerem (As modules).
Let p our ideal after lifting it to L, p=(2,1+−19)=2(1,21+−19)=2L. (There might be an abuse of notation)
We know that any element in L=Z[2α+1] can be written as a+b21+α. By direct computations we can see that the norm is 2⋅2=4
Again we have:
Finally, from previous set of excercises we found that ∣R/p∣=2 and from calculations above we find that
2=∣R/p∣=∣(R/p2)∣÷∣(p/p2)∣⇒∣(p/p2)∣=28=4. Since, the number of elements are different then R/p and p/p2 can't be isomorphic.
Finally, for seek of contradiction assume 2Rpcontains a power of the the maximal idea p, then as we showed in the last set of excercises that p2=2p⇒by repeated applicationpn+1=2np and for some power 2mp=pm+1=2Rp⇒integral domain2n−1p=Rp contradiction with the fact that 2Rp⊂p which is a proper ideal.
First of all notice that f(x)=x3−x−1 is irreducible over Z, consequently(Gauss lemma), irreducible over Q.
Well, for any a∈R,a≤1⇒f(a)<0. Also, a∈R,a≥2⇒f(a)>0. The only real solutions θ, by intermediate value theorem, is strictly between 1 and 2. Hence, it has no solutions in Z
Now to show the number ring R=Z[α] is Dedekind, it is sufficient to to show that each ideal is regular. The main tool that will be used here is the third statement in the Kummer-Dedekind theorem 3.1.
Let p a prime ideal of Z[α] lies over a prime number p. If f has multiple factors mod p then f and f′ has a common factor mod p. Consider the linear combination x3−x−1−3x(3x2−1)=−(32x+1) which has the root x=−23. Since it is multiple root then 0=f′(−23)=423 mod p iff p=23,x=. However, f(x)=(x−10)g+r where r is constant, hence r=989=f(10)=0⋅g+r but 432∣989=23⋅43, hence all ideals are regular.
If an ideal has a norm at most 30 then it lies over a prime numbers which is at most 30, i.e. [2,3,5,7,11,13,17,19,23,29].
Note: A prime ideal p can't lie over more than one prime p of Z due to Z[α]/p≅F[α] (Since α⋂Z=(p) some prime)
Now by trying all possible solutions for the polynomial x3−x−1 over primes in our list above.
Hence, 2,3,13 and 29 inert. Looking at the quotient by ideals as quotient of modules we find the norm of each is p3. Theorem 3.4
In other words, Z[α]≅Z×Z×Z as Z−module with 1,α,α2 as basis. Hence ∣Z[α]/(p)∣=p3 and since (p)⊆p⇒Z[α]/(p)⊇Z[α]/p. Where, p is a prime divisor of (p). We are now in a finite ring, by a simple counting argument, using quotients and irreducibility of the polynomial, one can conclude that N(p)=pdeg(p)
So, we exclude 13 and 23 since their norm are bigger than 30. By the same method, we enumerate all all norms that are less than 30 with the ideals.
Finally, in Z[α] we have 1=α3−α=α(α2−1)=α(α−1)(α+1). Hence, u=α+1 is unit and of infinte order because un=m+kα+hα2 where all coefficients are positive and they increase as n increases due to the fact a3=a+1.