Name: Ahmed Alharbi
Student Number: 1901974
Leiden University
Q.9
Let and which is Dedekind by theorem 3.19. For an ideal we have as discussed during the last lecture. Also, we have by the Third isomorphism thoerem (As modules).
Let our ideal after lifting it to , . (There might be an abuse of notation)
We know that any element in can be written as . By direct computations we can see that the norm is
Now since:
Again we have:
Finally, from previous set of excercises we found that and from calculations above we find that
Finally, for seek of contradiction assume contains a power of the the maximal idea , then as we showed in the last set of excercises that and for some power contradiction with the fact that which is a proper ideal.
Q.10
First of all notice that is irreducible over , consequently(Gauss lemma), irreducible over .
Well, for any . Also, . The only real solutions , by intermediate value theorem, is strictly between 1 and 2. Hence, it has no solutions in
Now to show the number ring 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 a prime ideal of lies over a prime number . If has multiple factors mod
then and has a common factor mod
. Consider the linear combination which has the root . Since it is multiple root then iff . However, where is constant, hence but , 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 can't lie over more than one prime of due to (Since )
Now by trying all possible solutions for the polynomial 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 . Theorem 3.4
In other words, as with as basis. Hence and since . Where, is a prime divisor of . We are now in a finite ring, by a simple counting argument, using quotients and irreducibility of the polynomial, one can conclude that
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 we have . Hence, is unit and of infinte order because where all coefficients are positive and they increase as increases due to the fact .