User:Michellepoliseno

HW 7

8.2.4

Let $$ R \ $$ be an Integral Domain. Prove that if the following two conditions hold, then $$ R \ $$ is a Principal Ideal Domain: (1) Any two nonzero a, b $$ \in \ $$ R have a gcd which can be written as $$ ra+sb \ $$ for some r,s $$ \in \ $$ R, and (2) If $$ a_1, a_2, a_3, \dots \ $$ are nonzero elements of $$ R \ $$ s.t. $$ a_i + 1 \ $$ divides $$ a_i \ $$ $$ \forall \ $$ i, then $$ \exists \ $$ a N > 0 s.t. $$ a_n \ $$ is a unit times $$ a_N \ $$ $$ \forall \ $$ n $$ \ge \ $$ N.

Let a, b $$ \in \ $$ R be nonzero and d=gcd(a,b). Then since d is a common divisor, (d) $$ \supset \ $$ (a) and (d) $$ \supset \ $$ (b) and thus (d) $$ \supset \ $$ (a,b) is an ideal. By condition (1), d=ra+sb, which implies conversely that (d) $$ \subset \ $$ (a,b). Note that if $$ I \ $$ is an ideal of $$ R \ $$ and $$ R \ $$ is generated by two elements, then $$ I \ $$ is generated by at most one element. Then by using induction, we deduce that if $$ I \ $$ is generated by $$ n \ $$ elements, then $$ a_1, a_2, a_3, \dots \ $$, then $$ I \ $$ is actually a principal ideal. Now suppose that n=3, where $$ I \ $$ is generated by (a, b) and c. Then $$ I \ $$ is the smallest ideal contained in (a,b) and c, so its the smallest ideal containing d and c, since d=gcd(a,d). Therefore $$ R \ $$ is a principal ideal domain once we know that every ideal $$ I \ $$ $$ \in \ $$ R is generated by a finite number of elements. The finite generation follows from condition (2). By arguing by contradiction, assume $$ I \ $$ is an ideal $$ \subset \ $$ $$ R \ $$ that cannot be generated by a finite set of its elements. Take a nonzero a $$ \in \ $$ $$ I \ $$, then $$ (a_1) \subset (a_1, a_2) \subset I \ $$ with strict inclusions. It follows that we get $$ (a_1) \subset (a_1, a_2) \subset (a_1, a_2, a_3) \subset \dots \subset I \ $$. Now, we know each of the ideals $$ (a_1, a_2, \dots, a_n) \ $$ is principal. Let $$ (a_1, a_2, \dots, a_n)= (r_n) \ $$. Then $$ r_2 | r_1 \ $$ and $$ r_3 | r_2 \ $$ and so on. The quotients $$ r_n | r_n+1 \ $$ are non-units because of the strict inclusions. This is in contradiction to condition (2), which says that $$ \exists \ $$ N s.t. $$ r_N | r_n \ $$ is a unit for n $$ \ge \ $$ N.

8.2.8

Prove if $$ R \ $$ is a Principal Ideal Domain and D is a multiplicatively closed subset of $$ R \ $$ then $$ D^-1 R \ $$ is also a Principal Ideal Domain.