1) A nonzero ideal in a principle ideal domain is maximal if and only if it is prime.
Proof- Let R be a principle ideal domain and I be a nonzero ideal in R.
Assume that I is maximal. Then is a field, which means that is an integral domain. Hence, I is prime.
Assume that I is a prime ideal. Since R is a PID, we can write for some nonzero . Then we know that the element a is prime and hence irreducible. Then since a is irreducible, is maximal in the set of all proper principle ideals of R. But since every ideal in R is principle, is maximal in R.
2) An integral domain R is a unique factorization domain if and only if every nonzero prime ideal in R contains a nonzero principle ideal that is prime.
Proof- In order to prove this, we need a lemma: In a UFD, irreducible elements are prime.
Subproof- Let R be a UFD and be irreducible. Then c is a nonzero nonunit. Now assume that for some . Then we can write for some . If or then we are done trivially so assume that a, b are nonzero. If either a or b is a unit, then either or which gives that or so assume that a, b are nonunits as well. If then since R is an integral domain we would have either a or b is zero, a contradiction. If x is a unit then since c is irreducible, a is a unit or is a unit. We know a is not a unit so that would have to be a unit. Then so that c is prime. Now assume that x is not a unit. Being in a UFD means that we can now factor a, b, and x into a product of irreducibles. Let . We have that . Both sides are products of irreducibles and so by the definition of UFD we must have that and that any element on one side must be an associate to some element on the other side. Thus c is an associate of some or . In the first case we get that for some unit . Then and similarly if c is an associate of some then . Thus c is prime. Now using the lemma we can do the proof:
Assume that R is a UFD. Let I be a nonzero prime ideal in R. Then there exists a nonzero element . This element cannot be a unit or else I would be all of R and this is a contradiction since prime ideals must be proper ideals. So then c is a nonzero nonunit. Since we are in a UFD we can write where each is irreducible. Then by the lemma we have that each is in fact prime. Therefore is a prime ideal contained in I.
Assume that every nonzero ideal in R contains a nonzero principle ideal that is prime. Let S be the set of all units in R and all products of prime elements of R. Note that S is a multiplicative set since a product of units is a unit and a product of a product of primes is a product of primes. Also, if is prime and is a unit, then we need to know that is prime or a unit as well. Suppose it is not a unit, then suppose . But p is prime so then or . In the first case we are done and if . Therefore S is closed. We want to show that . This will give that every nonzero nonunit is a product of primes and hence a product of irreducibles since we are in an integral domain. Suppose to the contrary that there exists some nonzero nonunit . Then we must have that since S is also closed under divisors, i.e. whenever a product . Next we need a few facts: 1) Since is an ideal, it is contained in a maximal ideal. 2) The set of all ideals of R which are disjoint from S contains a maximal element, and that element is prime. Therefore since is an ideal disjoint from S, it is contained in a maximal ideal, P, which is also a prime ideal and also disjoint from S. However, by assumption this prime ideal P contains a principle ideal which is prime, and therefore contains a prime element, say p. But S contains all prime elements so this contradicts the disjointness. Therefore S consists of units and irreducible elements and hence R is a UFD.
6) (a) If a and n are integers, , then there exist integers q and r such that where .
(b) the Gaussian integers form a Euclidean domain with .
Proof- a) Let with . By the division algorithm there exists such that where . Then . Pick and . Then , where . This ends the problem
b) Consider where . We must show that this is a Euclidean Ring. First assume that and . Then with some work we can show that since . Now let and , . Then we can find integers such that , with . Let . If r is zero we are done so assume r is nonzero. Then . So when for there exists such that either or and . Then just replace x with in the previous proof. Find such that with either or and . Set . Then . Then so we have a Euclidean domain.
7) What are the units of Gaussian integers ?
Proof – Let . We claim that and show it by subset inclusion. Let . Then there is such that . By exercise 6(b) we have that . Since we can only have the case that . But then says that either and or and . Thus or . Therefore we have inclusion. The other inclusion is trivial to check.
9) Let R be a UFD and d a nonzero element of R. There are only a finite number of distinct principal ideals that contain the ideal .
Proof- If d is a unit then we are done trivially, so assume d is a nonunit. Then since we are in a UFD we can write where is irreducible. Let be any element satisfying . Then k cannot be 0 and if k is a unit then it generates all of R so that is the only principle ideal which contains (d). So assume that k is a nonunit. Then we can write with each irreducible. Now so that for some . If then d would be zero also, which is a contradiction. If x is a unit then which tells us that is the only principle ideal containing . Therefore we can assume x is a nonunit. Then we can write for some irreducibles . Then we have since we are in a UFD. Note that for each we have and each is irreducible so that each is maximal. Thus is contained in m maximal ideals. If there were infinitely many such principal ideals we would have to have infinitely many m’s which satisfy . But and l is fixed so this is impossible. Hence there are only finitely many k satisfying the condition.
10) If R is a UFD and are relatively prime and then .
Proof- First note that if then have no irreducible factors in common. This means that if with irreducible, then no is an associate of . For if they were associates, then we could say where u is a unit, then . But since and gcd(a,b) = 1 this would give that which means that is a unit, contradicting that is irreducible. Finally, assume that . Then for some . Factoring into products of irreducibles we have that . Now cannot be associate of any so each must be an associate of some . Therefore so without loss of generality say that is an associate of for . This tells us that for . Thus we can say . But then we can multiply by what we need to get all of c on one side, so that .
11) Let R be a Euclidean ring and . Then a is a unit iff .
Proof- Assume that a is a unit. Then . Therefore .
Assume now that . There must exist such that where either r = 0 or . Note that , so we cannot have that . Therefore r = 0 and hence so that a is a unit.