Let be a squarefree integer greater than 3, and let .

- Prove that , , and are irreducible in .
- Prove that is not a unique factorization domain.
- Give an explicit ideal in that is not principal.

We begin with some general remarks.

Let be an integral domain. A divisibility norm (nonstandard term) on is a mapping such that for all . That is, is a multiplicative semigroup homomorphism, or a poset homomorphism with respect to divisibility. We say that is bounded if and . We can see that if is a bounded divisibility norm on a domain and is a unit with inverse , then , so that .

Let be squarefree greater than 3. We claim that is a bounded divisibility norm. To see this, let and .

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So is a divisibility norm. Certainly is bounded because and . In particular, if is a unit in , then . Since , we have , and then . So the units in are precisely . Conversely, if , then is a unit.

- Suppose . Then . Suppose . With , we have . Since , we have . Then , a contradiction. So either , and is a unit, or so that is a unit. Thus is irreducible in .
Suppose . Then ; we may assume without loss of generality that , with strictness because is squarefree. Suppose . Then divides properly. Then , and thus divides . Since is squarefree, and thus is a unit. So is irreducible.

Suppose . Then . Letting , divides . If , then since we have a contradiction since is too large. If , then divides , and we have . in either case, we have , so that and thus is a unit. Suppose , and let . Then . If , we have a contradiction. If , then in particular, . Then is a unit. If , then we have . In this case we have , a contradiction. Thus is irreducible.

- First, we prove the following lemma.
Lemma: For a squarefree integer greater than 3 and , divides if and only if divides . Proof: Note that . Conversely, .

Suppose is a composite integer; say where . Using the lemma, divides , but does not divide or . Thus is not prime in .

Now suppose is a prime integer; since , is a composite integer. Say with . Note that . But if , then we have , a contradiction. Thus is not prime in .

- Suppose is prime. Since , we have , and moreover is odd. Suppose the ideal is principal, with . Note that , so that divides . If , then is larger than 4. Thus , and since divides 4, up to units we have . If , then we have for some , . Multiplying this out and comparing coefficients, we have and . We wish to show that these equations have no solution in the integers. To see this, reduce both equations mod 2. Since is odd, we have the sum simultaneously congruent to 0 and 1 mod 2, a contradiction. Thus . Suppose instead that . But then for some , a contradiction. Thus the ideal is not principal after all.
Now suppose is composite; let , where is prime. Suppose the ideal is principal, with . Then divides . If , then is too large. If , then divides , and we have . But now, divides ; this contradicts either the compositeness or the squarefreeness of . Thus we have , so that divides . Since is squarefree, , and thus . In particular, for some and , we have . Simplifying and comparing coefficients, we see that ; this is a contradiction because divides . Thus is not principal after all.

## Comments

I think there are some typos in the third paragraph of part two. For example, I’m fairly certain the last bit should say that n isn’t prime, etc.