If we have that there exists such that and so and this last one has a factorization in integers since it's even.
If and odd then and this one is even and so it has non-trivial factorization in integers. If is even take
Let be an integer (square free, not equal to one)
Prove that if or then is not a UFD.
I know that 2 is not a prime in . I also know that if d is one mod three we can rewrite it as the sum of 2 squares. But I don't know how to come up with explicit d. For I don't have a handy characterization (sum of 2 squares or something like that).
you forgot that a domain is not a UFD if some element can be written as product of irreducible elements in more than one way.
1) we have and thus if then won't be integrally closed and hence it cannot be a UFD.
2) if then is an irreducible element of (very easy to see!) so if was a UFD, then would have to be prime. now choose such that is an even number.
so and thus either or which is obviously impossible. Q.E.D.
Remark: clearly if and only if so that "either ... or" statement in 2) is not necessary and we could have just said
Another example of an integral domain that is not U.F.D is
, where F is a field. The proof of showing that K is not U.F.D is similar to the above proof.
We see that is irreducible in K, but it is not a prime element of K.
If it were a prime element, implies , which is impossible.