A non-UFD where prime=irreducible

474 Views Asked by At

It is easy to see that in an atomic domain (where every element factors into irreducibles), we have that all irreducibles are prime iff the domain in question is an UFD.

I think it is not true for a general (commutative, unital) ring: if we consider the ring of eventually constant sequences of integers, we can find there elements which do not decompose (in fact, any sequence which is not eventually $0,1$ or $-1$ will be like that), and irreducible elements are those which have one prime coordinate, and all others $0,1$ or $-1$.

But what about domains? Can there be a domain where all irreducibles are prime, but which is not a unique factorisation domain? Better yet, is there a domain where there are no irreducible elements?

1

There are 1 best solutions below

0
On BEST ANSWER

Prime = irreducible in, e.g., GCD domains. (In fact, an integral domain is a UFD if and only if it is a GCD domain satisfying the ascending chain condition on principal ideals.) Now let's give an example of GCD domain which is not a UFD: $\mathbb Z+X\mathbb Q[X]$. (For instance, this ring doesn't have ACC on principal ideals: $(X)\subset(\dfrac{1}{2}X)\subset(\dfrac{1}{2^2}X)\subset\cdots$, so it's not a UFD.)

Edit. In the linked question the answers provide examples of rings having no irreducible elements. Instead, $\mathbb Z+X\mathbb Q[X]$ has irreducible elements: for instance, $2$ is irreducible.