I have recently stumbled upon a standard proof that if $R$ is a Euclidean ring, then $R$ is a PID. But in the proof they first show that $R$ is a principal ideal ring with identity. But then they deduce that thus $R$ is a PID without further explanation. In particular, I cannot see why $R$ is an integral domain. It has identity, commutative, but cannot see why it does not have zero divisors. (Proof from Hungerford's Algebra book)
2026-03-26 04:29:30.1774499370
Does being principal ideal ring with identity implies PID?
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No. There are lots of principal ideal rings (with identity) that are not domains.
For example, $F_2[x]/(x^2)$ where $F_2$ is the field of two elements, or even more simply $\mathbb Z/4\mathbb Z$.
Also $F_2\times F_2$, since I have an ongoing thing where I try to use this whenever it applies.