Does being principal ideal ring with identity implies PID

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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)

Best Answer

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.