TITLE: Definition of prime element in a Euclidean ring does not make sense. Herstein - Topics in Algebra
QUESTION [5 upvotes]: Herstein's Definition: In the Euclidean ring $R$, a nonunit $\pi$ is said to be a prime element of $R$ if whenever $\pi=ab$, where $a,b \in R$, then one of $a$ or $b$ is a unit in R.
$\mathbb Q$ is a field and hence a Euclidean ring. In $\mathbb Q$ every element except $0$ is a unit. So there are no prime elements in $\mathbb Q$?

REPLY [2 votes]: Yes there are no primes in a field. $5$ is a prime in $\mathbb{Z}$ but not in $\mathbb{Q}$
More generally if you have two rings $A\subset B$, being a prime in $A$ does not imply being a prime in $B$. As an example take $\mathbb{Z}\subset \mathbb{Z}[\sqrt{-2}]$. Then we have $3$ is a prime in $\mathbb{Z}$ while $3=(1+\sqrt{-2})(1-\sqrt{-2})$ is not a prime in $\mathbb{Z}[\sqrt{-2}]$