let $R$ be a Ring, show that p is a prime element if and only if (p) is a prime ideal in $R$.

Question

let $R$ be a Ring, show that p is a prime element if and only if (p) is a prime ideal in $R$.

I am puzzled as to what the definition of a prime element would be in a general ring. I thought divisibility is only well defined in integral domains.

Answer 1

This isn't true in general: $0$ isn't considered a prime element in $\mathbb Z$ (or any ring, for that matter), but it generates a prime ideal.

For a commutative ring and a nonzero element $p$, it is true that $(p)$ is a prime ideal iff $p$ is a prime element: one just has to check the definitions.

The choice to use commutativity is just an expediency to stick with the normal definition of "prime ideal" in commutative algebra. Prime ideals of noncommutative rings are slightly more complicated to explain in terms of elements.

I thought divisibility is only well defined in integral domains.

For a commutative ring, we usually say "$a$ divides $b$ if there exists a $c$ such that $ac=b$." Where does being a domain come into it? Asking for it to be a domain buys us other nice things, but it has nothing to do with the definition of divisibility. Divisibility is nothing more than the containment partial-order on the principal ideals, which is always there.

For noncommutative rings one can do something similar but you'd have to take side into consideration. That is not usually something authors take time to do in introductory texts.