Prove or disprove when for any r,x $\in$ R and rx $\in$I where I is an ideal, then x $\in$ I?

Question

Recall the definition of ideal I in a ring R;

• I is a subgroup of R under addition

• For any x $\in$ I and any r $\in$ R, rx $\in$ I and xr $\in$ I

My question is : Change the order

• If for any r,x $\in$ R and rx $\in$I where I is an ideal,
• Can we say that x $\in$ I? If Yes, please give a short proof, if not, give me a counterexample. Thanks

Answer 1

This is an excellent question because it leads to the extremely important notion of a prime ideal. A proper ideal $\mathfrak p$ of a commutative ring $R$ is said to be prime when this condition $rx \in \mathfrak p$ implies $r \in \mathfrak p$ or $x \in \mathfrak p$, exactly as you stated. I can then answer your question by providing an example of a non-prime ideal. Indeed, the name "prime" is appropriate. Consider the ring $\mathbb Z$ and the ideal $4\mathbb Z \subseteq \mathbb Z$. $4 = 2 \cdot 2 \in 4 \mathbb Z$ but $2 \notin \mathbb Z$.

The fact that $4$ has a nontrivial prime factorization into $2^2$ is exactly why this works. Here's an exercise for you if you'd care to try it: prove that if $n \geq 2$ that $n \mathbb Z$ is a prime ideal if and only if $n$ is prime.