Let $R$ be a commutative ring. An ideal $I$ is called prime if whenever $ab\in I$ then $a\in I$ or $b\in I$. I want to show that $I$ is prime if whenever $JK\subseteq I$, then $J\subseteq I$ or $K\subseteq I$. I want to show a third equivalent condition, so that I have:
- $I$ is prime.
- Whenever $XY\subseteq I$, then $X\subseteq I$ or $Y\subseteq I$ for subsets $X,Y\subseteq R$, $XY$ is the complex product $\{xy\mid x\in X,y\in Y\}$.
- Whenever $JK\subseteq I$, then $J\subseteq I$ or $K\subseteq I$ for ideals $J,K\subseteq R$, $JK$ is the ideal product.
My attempt is as follows:
$(1)\implies (2)$: Suppose $X,Y\not\subseteq I$ but $XY\subseteq I$. Let $a\in X-I$, $b\in Y-I$. Then $a,b\notin I$ but $ab\in XY\subseteq I$.
$(2)\implies(3)$: Let $JK\subseteq I$. Choose generating sets $(X)=J$, $(Y)=K$. Then $(XY)=(X)(Y)=JK\subseteq I$, so $XY\subseteq I$, so $X\subseteq I$ or $Y\subseteq I$, so $J\subseteq I$ or $K\subseteq I$.
$(3)\implies(1)$: Suppose $I$ is not prime. Let $a,b\notin I$ such that $ab\in I$. Now take $J=(a)$, $K=(b)$. Then 3. is not satisfied.
Is this proof correct?
The proof looks fine, clear and well-written. My only suggestion is that you could simplify $(2)\implies(3)$ by just letting $X=J$ and $Y=K$ and then applying $(2)$. If you want to be clear about where you are using $(2)$, just say so explicitly in the proof. It's conceptually simpler than finding generating sets.