Is prime ideal equivalent to cannot be factored into 2 proper ideals?

Question

Is prime ideal equivalent to cannot be factored into 2 proper ideals? Sorry if this is a stupid question I couldn't find this in the list of characterizations for prime ideals on Wikipedia.

Answer 1

No, this is not an equivalence. The ideal $(x,y^2)\subset k[x,y]$ for $k$ a field cannot be factored as a product of ideals, but is clearly not prime as $y\cdot y=y^2\in(x,y^2)$ but $y\notin (x,y^2)$. If $I,J$ are two proper ideals so that $IJ=(x,y^2)$ then we must have that $\sqrt{I}=\sqrt{J}=(x,y)$. But $IJ\subset \sqrt{I}\sqrt{J} = (x,y)^2$, so $x\notin IJ$ and $IJ\neq (x,y^2)$.