$I=(2,X)$ and $J=(3,X)$ ideals, why $IJ=(6,X)$?

Question

I'm still not 100% comfortable with ideals.

How do I show that for $I,J\in\mathbb{Z}[X]$ with $I=(2,X)$ and $J=(3,X)$, we have $IJ=(6,X)$?

Answer 1

$IJ=(6,2X,3X,X^2)$.

Now, since $2$ and $3$ are relatively prime, $(2X,3X)=(X)$. Hence, $IJ=(6,X,X^2)$, and because $X^2\in(X)$, $IJ=(6,X)$.