Prove that $I^2=2I$ as ideals in $Z[\sqrt { −3}]$

Question

Let $I$ be the ideal generated by $2$ and $1+ \sqrt {−3}$ in the ring $Z[\sqrt { −3}]$.

Prove That $I^2=2I$.

We have $2I=2(2,1+ \sqrt {−3} )=(4,2+2 \sqrt −3 )$ $I^2=(2,1+ \sqrt {−3} )(2,1+ \sqrt {−3} )=(4,2+2 \sqrt {−3}, -2+2 \sqrt{−3} )$

Now I am convinced that if we have 3 ideals $(a),(b),(c)$ then $(a,b,a-b)=(a,b)$ because $(a-b)\subset (a,b)$.

Which gives $I^2= (4,2+2 \sqrt {−3})=2I$.

Is that all correct? Thank you for your feedback.