Trong Z {\displaystyle \mathbb {Z} } ta có

( n ) ∩ ( m ) = bcnn ⁡ ( n , m ) Z {\displaystyle (n)\cap (m)=\operatorname {bcnn} (n,m)\mathbb {Z} }

Đặt R = C [ x , y , z , w ] {\displaystyle R=\mathbb {C} [x,y,z,w]} và I = ( z , w ) ,   J = ( x + z , y + w ) ,   K = ( x + z , w ) {\displaystyle I=(z,w),{\text{ }}J=(x+z,y+w),{\text{ }}K=(x+z,w)} . Thế thì,

• I + J = ( z , w , x + z , y + w ) = ( x , y , z , w ) {\displaystyle I+J=(z,w,x+z,y+w)=(x,y,z,w)} và I + K = ( z , w , x + z ) {\displaystyle I+K=(z,w,x+z)}
• I J = ( z ( x + z ) , z ( y + w ) , w ( x + z ) , w ( y + w ) ) = ( z 2 + x z , z y + w z , w x + w z , w y + w 2 ) {\displaystyle IJ=(z(x+z),z(y+w),w(x+z),w(y+w))=(z^{2}+xz,zy+wz,wx+wz,wy+w^{2})}
• I K = ( x z + z 2 , z w , x w + z w , w 2 ) {\displaystyle IK=(xz+z^{2},zw,xw+zw,w^{2})}
• I ∩ J = I J {\displaystyle I\cap J=IJ} trong khi I ∩ K = ( w , x z + z 2 ) ≠ I K {\displaystyle I\cap K=(w,xz+z^{2})\neq IK}