The moment of inertia of a ring about an axis perpendicular to its plane is $I_z = MR^2$. By the perpendicular axis theorem, $I_z = I_x + I_y$. Since for a ring $I_x = I_y$ (moment of inertia about a diameter), $MR^2 = 2I_{dia}$. Therefore, $I_{dia} = (1/2)MR^2$.