Angular momentum of a purely rotating body about any axis It's proved in my K&K mechanics textbook that in pure rotation about an axis passing through the body  it's angular momentum is $I\omega$.
What about if I want to find the angular momentum about an axis outside of it? Is it going to be the same?
In the special case of  the body  rotating about it's center of mass ,will it's angular momentum be the same for all axis?
My attempt to prove that for the above special case the angular momentum will be the same

Let the angular momentum about the centre of mass be $L_0$ and about another axis be $L$
Then we can write$\begin{aligned} \vec{L} &=\sum R_{i} \times m_{i} v_{i} \\ &=\sum\left(R_{0}+r_{i}\right) \times m_{i} v_{i} \\ &=\sum R_{0} \times m_{i} v_{i}+\sum r_{i} \times m_{i} v_{i} \\ &=\left(R_{0} \times \sum m_{i} v_{i}\right)+L_{0} \end{aligned}$
We know that the velocity of centre of mass is zero therefore  $\sum m_{i} v_{i}=0$
$\Rightarrow \quad \vec{L}=\vec{L}_{0}$
Where am I going wrong then?
 A: No, it won't be the same. Always remember to specify the axis with respect to which you are evaluating the moment of inertia. Usually we calculate the moment of inertia with respect to the center of mass which is the smallest possible of your rigid body. So it would be more correct to write $I_0$ for the moment of inertia or $I_{CM}$. There is a theorem that can answer to your question.
The Huygens-Steiner theorem, or parallel axis theorem, allows to calculate the moment of inertia of a solid with respect to an axis parallel to the one passing through the center of mass avoiding in many cases (where there is a symmetrical structure) the laborious direct calculation.
The moment of inertia with respect to an axis $a$, parallel to another $c$ passing through the center of mass, is obtained by adding to the initial moment of inertia with respect to $c$ the product between the mass of the body itself and the square of the distance between the axes $c$ and from $a$. $$I_z=I_{cm}+Md^2$$
So if you have different moment of inertia, at the same angular velocity, you will have different angular momentum.
With this formula you can also see that the smallest moment of inertia is when $d=0$ so the moment of inertia respect to the axes through the CM: $I_{CM}$.
Hope this can help you.
A: No,  it wouldn't not be same as Moment of Inertia will change .
$\omega$ is same for every particle in a rigid body .
Hence , $ L_{axis}=I_{axis}\omega$ for a rigid body in pure rotation.
