Is angular momentum equal to Inertia $\times \omega$ and why?

$\text{Explanation:}$
Angular momentum is also known as the moment of momentum
consider a rigid body rotating about a fixed axis with an angular velocity of $\omega$, whose moment of inertia is I
the definition of I is the sum of the products of the mass and the square of the perpendicular distance to the axis of rotation of each particle in a body to the axis of rotation.
i.e
$I=\sum m{r}^2$

consider the rigid body is made up of particles each of mass m

then the moment of momentum ( call it L) of each particle

$L=momentum\times r$

where r is the perp. distance from the axis

$\therefore L=mvr$

where v is the linear (tangential) velocity

but for angular motion

$v=r\omega$

so

$L=mrr\omega =m{r}^2\omega$

For the total angular momentum we have to sum up all the particles

$L_T=\sum m{r}^2\omega$

$L_T=\omega \sum m{r}^2$

$L_T=I\omega$

so angular momentum is moment of inertia $\times$ omega in vectors the formula is

$\vec{L}=\vec{r}\times m\vec{v}$