MOV-attack on ecc: Time complexity and example

There already is this pretty big post about the MOV-attack. It states, that the discrete logarithm problem on elliptic curves can be transformed to a discrete logarithm problem over a finite field. This is possible by using a mapping to a field extension.

The attack is not useful, if the resulting field extension $$F_{p^m}$$ has a large $$m$$. But if $$m$$ is small, e.g. $$m = 2$$, the complexity of solving the DLP is singificantly smaller than solving the ECDLP. ( e.g. if $$m=2$$ the security of a $$256$$ Bit ECC can be reduced to $$60$$ Bit )

My questions:

1. Why is solving the DLP instead of ECDLP easy, when $$m$$ is small? How to compute the resulting security Bits?
2. Can someone provide me a simple example including the important steps, in which the MOV attack is useful? ( Input: ECC with $$Q = n\cdot P$$ -> contructing the bilinear function $$e$$ -> using $$e$$ to transform the ECDLP to DLP -> solving DLP )? The main points I don't understand are constructing $$e$$ and how to solve the DLP faster than ECDLP.
• There's a step by step example of a MOV attack here, and here is the complexity to solve the discrete logarithm over $\mathbb{F}_{p^m}$ (replace $p$ by $p^m$ everywhere, it still applies, for the most part). – Samuel Neves Aug 6 at 19:25