Reduced density operator of a maximally entangled state Is the reduced density operator of a maximally entangled pure state always maximally mixed (trace being half)?
I test it on 4 bell state and this claim is true. I wonder why and can the degree of entanglement of pure state be revealed by the trace of the reduced density operator?
 A: Yes and no (for bipartite systems). 
For your first question, if you don't take it as a definiton (which is often done), you can prove it by noting that a maximally entangled (pure) state has the Schmidt decomposition $\sum_{i=1}^d 1/\sqrt{d} |i\rangle\otimes |i\rangle$ is an appropriate basis. The partial trace is the identity.
Your second question is trivial, because the reduced density matrix must have unit trace for normalized density matrices by the definition of the partial trace. In other words: Reduced density matrices are always states! If you just want to take the reduced density matrix and ask whether we can define a measure by comparing it to the identity matrix, this is indeed possible via majorization (see here for an article exploring a bit about majorization). A special case is the idea behind the so-called von Neumann entropy measure for entanglement.
For multipartite system, the answer is completely different, because the notion of a "maximally entangled state" is problematic, as there are different classes of entanglement each with their own maximally entangled set.
