The antiferromagnetic (AFM) s=1 chain with biquadratic exchange coupling has recently been the subject of considerable attention, and some controversy. Theoretical arguments by Affleck that the model has dimerized character associated with a small spectral excitation gap have not received unanimous support from numerical calculations. For finite systems the field-dependent dispersion spectra are interesting and informative concerning the nature of the model. The excitations divide into two classes having very different character, corresponding to SzT even and SzT odd. Reflection symmetry about k=π/2 is apparent if states with a given total spin ST, rather than SzT, are plotted as a function of wavevector k, consistent with the Affleck prediction of dimerized character. The high degeneracy of the eigenstates reflects the existence of new symmetry operators which (a) shift the SzT value, (b) change wave vector k to π/2−k, and (c) shift the total spin quantum ST, of the energy eigenstates. Parkinson has recently observed that finite N states with SzT even and SzT≥2 map into states of the finite N s= 1/2 XXZ model at a special value of uniaxial anisotropy. Barber and Batchelor (unpublished) have shown the existence of an analytic mapping between states of the s=1 biquadratic chain and the 2D 9-state Potts model. They further show that all states of the Parkinson s= 1/2 XXZ model map into states of the s=1 biquadratic model. Hence, they find the ground state and first excited state energies exactly, and verify the Affleck picture. Their approach falls short of establishing complete integrability, however. Numerical studies on finite chains indicate that the T=0 magnetization curve is determined by the lowest-energy states for given SzT even, which are states of the s= 1/2 XXZ chain. Hence the magnetization curve of the s=1 AFM biquadratic chain is the same as the analytically known magnetization curve of the s= 1/2 XXZ chain!