 Solution

The same pole-order argument provides a basisFor $n\ge 0$ the first set contains $\lfloor n/2\rfloor +1$ elements, and the second contains $\max (0,\lfloor (n-3)/2\rfloor +1)$. Distinct pole orders imply linear independence, and the coordinate-ring expression proves spanning. HenceFor $n<0$ a function in the space would have no poles anywhere and would vanish at infinity; a regular function on a projective integral curve is constant, so it must be zero. Equivalently the coordinate-ring pole argument first makes it constant, then the vanishing condition excludes nonzero constants.