Showing that a certain prime is a Weierstrass point

Question

This problem is taken from the exercises in Number Theory in Function Fields by M. Rosen, chapter 6, page 76:

Suppose that $\omega\in \Omega_K (0)$ and has a zero $P$ of degree 1, and that $ord_p (\omega)\geq g$. Show that $P$ is a Weierstrass point.

Here $K$ is some function field, over a ground field $F$. $\Omega_K(0)$ is the set of functionals on the adele ring of $K$ which vanishes on $K$ and on adeles $\xi=(x_P)$ such that $v_P(x_P)\geq 0$ for all prime $P$. $ord_p (\omega)$ refers to the order at $P$ of the divisor associated to $\omega$: we look at sub-vector spaces of the adeles, of the form $A_K (D)=\{\xi=(x_P):x_P\geq -ord_P(D),\forall p\}$. The divisor $(\omega)$ is the greatest divisor $D$ such that $\omega$ vanishes on $A_K (D)$. In general the book denotes $\Omega_K(D)$ the space of functionals vanishing on $K$ and on adeles $\xi\in A_K(D)$.

I tried to do this: A point is a non-Weierstrass point if all its gaps are consecutive at the beginning. Because $P$ is of degree 1, this happens iff $l(gP)=1$. We know that the Riemann-Roch space $L(gP)\cong\Omega((\omega)-gP)$. So, if we could show that this space of differentials has dimension$>1$ we finish. We know that $(\omega)>(\omega)-gP$, hence, $A_K(\omega-gP)\subset A_K(\omega)$ and so $\Omega_K(\omega)\subset \Omega_K(\omega-gP)$. If this inclusion was not proper, this is it- as $dim_F(\Omega_K(\omega))=1$ (This follows from Riemann-Roch).

I guess I have 3 questions:

1. Is this the right approach? Any help in proving this result would be great.

EDIT: To anyone who might be interested, I have a solution for this question - if I did not make any mistake - not in the direction I outlined above, though. We use Riemann-Roch to show that- $l(gP)=g-g+1+l((\omega)-P)$. Now, $l((\omega)-P)=dim_F(\Omega_K(P))$. Now we only have to show that the last space is non empty, which follows from the fact that it contains $\omega$. Note however that we assome $g\neq 0$. But it's OK since in the genus$=0$ case we have no non-Weierstrass points.

2. In general, how can you construct by hand a differential? I thought maybe we could do this using the local components decomposition for a differential, but how do you even find the primes of a function field? EDIT: actually I now understand that because the differentials are a 1 dimensional space over $K$, we only have to find one non zero differential and that will give us all the others. But how can we find such a differential, even in a particular cases (say the function field obtained by taking $y^2=x$)?

3. Searching around the site for similar question, and even because of the name "differential", I feel that there's some geometrical intuition to whole of this which I am missing. Is it true? If so, I'll be very glad for any help in seeing this geometrical meaning.