I'm reading example 8.20.3 in Hartshorne. If $X\subseteq \mathbb P^n$ is a hypersurface of degree $d$, then $\omega_Y\cong \mathcal O_{\mathbb P^n}(d-n-1)|_Y:= \mathcal O_Y(d-n-1)$. Then we have the geometric genus $p_g = dim_k\Gamma(Y,\mathcal O_Y(d-n-1))$.
My question:
(The 1st and 3rd highlight)Hartshorne seems to use the following thing: $p_g>0$ if and only if $d-n-1\geq 0$. I don't know how to get it. I know $\mathcal O_{\mathbb P^n}(k)$ has no global sections if and only if $k<0$. But why it is also true for $\mathcal O_{Y}(k)$?
(The 2nd highlight) Harshorne seems to use this fact: If $Y$ is a plane curve of degree $d$, then the degree of the line bundle $\mathcal O_Y(1)$ is $d$. I'm wondering how to show it.
To answer the first question: consider the exact sequence $$ 0 \to I_Y(k) \to \mathcal O_{\mathbb P^2}(k) \to \mathcal O_Y(k) \to 0. $$
We can identify $I_Y \cong \mathcal O_{\mathbb P^2}(-d)$, so after taking the l.e.s. we see that surjectivity of $H^0(\mathcal O_{\mathbb P^2}(k)) \to H^0(\mathcal O_Y(k))$ follows from vanishing of $H^1(\mathcal O_{\mathbb P^2}(k-d))$. You need to work out the numerics now to see that $k-d$ is not so negative that it has higher cohomology. If this is hard, I can add more detail when I have more time later.
To answer the second question, recall that a section of $\mathcal O_{\mathbb P^2}(1)$ vanishes along a line $L$ in $\mathbb P^2$. Thus the restriction of this section to $Y$ vanishes at the $d$ points (with multiplicity) of $Y \cap L$.