Let $X$ be the d-uple embedding of $P^1$ to $P^d$, for any $d\geq1$. We call $X$ the rational normal curve of degree $d$.
Hartshorne's says that if $X$ is any curve of degree $d$ in $P^n$, with $d\leq n$,which is not contained in any $P^{n-1}$, then in fact $d=n$, the genus $g(X)=0$, and $X$ differs from the rational normal curve of degree $d$ only by an automorphism of $P^d$.
I think it's quite difficult to give the proofs. And I try to compute the genus of rational normal curve, which should be zero. If $d=2$, then $X$ is a plane curve of degree 2. And according to a classical result, $g(X)=\frac{1}{2} (d-1)(d-2)=0$. We get the conclusion.
But when I try to compute the case of $d=3$, thing is getting out of my hands. We need to compute $H^1(X,\mathcal{O}_X)$, where $\mathcal{O}_X=Spec\frac{k[x,y,z,w]}{(y^2-xz,z^2-yw,xw-yz)}$. I guess we can use Cech cohomology to compute it, but it's really complicated. So, I hope someone could help me out here.
And if anyone could show me proofs to other statements, I would be grateful.
First I'll show that $d=n$. We can prove the following more general result: Let $X\subset \mathbb{P}^n$ be a variety not contained in a hyperplane. Then $\deg(X)\geq 1+\mathrm{codim} X $. Proof: We proceed by induction on codimension. If the codimension is $1$, then the result follows from the fact that $X$ is not in a hyperplane. Now if the codimension of $X$ is at least $2$, project from a general point of $X$. This will reduce the codimension by $1$ and the degree by at least $1$, so we're done by induction. Now applying this to the case of $X$ a curve, we get that $\deg(X)\geq 1+n-1=n$. The assumption that $\deg(X)\leq n$ then forces $d=n$. So now we know that our curve is of degree $n$ in $\mathbb{P}^n$. Now I want to essentially do the argument from above in a more refined way. Take a point $p\in X$ and project from it to get a rational map $\pi:\mathbb{P}^n\dashrightarrow\mathbb{P}^{n-1}$. Then any hyperplane in $\mathbb{P}^{n-1}$ is of the form $\pi(H)$ where $H$ is a hyperplane in $\mathbb{P}^n$ which passes through $p$. The general such hyperplane $H$ will intersect $X$ in $n$ distinct points because $X$ is of degree $n$ in $\mathbb{P}^n$. One of those points is $p$, so therefore $\pi(H)$ will intersect $\pi(X)$ in $n-1$ points. Hence, the image $\pi(X)$ has degree $n-1$. Continue the process until we get to a conic in $\mathbb{P}^2$, which is rational, by the degree-genus formula you used in your question.