I have been stuck on this exercise in Vakil's notes (and moved on hoping it would come to me later), and it seems to be useful for other results (for example, when expressing curves as complete intersections). Can somebody tell me what I'm missing?
In 18.6.Q, Vakil asks to show that, if $X$ is a complete intersection in projective space of positive dimension, then $h^{0}(X,\mathscr{O}_X)=1$. The obvious thing to try is use induction, say $X$ is a complete intersection and $D$ is an effective cartier divisor on $X$, and consider \begin{align*} 0\rightarrow \mathscr{O}_X(-D)\rightarrow \mathscr{O}_X\rightarrow \mathscr{O}_{D}\rightarrow 0 \end{align*} and take the long exact sequence in cohomology and hope that it shows $H^{0}(X,\mathscr{O}_X)\rightarrow H^{0}(X,\mathscr{O}_D)$ is surjective, but I don't see how to proceed. It's possible I'm just being stupid since I'm working in isolation.
For what it's worth, Exercise 18.5.B seems to ask us to show that an effective ample divisor on a projective scheme with dimension at least 2 (and with enough assumptions to use Serre duality) is connected. Vakil says to first reduce to the case of very ample by taking a tensor power and then show $h^0=1$ by taking the long exact sequence and applying Serre duality and Serre vanishing, but to my uneducated mind, it seems like the current task is even more general. I don't even know how to compute $h^0$ before taking the tensor power, but it seems like the current exercise would imply it is 1.
(I would log in, but I'm in China and the log in page is being funny. I'm the same guy as https://math.stackexchange.com/users/34943/dtseng)
As you write, the problem is to show that if $X$ is a variety of dimension at least two, then any ample divisor on $X$ is connected. This is usually proved via the Lemma of Enriques--Severi--Zariski, which is in turn is (following Serre) proved via Serre duality.
The key trick is that if $\mathcal O(-D)$ is the ideal sheaf of $D$, then its $n$th tensor power cuts out the same locus set-theoretically, so we can replace $D$ by $nD$, not change anything.
Then $H^1(X,\mathcal O(-nD))$ is dual to $H^{d-1}\bigl(X, \omega_X (nD) \bigr),$ where $d$ is the dimension of $X$. If $n$ is large enough, this vanishes, by Serre vanishing (and this is where we use that $d \geq 2$).
(Note that it's crucial to take the tensor power of $\mathcal O(-D)$ before computing any cohomology.)
Added: If you don't want to argue via Serre duality, an alternative is to just proceed by induction on the codimension of the complete intersection.
Begin with the case of a hypersurface $S_d$ of degree $d$: you have $$0 \to \mathcal O(-d) \to \mathcal O_{\mathbb P^n}\to \mathcal O_{S_d} \to 0,$$ and passing to cohomology and known vanishing results on cohomology of projective space, you get the desired result.
Now suppose we intersect $S_d$ with $S_e,$ a hypersurface of degree $e$, to obtain a codim'n $2$ complete intersection $X$.
Then we have
$$0 \to \mathcal O_{S_d}(-e) \to \mathcal O_{S_d} \to \mathcal O_{X} \to 0,$$
and now you have to show that $H^1(\mathcal O_{S_d}(-e)) = 0$. But you can compute this by twisting the first short exact sequence by $-e$, and passing to cohomology, and again using known vanishing results. (These will apply provided the projective space has dimension at least $3$; the key fact is just that the cohomology of all the $\mathcal O(n)$ vanish in degrees strictly between the top and the bottom.)
Now let's continue. At each stage you ultimately get back to cohomology of some $\mathcal O(n)$ on projective space of one degree higher, and eventually, when this degree reaches the top, the argument breaks down --- this is exactly the point at which you reach a complete intersection of dimension zero.