TITLE: Why separating tangent vectors?
QUESTION [10 upvotes]: Let $X$ be a projective scheme over an algebraically closed field.
It is a well-known theorem that a line bundle $L$ defines a closed embedding into projective space if and only if it separates points and tangent vectors, which means:
(1) For any distinct closed points $P, Q \in X$ there is an $s \in \Gamma(X,L)$ with the property that $s\in m_PL_P$ but $s \not\in m_QL_Q$.
(2) For every closed point $P \in X$ the set $\{s \in  \Gamma(X,L)|s_P \in m_PL_P\}$ spans the vector space $m_PL_P/m_P^2L_P$.
Now in Hartshorne (Prop. IV 3.4.) this criterion is applied in the following situation: One wants to show that every curve $C$ can be embedded into $\mathbb{P}^3$. First one embedds the curve $C$ into an arbitrary $\mathbb{P}^n$. Then one chooses some $O \in \mathbb{P}^n$ andprojects the curve down from $O$ into $\mathbb{P}^{n-1}$.
Then one wants to show that this projection map is a closed immersion if and only if
(a) $O$ does not lie on any secant of $C$
(b) $O$ does not lie on any tangent of $C$.
At this point Hartshorne applies the above criterion. But at this point I do not understand why the fact that $O$ does not lie on any tangent implies $(2)$ of the above criterion. I understood why $(a)$ implies $(1)$ but why does $(b)$ yield that the line bundle separates tangent vectors

REPLY [6 votes]: [I assume $C$ is smooth]
The point is this: $m_PL_P/m_P^2L_P$ is $1$-dimensional and hence generating it is equivalent to finding a single element that is not zero. The statement that all sections that are in $m_PL_P$ are also in $m_P^2L_P$ is equivalent with the statement that every member of the corresponding linear system that contains $P$, also contains the tangent line to $C$ at $P$. I think you should be able to finish from here.
[A modification of this idea actually works in higher dimensions, so the first sentence is not really necessary, but makes it easier in the curve case.]