Closed immersions and complete linear systems Let $X$ be a local complete intersection subscheme in $\mathbb{p}^n$ for some integer $n>0$. Denote by $i:X \to \mathbb{P}^n$ the induced closed immersion, $\mathcal{O}_X(1):=i^*\mathcal{O}_{\mathbb{P}^n}(1)$ and $N:=h^0(\mathcal{O}_X(1))$. Does there exist a closed immersion of $X$ into $\mathbb{P}^N$ via the complete linear system $H^0(\mathcal{O}_X(1))$, which factors through $i$?
 A: I always get confused about what the phrase "factors through" means, so let me describe the setup and you can decide if this is what you wanted or the opposite.
We have an exact sequence of sheaves on $\mathbf P^n$ of the form
$$ 0 \rightarrow I_X(1) \rightarrow O_{\mathbf P^n} (1) \rightarrow O_X(1) \rightarrow 0 $$ 
where $I_X$ is the ideal sheaf of $X$. Let's assume $X$ isn't contained in a hyperplane. (If it is, replace $n$ by $n-1$ and repeat). Then the first sheaf has no global sections, so taking cohomology above we get the exact sequence
$$ 0 \rightarrow H^0 (O_{\mathbf P^n} (1)) \rightarrow H^0(O_X(1)) \rightarrow H^1(I_X(1))$$
So $H^0 (O_{\mathbf P^n} (1))$ is a subspace of $H^0(O_X(1))$. 
Suppose we choose a basis $\{s_0,\ldots, s_N\}$ for $H^0(O_X(1))$ such that $\{s_0,\ldots, s_n\}$ is the standard coordinate basis for $H^0 (O_{\mathbf P^n} (1))$. Then the immersion $X \rightarrow \mathbf P^N$ you want is given simply by $x \mapsto [s_0(x), \ldots, s_N(x)]$, while your given map $i: X \rightarrow \mathbf P^n$ is just$x \mapsto [s_0(x), \ldots, s_n(x)]$.
So there is a diagram
$$X \rightarrow \mathbf P^N \dashrightarrow^\pi \mathbf P^n$$
where $\pi$ is projection away from a linear subspace (or if you like, forgetting the last $N-n+1$ coordinates) and the composite is equal as a rational map to $i$. 
To me this situation would be described as saying that $i$ factors through the new immersion, rather than the other way around.
Remark: You might wonder if I'm making a mountain out of a molehill here, but I claim I am not: there are indeed subvarieties $X \subset \mathbf P^n$ for which $H^1(I_X(1)$ is nonzero, and so the factorisation is nontrivial. Indeed, the property that $H^1(I_X(1)=0$ is interesting enough to get its own name: such an $X$ is called linearly normal. Searching this site for that phrase will give examples of subvarieties which fail to be linearly normal.
