We consider a variety $X$ over $k$. I have a question on an statement from wikipedia deals with interpretations of normality in algebraic geometry. It says:
An older notion is that a subvariety $X$ of projective space is linearly normal if the linear system giving the embedding is complete. Equivalently, $X \subset \mathbb{P}^n_k$ is not the linear projection of an embedding $X \subset \mathbb{P}^{n+1}_k$ (unless $X$ is contained in a hyperplane $\mathbb{P}^n$). This is the meaning of "normal" in the phrases rational normal curve and rational normal scroll.
By definition a linear system $L$ is complete if there exist a divisor $D \subset C$ such that $L= \vert D \vert$.
Recall: $\vert D \vert:= \{ (f) + D \ \vert \ f \in \mathcal{O}_X(D) \}$.
Futhermore a map $f: X \to \mathbb{P}^n$ induced by a divisor $D$ is completly determined by data $\{s_i \in \mathcal{O}(D) \ \vert i=0,1,...,n \}$ and these sections are considered as pullbacks of the canonical generators $X_i$ of $ H^0(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(1))=kX_0 \oplus kX_1 \oplus ... \oplus kX_n$; i.e. $s_i = f^*X_i$.
Denote by $i_n: X \subset \mathbb{P}^n$ the embedding of $X$ into projective space above. Using definition above we can indentify $\vert D \vert$ with projective space $(H^0(X,D) - \{0\})/k^*$.
The definition of completeness of linear system $L$ above is equivalent to the statement the global sections $s_i = i_n^*X_i$, which generate in general only the linear $k$-subspace $<\pi^*X_i \vert i=0,1,2>_k$ of $k$-space $H^0(X, \mathcal{O}_X(D_0))$ for certain divisor $D_0 \subset X$, in case of completeness generate the whole $H^0(X, \mathcal{O}_X(D_0))$.
Question: Why is it equivalent that the subvariety $X$ of projective space $\mathbb{P}^n_k$ is linearly normal to the claim that $X \subset \mathbb{P}^n_k$ is not the linear projection of an embedding $X \subset \mathbb{P}^{n+1}_k$ (unless $X$ is contained in a hyperplane $\mathbb{P}^n$)? Moreover, can the assumption on $X$ to be a variety replaced by $k$-scheme ?
A couple words about the (linear) projection $\mathbb{P}^{n+1} \dashrightarrow \mathbb{P}^n$: it's a well known construction: Let $p \in \mathbb{P}^{n+1}$ and $\mathbb{P}^n \cong H \subset \mathbb{P}^{n+1}$ a hyperplane not containing $p$. This induces the projection $\pi_{p,H}: \mathbb{P}^n \dashrightarrow \mathbb{P}^{n-1}$. It's only a rational map, but a sophisticated choise of $p$ and hyperplane allows that the restriction of $\pi_{p,H}$ to $X$ becomes a morphism.
Ideas: In case of $X$ a variety I think I have a solution. The problem reduces obviously to following claim: Let $i_{n+1}:X \subset \mathbb{P}^{n+1}_k$ defined (compare with my remarks above) by $n+1$ sections /pullbacks of $ X_i \in H^0(\mathbb{P}^{n+1}, \mathcal{O}_{\mathbb{P}^{n+1}}(1))$: $s_i = f^*X_i$.
Claim: if the $s_i$ are not linear independent, then $X$ lies in a hyperplane $H \cong \mathbb{P}^n$.
For varieties it’s fine : a map from $X$ to projective space $\mathbb{P}^{n+1}$ is defined by $x \mapsto [s_0(x) : s_1(x) :... :s_n(x)]$ and if the $s_i$ are not linear independent we obviously can factorize it through a hypersurface. e.g. if $s_n= \sum_{i=0}^{n-1}a_i s_i$ then obviously the map factorize through the map $x \mapsto [s_0(x) : s_1(x) :... :s_{n-1}(x)]$.
Now, if we think about the second part of my question we can ask if the Claim holds also for $k$-schemes instead of $k$-varieties, we observe that although maps to $\mathbb{P}^{n+1}$ are still uniquely determined by pullbacks $s_i$, we can’t the morphism $X \to \mathbb{P}^{n+1}$ simply "write down" as $x \mapsto [s_0(x) : s_1(x) :... :s_n(x)]$, that's only a suggestive notation, that is the linear algebra argument as above might be too fishy.
Is the Claim for $k$-schemes nevertheless true and which formal argument can be used to reason it?