Let $X\longrightarrow S$ be a morphism of (in my case: Noetherian) schemes. Then $X$ is called projective over $S$ if $X$ is isomorphic to a closed subscheme of $\mathbb{P}^{n}_{S}$.
By a slightly more general definition of projectivity, $X\longrightarrow S$ is called projective if $X$ is a closed subscheme of $\mathbb{P}(E)$ for some vector bundle $E$ over $S$.
The claim is: An example of a morphism that satisfies the more general definition can be constructed as follows: Let $S$ be a Noetherian scheme with non-trivial Picard group. Take a non-trivial line bundle $E$ over $S$ and let $X=\mathbb{P}(E)$.
Question: Although $\mathbb{P}(E)\ncong\mathbb{P}^{1}_{S}$, this does not imply (in my opinion) that $\mathbb{P}(E)$ is not isomorphic to a closed subscheme of $\mathbb{P}^{1}_{S}$.
How do I prove this? Do I need extra conditions, or does Noethericity suffice?
Question: "Although P(E)≆P1S, this does not imply (in my opinion) that P(E) is not isomorphic to a subscheme of P1S. How do I prove this? Do I need extra conditions, or does Noethericity suffice?"
Answer: If $S$ is any scheme and $F:=\mathcal{O}_S\{e_0,..,e_n\}$ is a free rank $n+1$ module on $S$ it follows
$$\mathbb{P}(F^*) \cong \mathbb{P}^n_S$$
is projective space over $S$. If $E$ is a locally trivial (non-trivial) $\mathcal{O}_S$-module of rank $n+1$ you may construct $\pi:\mathbb{P}(E^*) \rightarrow S$. A local trivialization $U_i$ of $E$ gives $E_{U_i} \cong \mathcal{O}_{U_i}^{n+1}$ hence $\pi^{-1}(U_i) \cong \mathbb{P}^n_{U_i}$. Since $E$ is non-trivial and $F$ is trivial, certainly there is no isomorphism
$$\mathbb{P}(E^*) \cong \mathbb{P}(F^*)$$
in general. There could however be a closed immersion
$$i: \mathbb{P}(E^*) \rightarrow \mathbb{P}^N_S$$
for some large $N$. This is Theorem II.7.1 and 7.2 in Hartshorne: If $S:=Spec(A)$, there is a closed immersion $i:X:=\mathbb{P}(E^*) \rightarrow \mathbb{P}^n_S$ iff there is an invertible sheaf $L\in Pic(X)$ satisfying the two conditions $(1),(2)$ in Prop 7.2.
Example: In particular if $Pic(X)=(0)$ there can be no such embedding. For this reason there can be no closed immersion
$$i: \mathbb{A}^n_k \rightarrow \mathbb{P}^n_k$$
since there is no linebundle $L \in Pic(\mathbb{A}^n_k)$ satisfying Proposition II.7.2, (1) and (2).
Example: If $k$ is a field and if $A$ is a finitely generated $k$-algebra it follows $X:=Spec(A[x_0,..,x_n])$ is not projective, ie there is no closed immersion
$$i: X\rightarrow \mathbb{P}^n_A$$
for any $n\geq 1$. If there was such a closed embedding it would follow from HH.Thm.II.5.19 that $H^0(X, \mathcal{O}_X):=A[x_0,..,x_n]$ was a finitely generated $A$-module - a contradiction.