Expressing $\mathbb{P}_A^n$ as $n+1$ glued copies of $\mathbb{A}_A^n$ -- what is the inclusion map? I wish to define an inclusion morphism $\varphi_i:\mathbb{A}_A^n\to\mathbb{P}_A^n$, where $\mathbb{A}$ and $\mathbb{P}$ are defined as schemes, and $A$ is some commutative ring. I'm struggling to define the mapping of topological spaces. Here are my thoughts so far:
I define $A_i:=A[x_0,\ldots,\hat{x_i},\ldots,x_n]$ for convenience, and set $U_i:=\mathrm{Spec}\ A_i=\mathbb{A}_A^n$. I also set $S=A[x_0,\ldots,x_n]$ as a graded ring, defining $S_+$ and $S_d$ as usual.
For any prime ideal $\mathfrak{p}\subset A_i$, I have to define $\varphi_i(\mathfrak{p})$ as some homogeneous prime ideal of $S$ not containing$S_+$. I have a few thoughts about possible ways to do this, but I'm not sure which one is the correct one, and I don't see any obvious way to check.
Firstly, for $\mathfrak{p}=\langle f_1(x_0,\ldots,\hat{x_i},\ldots,x_n),\ldots, f_m(x_0,\ldots,\hat{x_i},\ldots,x_n)\rangle$, I might define $$\varphi_i(\mathfrak{p})=\left\langle x_i^{e_1}f_1(\frac{x_0}{x_i},\ldots,\hat{\frac{x_i}{x_i}},\ldots,\frac{x_n}{x_i}),\ldots, x_i^{e_m} f_m(\frac{x_0}{x_i},\ldots,\hat{\frac{x_i}{x_i}},\ldots,\frac{x_n}{x_i})\right\rangle$$ where $e_j$ is minimal such that $$x_i^{e_j} f_j(\frac{x_0}{x_i},\ldots,\hat{\frac{x_i}{x_i}},\ldots,\frac{x_n}{x_i})\in S.$$
Alternatively, I might define $$\varphi_i(\mathfrak{p})=\{x_i^a f(\frac{x_0}{x_i},\ldots,\hat{\frac{x_i}{x_i}},\ldots,\frac{x_n}{x_i}) \bigg|\  f\in\mathfrak{p},a\in\mathbb{N}\}\cap S.$$
Unfortunately, I'm having a hard time showing that either of these definitions give a prime ideal not containing $S_+$, and furthermore, both of them are incredibly clunky to work with. Is there any better way of going about this?
 A: Denoting homogeneous spectrum by $\text{Proj}$, we want to describe the embedding$$\mathbb{A}^n = \text{Spec}\,A_i \to \text{Proj}\,S = \mathbb{P}^n$$sending prime ideals of $A_i$ to homogeneous prime ideals of $S$. Consider the homomorphism $g: S \to A_i$ which simply substitutes $1$ for $x_i$. This induces a morphism$$\mathbb{A}^n = \text{Spec}\,A_i \to \text{Spec}\,S = \mathbb{A}^{n + 1},$$which sends a prime ideal $I$ to the (prime) ideal $g^{-1}(I)$. This is not quite (yet) what we want. Compose this morphism with the map $\text{Spec}\,S \to \text{Proj}\,S$ which sends a prime ideal $J$ to the (prime) ideal generated by all the homogeneous elements of $J$. Then the composite$$g^*: \text{Spec}\,A_i \to \text{Proj}\,S$$sends each $I$ to$$I^* := \text{ ideal generated by all the homogenizations }F = x_i^{\deg f}f(x_0, \ldots, \hat{x}_i, \ldots, x_n) \text{ for each }f \text{ in }I.$$(This is essentially your second definition.) If some $I^*$ were to contain $S_+$, then since $I^*$ is contained in $g^{-1}(I)$, the latter would contain $S_+$, and then $I = g(g^{-1}(I))$ would contain $g(S_+)$. But $S_+$ contains $x_i$, and so $g(S_+)$—hence $I$—contains $1$. Then $I = A_i$, a contradiction.
