I am currently trying to understand the construction in the Stacks project of the Segre embedding: https://stacks.math.columbia.edu/tag/01WD.
They use the correspondence between invertible sheaves and morphisms to $\mathbf{P}^n$, but very quickly say the following:
"The induced morphism $\phi$ has the property that .."
Also, in the comments it is noted that there is some error.
Could someone clarify the last part of the proof given there ? Especially, how do they so quickly conclude that $\phi^{-1}(D_{+}(Z_{i+(n+1)j})) = D_+(X_i)\times D_+ (Y_j)$?
Any help is much appreciated.
Question/commenty: "Any help is much appreciated."
Answer: Let $k$ be a field, the ring of integers or any commutative unital ring. An "elementary" approach using an open cover may be done as follows: Let $V:=k\{e_0,e_1,e_2,e_3\}$ and $V^*:=k\{z_0,z_1,z_2,z_3\}$ with $z_i:=e_i^*$ the dual basis.
Let us for simplicity study the case of a product of two projective lines
$$t: \mathbb{P}^1 \times_k \mathbb{P}^1:=S \rightarrow \mathbb{P}^3\cong \mathbb{P}(V^*):=Y$$
"defined" by the following formula:
$$ t(a_0:a_1)(b_0:b_1):=(a_0b_0: a_0b_1:a_1b_0:a_1b_1).$$
Let $X:=Im(t) \subseteq Y$. You may define this map locally using the open cover $D(x_i)\times D(y_j)$ and $D(z_i)$ and maps of coordinate rings, where $S$ has local coordinates $x_0,x_1,y_0,y_1$ and $X$ has local coordinates $z_0,z_1,z_2,z_3$. At the level of graded rings you may consult Hartshorne Ex.II.5.11.
At $D(z_0)\cap X:=X_0$ you may define an inverse map "coordinatewise" as follows:
$$s_0: X_0 \rightarrow D(x_0)\times D(y_0)$$
by
$$s_0(1,u,v,w):=(1, v) \times (1,u).$$
At $D(z_1)\cap X:=X_1$ you define
$$s_1(u,1,v,w):= (1,w)\times (u,1) \in D(x_0)\times D(y_1) \subseteq S.$$
At $D(z_2)\cap X:=X_2$ you get
$$s_2(u,v,1,w):=(u,1)\times (1,w) \in D(x_1)\times D(y_0)$$
and at
$D(z_3)\cap X:=X_3$ you get
$$s_3(u,v,w,1):=(v,1)\times (w,1)\in D(x_1)\times D(y_1).$$
Let $X':=Z(z_0z_3-z_1z_2) \subseteq Y$. You may check that the above maps glue to define an inverse map
$$s: X' \rightarrow S$$
with the property that $ s \circ t = t \circ s =Id$. Hence $X:=Im(t) \cong Z(z_0z_3-z_1z_2)$ is the hypersurface defined by the polynomial $f:=z_0z_3-z_1z_2 \in k[z_0,z_1,z_2,z_3]$. Hence the map $t$ is a "closed immersion"
$$t: \mathbb{P}^1 \times_k \mathbb{P}^1 \rightarrow Z(f) \subseteq \mathbb{P}^3.$$
When you write the above out in terms of localizations $k[x_i/x_j, y_k/y_l]$ and $k[z_i/z_j]$ the result holds over any commutative unital ring $k$.
Example: The map $s_0$. The map $s_0$ corresponds to the map
$$S_0: k[\frac{x_1}{x_0},\frac{y_1}{y_0}] \rightarrow k[\frac{z_1}{z_0},\frac{z_2}{z_0},\frac{z_3}{z_0}]/(\frac{z_3}{z_0}-\frac{z_1}{z_0}\frac{z_2}{z_0}) $$
defined by
$$S_0(\frac{x_1}{x_0}):= \frac{z_2}{z_0}, S_0(\frac{y_1}{y_0}):=\frac{z_1}{z_0}.$$
This gives a well defined map $X'_0 \rightarrow D(x_0)\times D(y_0)$ where $X'_0:=X' \cap D(z_0)$. You must verify that there are maps $S_i$ defined on $X'_i$ glueing to give a well defined map $s: X' \rightarrow S$. You may check that $Im(S) \subseteq X'$ and that you get well defined maps $s,t$ as claimed above and that $s,t$ are inverses of each other.
Example: The stacks project approach uses invertible sheaves: Let $V^*:=k\{z_0,z_1,z_2,z_3\}$ and let $\pi: S \rightarrow Spec(k)$ be the structure map. There is a 1-1 correspondence between surjections
$$\pi^*(V^*) \rightarrow L \rightarrow 0$$
with $L \in Pic(S)$ and maps $\phi_L: S \rightarrow \mathbb{P}(V^*)\cong \mathbb{P}^3_k$. Hence instead of constructing the Segre embedding explicitly they construct a line bundle quotient $L\in Pic(S)$ explicitly. The above approach is more elementary. The approach using invertible sheaves is an application of the following result:
Proposition 1: Let $g:Y \rightarrow X$ be a map of schemes and let $E$ be a finite rank locally trivial $\mathcal{O}_X$-module with structure map $\pi: \mathbb{P}(E^*) \rightarrow X$. There is a 1-1 correspondence between quotients $g^*(E^*) \rightarrow L$ with $L\in Pic(Y)$ and maps $\phi_L: Y \rightarrow \mathbb{P}(E^*)$ with $\pi \circ \phi_L =g$.
This proposition has the following consequence: define the functor
$$F: Sch(X) \rightarrow Sets$$
where $Sch(X)$ is the category of schemes over $X$ and morphisms over $X$, and $Sets$ is the category of sets. Let
$$F(g: Y \rightarrow X):=\{\phi_L:g^*(E^*) \rightarrow L: L\in Pic(Y)\}/\cong$$
where $(L, \phi_L) \cong (L', \phi_{L'})$ iff there is an isomorphism $L \cong L'$ making the obvious diagram commute. It follows from Proposition 1 that $F$ is a representable functor and that there is an isomorphism
$$F(-) \cong h_{\mathbb{P}(E^*)}(-)$$
of functors. Hence the projective space bundle $\mathbb{P}(E^*)$ represents $F$.
You may view $S\cong \mathbb{P}(\mathcal{O}\oplus \mathcal{O})$ hence $Pic(S) \cong \mathbb{Z} \oplus \mathbb{Z}$. Hence any invertible sheaf $L\in Pic(L)$ is on the form $p^*\mathcal{O}(m)\otimes q^*\mathcal{O}(n)$ with $p,q$ the projection maps. It seems there is a quotient map
$$\phi_t:\pi^*(V^*) \rightarrow L(1,1):=p^*(\mathcal{O}(1))\otimes q^*(\mathcal{O}(1)) \rightarrow 0$$
corresponding to the Segre embedding $t$, and that this is the map constructed on the Stacks homepage. You must check that this map agrees with the map defined above. Let $C:=\mathbb{P}^1_k$. By the Kunneth formula you get an isomorphism
$$H^0(S, p^*\mathcal{O}(1)\otimes q^*\mathcal{O}(1)) \cong H^0(C, \mathcal{O}(1)) \otimes_k H^0(C, \mathcal{O}(1)) \cong$$
$$k\{x_0\otimes y_0, x_0\otimes y_1, x_1\otimes y_0, x_1\otimes y_1\}.$$
The map $\phi_t$ is the following map:
$$\phi_t(z_0):=x_0\otimes y_0, \phi_t(z_1):=x_0\otimes y_1, \phi_t(z_2):=x_1\otimes y_0, \phi_t(z_3):=x_1\otimes y_1.$$
Let $s_i:=\phi_t(z_i)$. As an example, using Prop. HH.II.7.2 you get maps
$$(\phi_t)_j: k[z_i/z_j] \rightarrow H^0(S_{s_j}, \mathcal{O}_{S_{s_j}})$$
defined by
$$(\phi_t)_j(z_i/z_j):=s_i/s_j.$$
Here you must give a definition of the element $s_i/s_j\in H^0(S_{s_j}, \mathcal{O}_{S_{s_j}})$. The sections $s_i,s_j \in H^0(S, L(1,1))$ but you can construct a well defined "quotient"
$$ s_i/s_j \in H^0(S_{s_j},\mathcal{O}_{S_{s_j}}).$$
You must prove that $(\phi_t)_j$ is surjective for $j=0,..,3$ and use II.7.2. Here you must prove that the element $s_i/s_j$ exists and that the maps $(\phi_t)_j$ agree on overlaps and glue to give the Segre morphism
$$t: \mathbb{P}^1 \times_k \mathbb{P}^1 \rightarrow \mathbb{P}(V^*)$$
with the property that $t^*(\mathcal{O}(1))=L(1,1)$.
Note that to give a surjective map $\phi_t: \pi^*(V^*) \rightarrow L$ for some invertible sheaf $L$ is equivalent to specifying 4 global sections of $L$ generating $L$ locally.
It is an instructive exercise to construct the Segre embedding using the above elementary approach with an open cover AND Proposition 1. Then you understand the fact that projective space (or a projective bundle) is a parameter space - it represents a certain functor. The proof of Proposition 1 may be found in Hartshorne Prop.II.7.12. Proposition HH.II.7.2 proves the map $\phi_t$ gives rise to a closed embedding.
The approach above generalize to arbitrary Segre embeddings: There is for any $T$ a closed embedding
$$t_{n,m}:\mathbb{P}^n_T \times_T \mathbb{P}^m_T \rightarrow \mathbb{P}^{nm+n+m}_T$$
There is a similar approach to the "Veronese embedding":
An explicit construction of the "Veronese embedding".