The tensor product of rings is in Hartshorne's book Thm. II.3.3 used to give a general construction of the fiber product of schemes. Given two schemes $X,Y$ over a scheme $S$ we define the fiber product $X\times_S Y$ (if it exists) using universal properties (see HH, Chapter II.3). It is not completely clear that the fiber product exists in general.
Example. If $S:=\mathrm{Spec}(R), X:=\mathrm{Spec}(A), Y:=\mathrm{Spec}(B)$ where $R$ is a commutative unital ring and $A,B$ are commutative unital $R$-algebras it follows
F1. $X\times_S Y\cong \mathrm{Spec}(A\otimes_R B)$.
Using the "universal property" of $\otimes$ we may show that the fiber product in F1 is given by the tensor product of $A$ and $B$. In HH Thm. II.3.3
one proves that the fiber product exists in general.
Question 1. So tensoring is not only thought of as a restriction, but it is also thought of as an extension? Why do we need or when do we use base extensions?
Question 2. Geometrically, what are $R_1:=\mathrm{Spec}(\mathbb{C}[s]⊗_{\mathbb{Z}}\mathbb{C}[u,v]), R_2:=\mathrm{Spec}(\mathbb{C}[s]⊗_{\mathbb{R}}\mathbb{C}[u,v])$ and $R_3:=\mathrm{Spec}(\mathbb{C}[s]⊕\mathbb{C}[u,v])$?
Example. In general if $\pi: X \rightarrow S$ is any morphism and if $i:U \rightarrow S$ is an open subscheme it follows the fiber product $U\times_S X\cong \pi^{-1}(U) $ is the definition of the inverse image scheme of $U$ via the morphism $\pi$. If $j: \mathrm{Spec}(\kappa(s))\rightarrow S$ is a point it follows $\mathrm{Spec}(\kappa(s))\times_S X :=\pi^{-1}(s)$ (by definition) is the fiber of $\pi$ at $s$. With this definition it follows the inverse image $\pi^{-1}(U)$ and the fiber $\pi^{-1}(s)$ have a canonical scheme structure and satifies a universal property.
Example. Base change. Let $k:=\mathbb{R}$ be the field of real numbers and let $K:=\mathbb{C}$ be the field of complex numbers. Let $A:=k[t]/(t^2+1)$
Let $C:=\mathrm{Spec}(A)$ and let $S:=\mathrm{Spec}(k)$ with $\pi: C \rightarrow S$ the canonical morphism induced by the inclusion $k \rightarrow A$. Let $T:=\mathrm{Spec}(K)$ and let $i:=T \rightarrow S$ be the canonical map induced by the field extension $k \subseteq K$. The fiber product $T\times_S C$ is canonically isomorphic to
$C':=\mathrm{Spec}(B)$ where $B:=K[t]/(t^2+1)$ and in $K[t]$ it follows the polynomial
$f(t):=t^2+1$ equals $f(t)=(t-i)(t+i)$ and since the two maximal ideals $(t-i)$ and $(t+i)$ are coprime, it follows there is an isomorphism of rings $B \cong B_1 \oplus B_2$ (here we use the Chinese remainder theorem) where $B_1\cong K[t]/(t-i)\cong K$ and $B_2\cong K[t]/(t+i)\cong K$. It follows
F1. $\mathrm{Spec}(B)\cong C_1 \cup C_2 \cong \mathrm{Spec}(B_1)\cup \mathrm{Spec}(B_2) =\mathrm{Spec}(K) \cup \mathrm{Spec}(K)$.
By $\cup$ I mean disjoint union. Here we have used the fact that the spectrum of a direct sum of rings is the disjoint union of their spectra. Hence the fiber product is a disjoint union of two copies of $\mathrm{Spec}(K)$. The scheme $\mathrm{Spec}(B)$ is reduced ($B$ is a product of fields) and reducible. The original scheme $C$ was integral since $A:=k[t]/(t^2+1)\cong K$ is an integral domain, but when we make a field extension from $k$ to $K$ it becomes reducible.
Example. Finite extensions of number fields. If $K \subseteq L$ is a finite extension of algebraic number fields with rings of integers $\mathcal{O}_K \subseteq \mathcal{O}_L$ and $\mathfrak{q} \subseteq \mathcal{O}_K$ is a maximal ideal, there is always an equality $\mathfrak{q}\mathcal{O}_L =\prod_{i=1}^m \mathfrak{p}_i^{l_i}$ where $\mathfrak{p}_i \subseteq \mathcal{O}_L$ are maximal ideals. Here $\mathfrak{p_i}\neq \mathfrak{p}_j$ for $i\neq j$. It follows there is a direct sum decomposition of rings
$\mathcal{O}_L/\mathfrak{q}\mathcal{O}_L \cong D_1 \oplus \cdots \oplus D_m$
with $D_i \cong \mathcal{O}_L/\mathfrak{p}_i^{l_i}$. Let $C_K:=\mathrm{Spec}(\mathcal{O}_K)$ and $C_L:=\mathrm{Spec}(\mathcal{O}_L)$ with canonical morphism
$\pi: C_L \rightarrow C_K$.
The fiber of $\pi$ at a point $\mathfrak{q}\in C_K$ is by definition given by the ring
$\mathcal{O}_K/\mathfrak{q}\otimes_{\mathcal{O}_K} \mathcal{O}_L \cong D_1\oplus \cdots \oplus D_m$.
It follows the fiber of $\pi$ at $\mathfrak{q}$ is
$ \pi^{-1}(\mathfrak{q})\cong \mathrm{Spec}(D_1) \cup \cdots \cup \mathrm{Spec}(D_m)$.
The rings $D_i$ are Artinian local rings. The ring $D_i$ is non-reduced iff $l_i \geq 2$, hence non-reduced schemes appear in algebraic number theory when we calculate the fiber of the map $\pi$.
Question 1. We need tensor product when we calculate the fiber of a morphism. Sometimes the fiber has non-reduced structure and the tensor product gives a canonical scheme structure to the fiber of a map between affine schemes.
Question 2. The scheme in Q2.3 is the disjoint union $\mathbb{A}^1_{\mathbb{C}} \cup \mathbb{A}^2_{\mathbb{C}}$. The scheme in Q2.2 seems to be $\mathbb{A}^3_U$ where $U:=\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}$. The ring $U$ seems not to be a domain:
Eq1. $(i\otimes 1 -1\otimes i)(i\otimes 1+1\otimes i)=(i)^2\otimes 1 -1\otimes (i)^2=-1\otimes 1+1\otimes 1=0$.
It seems the following is true:
$\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C} \cong \mathbb{C}[t]/(t^2+1)$ and there is an equality $(t^2+1)=(t-i)(t+i)$ in the ring $\mathbb{C}[t]$.
The ideals $I_1:=(t-i)$ and $I_1:=(t+i)$ are coprime hence there is an isomorphism of rings
$\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C} \cong \mathbb{C} \oplus \mathbb{C}$.
Hence $\mathbb{A}^1_{\mathbb{C}} \times_{\mathbb{R}} \mathbb{A}^2_{\mathbb{C}} \cong \mathbb{A}^3_U \cong \mathbb{A}^3_{\mathbb{C}} \cup \mathbb{A}^3_{\mathbb{C}}$ is a disjoint union of two copies of affine 3-space over the complex numbers. Hence the fiber product of two irreducible schemes may no longer be irreducible.
Question 2.1: Let $R:=\mathbb{C}\otimes_{\mathbb{Z}}\mathbb{C}$. It seems there is an isomorphism
$\mathbb{C}[s]\otimes_{\mathbb{Z}} \mathbb{C}[u,v]\cong R[s,u,v]$. Hence
$\mathrm{Spec}(\mathbb{C}[s]\otimes_{\mathbb{Z}} \mathbb{C}[u,v]) \cong \mathbb{A}^3_T$ with $T:=\mathrm{Spec}(R)$. The ring $T$ is not an integral domain since Eq1 holds in $R$ as well.
Here is an other example: Let $S:=Spec(\mathbb{Q}[x])$ and let $k:=\mathbb{Z}$.
It follows $S\times_k S \cong Spec(R[x,y])$ where $R:=\mathbb{Q}\otimes_{\mathbb{Z}} \mathbb{Q}$. What is $R$?
There is an inclusion $\mathbb{Z} \subseteq \mathbb{Q}$ and $\mathbb{Q}$ is a torsion free $\mathbb{Z}$-module: For any element $s\in \mathbb{Q}$ it follows
the canonical map $\phi_s:\mathbb{Z}\rightarrow \mathbb{Q}$ defined by $\phi_s(m):=ms$ is an injection. But $\mathbb{Q}$ does not have a basis as left $\mathbb{Z}$-module and is not finitely generated.
It seems the following holds: For any element $z:=\frac{1}{p}\otimes \frac{1}{q}\in R$ there is the following equality:
$\frac{pq}{pq}z=\frac{1}{pq}(\frac{p}{p}\otimes \frac{q}{q})=\frac{1}{pq}(1\otimes 1)$.
Let $\rho: \mathbb{Q}\rightarrow R$ be defined by $\rho(r):=r\otimes 1$.
We may write any element in $w\in R$ as follows:
$w:=\sum_i \frac{p_i}{q_i}\otimes \frac{a_i}{b_i} =\sum_i \frac{p_ia_i}{q_ib_i}(1\otimes 1):=r(1\otimes 1)$ for a rational number $r$. Hence it seems $\rho$ is an isomorphism giving an isomorphism
$\mathbb{A}^1_{\mathbb{Q}}\times_{\mathbb{Z}} \mathbb{A}^1_{\mathbb{Q}} \cong
\mathbb{A}^2_{\mathbb{Q}}$.
In general for any multiplicative subset $S\subseteq A$, there is an isomorphism $S^{-1}A\otimes_A S^{-1}A\cong S^{-1}A$ of rings.
