Let $R$ is a commutative ring with unit. Consider two $R$-algebra $A_1$ and $A_2$, namely, two unit commutative rings with morphisms $\phi_1: R\to A_1$ and $\phi_2:R\to A_2$. Consider any a prime ideal $\mathfrak p\in \mathrm{Spec}(A_1\otimes_R A_2)$. Let $\mathfrak p_1$ is the set that consists of all elements $a$ such that $a\otimes 1\in \mathfrak p$, and $\mathfrak p_2$ is the set consists of all elements $b$ such that $1\otimes b\in \mathfrak p$. It is easy to prove $\mathfrak p_1\in \mathrm{Spec}A_1$ and $\mathfrak p_2\in \mathrm{Spec} A_2$. Then, my question is, does exist two prime ideal $\mathfrak p,\overline{\mathfrak p}$, will give the same $\mathfrak p_1$ and $\mathfrak p_2$? How to prove? Thanks!
OK, in fact, my question is to find a non-standrad but more direct method to prove $\mathrm{Spec}\,A_1\times_{\mathrm{Spec}\,R}\mathrm{Spec}\,A_2\simeq \mathrm{Spec}\,\big(A_1\otimes_R A_2\big)$.
The underlying set of $\mathrm{Spec}(A_1)\times_{\mathrm{Spec}(R)} \mathrm{Spec}(A_2)$ is not the cartesian product of $\mathrm{Spec}(A_1)$ and $\mathrm{Spec}(A_2)$: think for example about the case $R=k$, $A_1=A_2=k[x]$. For every irreducible affine plane curve, there's a point in $\mathrm{Spec}(A_1) \times_{\mathrm{Spec}(R)}\mathrm{Spec}(A_2)\cong \Bbb A^2_k$, but such points don't correspond to a pair of points in $|\Bbb A^1_k| \times |\Bbb A^1_k|$. But it seems that this is exactly what you're trying to prove.
Edit: For simplicity let's write $A_1=k[x]$ and $A_2=k[y]$. Consider for example the point $(x-y) \in \mathrm{Spec}(k[x,y])$, geometrically this corresponds to the diagonal in the affine plane. When we consider the intersections $(x-y) \cap k[x]$ and $(x-y) \cap k[y]$, we get the zero ideal both times. Geometrically, if you project the diagonal on the $x$-axis or the $y$-axis you get the whole axis. But if we consider instead the zero ideal $(0) \in \mathrm{Spec}(k[x,y])$ (geometrically, the whole affine plane), then you get the same ideals in $k[x]$ and $k[y]$.