Let $A_1$ and $A_2$ be finitely generated $k$-algebras, where $k$ is a field.
If we write $A_1 = k[X_1, \dots, X_m]/I$ and $A_2 = k[Y_1, \dots, Y_n]/J$ where $ I \subset k[X_1, \dots ,X_m]$ and $ J \subset k[Y_1, \dots ,Y_n]$ are ideals, and $m,n \in \mathbb N,$ we know that $A_1 \otimes_k A_2 \cong_k R $ where:
$$R = k[X_1,\dots,X_m,Y_1, \dots,Y_n ]/(I,J)$$
And we did that by defining a biadditive balanced function $\varphi: A_1 \times A_2 \rightarrow R$ to get an algebra homomorphism $A_1 \otimes_k A_2 \rightarrow R.$
Now, I want to give an example where $A_1$ and $A_2$ are integral domains but $A_1 \otimes_k A_2$ is not an integral domain.
I know that if $k$ is algebraically closed, then $A_1 \otimes_k A_2$ is always an integral domain. So I need to use $k$ which is not algebraically closed.
Could someone give me a help to create this example?
Let $A_1=A_2=\mathbb C$ and $k=\mathbb R$. Then, we have the isomorphism \begin{align} \mathbb C\otimes_\mathbb R\mathbb C&=\mathbb C\otimes_\mathbb R(\mathbb R[X]/(X^2+1))\cong\mathbb C[X]/(X^2+1)\\ &\cong\mathbb C[X]/(X-i)\times\mathbb C[X]/(X+i)\cong\mathbb C\times\mathbb C, \end{align} by the Chinese Remainder Theorem.
This is clearly not an integral domain, since $(0,1)\cdot(1,0)=(0,0)\in\mathbb C\times\mathbb C$.
In fact, the same construction works in general. Let $k$ be a non-algebraically closed field, so there is a monic irreducible polynomial $f(X)\in k[X]$ of degree $>1$. Let $K=k[X]/(f)$ be the splitting field of $f$, so that $K/k$ is a finite extension. Let $f(X)=\prod_{i=1}^n(X-\alpha_i)$ for $\alpha_i\in K$.
We then have the isomorphism $K\otimes_kK=k[X]/(f)\otimes_kK\cong K[X]/(f)$. Here, $X-\alpha_1,\prod_{i=2}^n(X-\alpha_2)\in K[X]/(f)$ are nonzero, but their product is $f=0$. Hence, $K\otimes_kK$ is not an integral domain.
EDIT (the explicit map):
$\mathbb C\otimes_\mathbb R\mathbb C\cong\mathbb C\times\mathbb C:u\otimes v\mapsto(uv,u\overline v)$, where $\overline v$ is the complex conjugate of $v\in\mathbb C$.