Proof that a Scheme is connected My question refers to a step in follwing thread: How do I prove a scheme is connected?
Here there is introduced a proof of following lemma:
Suppose $k$ is algebraically closed, and $K$ contains $k$, $A$ is a $k$ algebra, then the idempotents of $A\otimes_kK$ come from $A$.
In the proof he concludes that since $k$ is algebraically closed, $A/\mathfrak{p}_i\otimes_k K$ is a (integral) domain. 
Why does it holds? Especially why do we need that $k$ is algebraically closed to show that $A/\mathfrak{p}_i\otimes_k K$ is a domain?
 A: Claim: If $A,B$ are $k-$algebras $A,B$ are integral domains and $k$ is algebraically closed, then $A\otimes_k B$ is an integral domain.
Proof: Suppose $(\sum_i a_i\otimes b_i)(\sum_j a'_j \otimes b'_j) = 0 \in A\otimes_k B$, where we may choose $b_i,b_j$ such that $(b_i)$ and $(b'_j)$ form a $k-$linearly independent set. Replacing $A$ with the $k-$algebra generated by $a_i, a'_j$, we may assume that $A$ is finitely generated. Now for every maximal ideal $\mathfrak{m}$ of $A$, we have that $(\sum_i \bar{a_i}\otimes b_i)(\sum_j \bar{a'_j} \otimes b'_j) = 0 \in A/\mathfrak{m} \otimes_k B$. Since $k$ is algebraically closed we know that $A/\mathfrak{m} \simeq A$ and in particular we have WLOG that $\sum_i \bar{a_i}\otimes b_i = \sum_i \bar{a_i} b_i = 0 \in B$. Since $(b_i)$ is $k-$linearly independent we deduce that $\bar{a_i} = 0$ for all $i$. 
So we see that for every maximal ideal $\mathfrak{m}$, either $a_i \in \mathfrak{m}$ for all $i$ or $a'_j \in \mathfrak{m}$ for all $j$. This implies that every maximal ideal is either contained in $V(a_i)$ or $V(a'_j)$. Since $A$ is a finitely generated $k-$algebra, its closed points are dense. Since $A$ is an integral domain (in particular $\mathrm{spec}(A)$ is irreducible), this implies WLOG that $A = V(a_i)$, i.e. $a_i \in \bigcap_{\mathfrak{m}} \mathfrak{m}$. Since $A$ is Jacobson and reduced it follows that $a_i = 0$ for all $i$. 
