Several questions actually.

All rings and algebras are supposed to be commutative and with $1$ here.

*(1)* Let $k$ be a field, and let $A$ and $B$ be two $k$-algebras. I need a proof that if $A$ and $B$ are reduced (i. e., the only nilpotents are $0$) and $\operatorname{char}k=0$, then $A\otimes_k B$ is reduced as well.

The condition $\operatorname{char}k=0$ can be replaced by "$k$ is perfect", but I already know a proof for the $\operatorname{char}k>0$ case (the main idea is that every nilpotent $x$ satisfies $x^{p^n}=0$ for some $n$, where $p=\operatorname{char}k$), so I am only interested in the $\operatorname{char}k=0$ case.

Please don't use too much algebraic geometry - what I am looking for is a constructive proof, and while most ZFC proofs can be made constructive using Coquand's dynamic techniques, the more complicated and geometric the proof, the more work this will mean.

BTW the reason why I am so sure the above holds is that some algebraist I have spoken with has told me that he has a proof using minimal prime ideals, but I haven't ever seen him afterwards.

Ah, and I know that this is proven in J. S. Milne's *Algebraic Geometry* (version 6.01, Proposition 5.17 (a)) for the case $k$ algebraically closed.

*(2)* What if $k$ is not a field anymore, but a ring with certain properties? $\mathbb{Z}$, for instance? Can we still say something? (Probably only to be thought about once (1) is solved.)

*(3)* Now assume that $k$ is algebraically closed. Can we replace reduced by connected (which means that the only idempotents are $0$ and $1$, or, equivalently, that the spectre of the ring is connected)? In fact, this even seems easier due to the geometric definition of connectedness, but I don't know the relation between $\operatorname{Spec}\left(A\otimes_k B\right)$ and $\operatorname{Spec}A$ and $\operatorname{Spec}B$. (I know that $\operatorname{Spm}\left(A\otimes_k B\right)=\operatorname{Spm}A\times\operatorname{Spm}B$ however, but this doesn't help me.)

PS. All algebras are finitely generated if necessary.