reduced ⊗ reduced = reduced; what about connected? 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.
 A: Over $\mathbb Z$, there are several possibilities. 


*

*If both $A, B$ are torsion-free over $\mathbb Z$, then $A\otimes_{\mathbb Z} B$ is connected if and only if its generic fiber $(A\otimes_{\mathbb Z} \mathbb Q)\otimes_{\mathbb Q} (B\otimes_{\mathbb Z} \mathbb Q)$ is connected, and you are reduced to the case of algebras over a field. 


EDIT the above claim is incorrect. See a partial answer here.


*

*If $A$ (or $B$) have positive characteristic, then $A_{\rm red}$ is an algebra over a product of finite fields, then so is $A\otimes_{\mathbb Z} B$. Again you are reduced to the case of algebras over a field (if at least two finite fields involve really, then the tensor product is not connected. 

*The remainning case: if $A$ is noetherian, then ${\rm Spec}(A)$ has a flat part and a ''vertical part'': take the ideal $I$ of $A$ consisting in torsion elements. Then $A/I$ is flat and $mI=0$ for some non-zero integer $m$, and we have ${\rm Spec A}=V(I)\cup V(mA)$.
The same is true for $A\otimes B$. To have the connectedness, you want [EDIT: it is enough  (but far to be necessary) that] the flat and vertical parts to be connected and meet each other. Sorry, I don't have a simple statement.  
A: The answer to (2) is no. For example, for every prime $p$,
$$\mathbb{Z}[\sqrt{p}] \otimes_{\mathbb{Z}} \mathbb{F}_p = \mathbb{Z}[x]/(x^2-p) \otimes_{\mathbb{Z}} \mathbb{F}_p = \mathbb{F}_p[x]/(x^2)$$
is not reduced.
A: Ad (1): It is perfectly true that the tensor product of two reduced algebras over a perfect field is reduced. You can find a proof in Bourbaki's Algebra (Chapter V; §15; 4,5), but of course this venerable author is not especially known for his enthusiasm toward constructive mathematics .
Ad (2): I don't know.
Ad (3): The relation you are looking for is the simplest possible, namely 
$$Spec(A\otimes_k B)=Spec(A)\times_k Spec(B).$$
But that doesn't help because the product of connected schemes has no reason to be connected.
I can only give you the following sufficient condition for connectedness of tensor products of algebras.
Suppose $k \subset K$ is a separable extension of fields such that $k$ is algebraically closed in $K$. Such an extension is called (rather unimaginatively) a REGULAR extension.With this definition, we can state the 
Theorem: If $k \subset K$ is regular, then for every $k$-subalgebra $A$ of $K$ and every $k$-domain $B$ (not related at all to $K$),
the tensor product $A\otimes_k B$ is a domain and in particular has connected spectrum.
Here are examples of regular extensions :
a)Every purely transcendantal extension of $k$ is regular.
b)If $k$ is algebraically closed, every extension of $k$ is regular.
PS: Separable extension above means universally reduced and, of course, does not imply that the extension is algebraic.
A: In (1), A (resp. B) injects into the product of the residue fields at the minimal primes, so we can reduce to the case where A and B are fields which should be sufficiently well-known. Details are in Bourbaki, Algebra, ch.5 §15 no.2
(3) is answered in EGA IV, 4.5: the product of a connected scheme and a geometrically connected scheme is connected.
A: Let A and B finitely generated reduced k-algebras. Then A and B are coordinate rings of two affine closed sets X and Y: A=k[X] and B=k[Y]. It is true that
A ⊗k B=k[X]⊗kk[Y]=k[X×Y]
(See Shafarevich, Basic algebraic geometry, p. 25, example 4), and then A ⊗k B is finitely generated reduced k-algebra too.
