Given a field $k$ of characteristic $0$. For every $k$-bialgebra $A$, let $\mathrm{Prim} A$ denote the $k$-vector subspace of $A$ consisting of all primitive elements of $A$.


What conditions can we put on two $k$-bialgebras $A$ and $B$ to ensure that $\mathrm{Prim}\left(A\otimes B\right) = k\otimes \left(\mathrm{Prim}A\right) + \left(\mathrm{Prim} B\right)\otimes k$ ?


I haven't given this much thought, but I am not good at constructing counterexamples and it seems pointless to try proving anything here before having an "upper bound" on how far we can go. The only results I know about is that $k\otimes \left(\mathrm{Prim}A\right) + \left(\mathrm{Prim} B\right) \otimes k \subseteq \mathrm{Prim}\left(A\otimes B\right)$ always holds (for trivial reasons), and that if $A$ and $B$ are two connected graded cocommutative bialgebras, then $\mathrm{Prim}\left(A\otimes B\right) = k\otimes \left(\mathrm{Prim}A\right) + \left(\mathrm{Prim} B\right)\otimes k$ (as a consequence of Cartier-Milnor-Moore and Poincaré-Birkhoff-Witt).


It sounds rather natural to assume $A$ and $B$ to be cocommutative (after all, $\mathrm{Prim} A$ is always $=\mathrm{Prim}\left(A^c\right)$, where $A^c$ the greatest cocommutative sub-bialgebra of $A$), but I am not sure whether we can WLOG assume this to be so (maybe $\left(A\otimes B\right)^c$ is greater than $A^c\otimes B^c$ ?).



Primitive elements of a tensor product of bialgebras
Well first when we restrict to the case when $A, B$ are filtered (see Bourbaki for example), in this case $\log\_{\*}$ always converges at $Id$ (as $Id=e+I\_+$, $e$ being the unit for the convolution, it suffices to remark that $I\_+^{\*N}(h)=0$ for $N=N(h)$ large enough). 


Now, in the general case, you can adapt the following computation to the algebra $H(A\otimes B)$ generated by the primitive elements of $A\otimes B$ where the series of $\log\_{\*\_{12}}$ always converges.


For clarity, I note $A=A\_1,B=A\_2$ and and $e\_i=1\_{A\_i}\circ \epsilon\_i$.


Then
$$
\log\_{\*\_{12}}(I\_1\otimes I\_2)=\log\_{\*\_{12}}((I\_1\otimes e\_2)\*\_{12}(e\_1\otimes I\_2))=
$$
$$
\log\_{\*\_{12}}(I\_1\otimes e\_2)+\log\_{\*\_{12}}(e\_1\otimes I\_2)
$$  

as the two terms $(I\_1\otimes e\_2), (e\_1\otimes I\_2)$ commute. Now 
$\log\_{\*\_{12}}(I\_1\otimes e\_2)=\log\_{\*\_{1}}(I\_1)\otimes e\_2$ and 
$\log\_{\*\_{12}}(e\_1\otimes I\_2)=e\_1\otimes\log\_{\*\_{2}}(I\_2)$.


Which, in view of $\left(\mathrm{Prim}A\_1\right) \otimes k + k \otimes \left(\mathrm{Prim} A\_2\right) \subseteq \mathrm{Prim}\left(A\_1\otimes A\_2\right)$, proves that $Prim(A\_1\otimes A\_2)=Prim(A\_1)\otimes k+k\otimes Prim(A\_2)$. 


Which does your job. 


**Addition** : *To answer your first question*, $I\_1$ and $e\_2$ are morphisms of bialgebras so $I\_1\otimes e\_2$ maps $Prim(A\_1\otimes A\_2)$ into $Prim(A\_1\otimes A\_2)$ and then $H$ into $H$ (in fact the image of $H(A\_1\otimes A\_2)$ is a subbialgebra of $H(A\_1)\otimes k.1\_{A\_2}$). 


*To answer the second point.* For a bialgebra let us denote $I^+=Id-e$ (the complement projector of $e$) and $H(?)$ the subalgebra generated by the primitive elements. One has, with the morphism of bialgebras
$$
(I\_1\otimes e\_2) : H(A\_1\otimes A\_2) \rightarrow H(A\_1)\otimes k.1\_{A\_2} 
$$ 
the intertwining 
$$
(I\_1^+\otimes e\_2)\circ (I\_1\otimes e\_2)=(I\_1\otimes e\_2)\circ (I\_1\otimes I\_2)^+
$$
so, using series, we get 
$$
(\log\_{\*\_1}(I\_1)\otimes e\_2)\circ (I\_1\otimes e\_2)=(I\_1\otimes e\_2)\circ \log\_{\*\_{12}}(I\_1\otimes I\_2)
$$
This is because, as a general principle, the intertwining intertwines the convolution. Let, 
$$
\begin{matrix} 
A & \stackrel{\varphi}{\longrightarrow} & B \cr
\downarrow && \downarrow \cr
A & \stackrel{\varphi}{\longrightarrow} & B 
\end{matrix}
$$
with $\varphi$ a morphism of bialgebras and the down arrows $f,g$ such that $g\varphi=\varphi f$. Then, if the bialgebras are generated by primitive elements, if $f(1\_A)=0,g(1\_B)=0$ and if $S\in k[[x]]$ is a series, we have $S(g)\varphi=\varphi S(f)$. This is not difficult and argued in details (in particular the notion of summability and substitution) in my paper.


In conclusion, I think that
$$
Prim(A\_1\otimes A\_2)=Prim(A\_1)\otimes k.1\_{A\_2}+k.1\_{A\_1}\otimes Prim(A\_2)
$$ 
is true in full generality. One even does not have to suppose that $k$ is a field, only $\mathbb{Q}\subseteq k$ seems to be needed. 


Do not hesitate to question and comment if something is unclear or wrong. 


Regards


