Let $A,B$ be $k$-algebras with $k$ a commutative ring. Suppose that the free product $A \cdot_k B$ is a free associative $k$-algebra of rank $n$. Does it imply that $A,B$ are both free associative $k$-algebras?

Working this sort of "factorization" problems in several different settings, I realized that there is no mention in the literature (which I have access) in this case and I am not able to prove it.

EDIT: The question is false if is considered the usual tensor product.

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    $\begingroup$ I would be surprised if a free associative algebra can be written as a tensor product in any nontrivial way at all. Also, to be clear: Are $A$ and $B$ supposed to be $k$-algebras (and $A\otimes_k B$ a free associative $k$-algebra) for some commutative ring $k$? $\endgroup$ – Eric Wofsey Dec 9 '15 at 3:59
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    $\begingroup$ Your question is phrased as if the converse is true, but it's not: the coproduct in associative algebras is the free product, not the tensor product, and if $A, B$ are free then $A \otimes_k B$ will almost never be free. $\endgroup$ – Qiaochu Yuan Dec 9 '15 at 5:19
  • $\begingroup$ @eric wofsey: Yes. I will edit my question. $\endgroup$ – Binai Dec 9 '15 at 20:58
  • $\begingroup$ @Qiaochu Yuan: I am OK with the fact that the converse is not true. My questions are phrased the way I can write them in a direct form. It is not based on the validity of the converse. If you want, you can rewrite it. $\endgroup$ – Binai Dec 9 '15 at 21:04

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