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MacToDo 1.0 – Mac To Do Schedule For Outlook. Free – Mac Apps Addict.Q:

Why does my free product of noncommutative rings have zero intersection?

Let $A$ and $B$ be $\mathbb{Z}$-algebras. If we want to define the product $A \times B$ to be their $\mathbb{Z}$-algebra with the operation $(a,b)(a’,b’) = (aa’,bb’)$, it is common to require that $ab = ba$.
However, I believe it is also natural to ask for each $a \in A$ and $b \in B$, $ab = ba$ holds individually, and by the free product we then require that $a\star b = a\star_A b$, the image of $a$ under the canonical map $A \to A \star_A B$. (Does this last condition suggest a connection to free modules?)
Now, suppose $A$ and $B$ are (noncommutative) $\mathbb{Z}$-algebras satisfying $ab=ba$ for all $a \in A$ and $b \in B$. Does it follow that $A \star_A B = 0$?
Note: This is not a question about “free objects”. The ring of integers is free (and hence no assumptions of commutativity are made), but the free product would be the zero ring, and indeed any zero ring can be identified with a free object. I’m interested in the noncommutative case.

A:

I think the following should be a direct generalization of Richard Stanley’s comment on your previous question. Recall that there is a bijection between the category of $\mathbb{Z}$-algebras and the category of (graded) rings given by
$$ R\mapsto R[X]_{\ge 0}$$
and
$$ R\mapsto R[X]_{\le 0} $$
where $R[X]$ is the free algebra over $
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