Tensor product of rings: well-definedness of multiplication

Question

Let $R$ and $S$ be rings, and let $T$ be a ring with ring homomorphisms $\alpha: T \to R$, $\beta: T \to S$. (None of these rings are assumed to be commutative.) Since $R$ and $S$ contain subrings that are isomorphic to quotients of $T$, we may view them as $T$-bimodules. Let $\varphi : R \times S \to R \otimes_T S$ be the $T$-balanced map $\varphi(r, s) = r\otimes s$.

We want to turn the tensor product into a ring. The ring multiplication behaves exactly how you would expect on elementary tensors: $(r\otimes s) \cdot (r^\prime\otimes s^\prime) = (rr^\prime)\otimes(ss^\prime)$. We then extend it to sums of elementary tensors by distribution.

Unfortunately, I can't seem to convince myself that this is well defined. My first idea was to invoke the universal property of the tensor product to show that if $r^\prime \otimes s^\prime = r^{\prime\prime}\otimes s^{\prime\prime}$, then $(rr^\prime)\otimes (ss^\prime) = (rr^{\prime\prime}) \otimes (ss^{\prime\prime})$. To this end I defined $f: R \times S \to R \otimes_T S$ by $f(x, y) = (rx) \otimes (sy)$, where $r \in R$ and $s \in S$ are fixed. However, this does not seem to be $T$-balanced in general, so the universal property is of little help.

Another option is to instead define $f(x, y) = (rx) \otimes (ys)$. This map is $T$-balanced, but the induced map $R \otimes_T S \to R \otimes_T S$ only seems to show that $x \otimes y = x^\prime \otimes y^\prime \implies (rx)\otimes (ys) = (rx^\prime) \otimes (y^\prime s)$, which isn't quite what we want.

How can I prove that the multiplication is well defined (for elementary tensors and for arbitrary tensors)? Do I need to make any additional assumptions? (I'm willing to assume that $R$, $S$, and $T$ are finite-dimensional algebras over a field $k$, and that $\alpha$ and $\beta$ are $k$-linear. If absolutely necessary I might also be willing to assume that $T$ is commutative. However, I would prefer not to assume that $R$ and $S$ are $T$-algebras, i.e. I don't want to assume that $\operatorname{Im}\alpha \subseteq Z(R)$ or $\operatorname{Im}\beta \subseteq Z(S)$.)

Answer 1

This doesn't work, and I wouldn't expect there to be any assumptions that make it work that are signficantly weaker than assuming $R$ and $S$ are $T$-algebras.

In particular, let $r\in R$ and $t\in T$ and consider the following calculation, which is a version of the Eckmann-Hilton argument: $$r\alpha(t)\otimes 1=r\otimes\beta(t)=(1\otimes\beta(t))(r\otimes 1)=(\alpha(t)\otimes 1)(r\otimes 1)=\alpha(t)r\otimes 1.$$ Each step here either uses the fact that $\otimes$ is balanced or your definition of multiplication. So, this shows that if your multiplication is to be well-defined, then all the commutators of the image of $T$ with elements of $R$ must be killed when you form the tensor product $R\otimes_T S$. This is certainly not true in general; for instance, if $S=T$ and $\beta$ is the identity so that $R\otimes_T S\cong R$. Similarly, all the commutators of the image of $T$ with elements of $S$ must also die in the tensor product. That is, that even if $R$ and $S$ were not $T$-algebras, $R\otimes_T S$ has to be a $T$-algebra (i.e., the image of $T$ in it must be central) in order for your multiplication to be well-defined. Another way to think about this is that if you want to just formally declare that your multiplication is well-defined, this amounts to imposing relations on $R$ and $S$ that turn them into $T$-algebras (i.e., modding out all the commutators with elements in the image of $T$) so that you are just taking the usual tensor product of $T$-algebras.