Topologizing the Matrix Algebra over a Group Ring Suppose that $G$ is some group whose complex group ring is denoted as $\Bbb{C}G$. This group ring is endowed with a specific and technically defined topology on it, so I don't want to get into the details of it. Indeed, I am still in the process of studying it; e.g., the only thing I know about it as of now is that it is Hausdorff. 
I want to construct a topology on $M_k(\Bbb{C}G)$ and use the topology on $\Bbb{C}G$ in this construction.  I know that it is isomorphic to $M_k(\Bbb{C}) \otimes \Bbb{C}G$, so I was thinking that, perhaps, this could be done by topologizing the tensor product with the topology from each factor contributing to it, and then transfer the topology onto $M_k(\Bbb{C}G)$. The one factor $\Bbb{C}G$ will have on it the complicated topology I've already alluded to, while $M_k(\Bbb{C})$ has any norm topology (they're all equivalent, since it is a finite dimensional space). Basically, my question is, is there a natural way of topologizing the tensor product when each factor has its own topology? 
 A: The way I would call "natural" here is to endow $M_k(\mathbb CG)$ with entrywise convergence: $A_n\to A$ if and only if $A_n(k,j)\to A(k,j)$ for all indices $k,j$. 
The above topology satisfies the obvious requirement that the natural embeddings of $\mathbb CG$ as a subalgebra of $M_k(\mathbb CG)$ (either as matrices with a single nonzero diagonal entry, or as constant diagonals) preserve the original topology on $\mathbb CG$. 
A: I would say that $R \otimes_{\mathbb C} R'$ is a quotient of the free group $\mathbb{C}(R \times R') $, so it inherits a natural topology (it is a union of a lot of $\mathbb{C} ^n$`s!). One can verifies that this is still a topological ring.
Essentially, things "close to" $\sum  x_i \otimes y_j $ are elements like $\sum x_i' \otimes y_j' $ with each $x_i', y_j'$ respectively close to $x_i, y_j$. But don't forget that an open must contain the whole class of equivalence of $\sum x_i \otimes y_j$! For further details ask me in comments :)
Your case is simpler: $M_n(R) \simeq R^{n^2}$ as abelian groups, so it is endowed with the product topology. This turns ths ring into a topological ring, because operations are polynomial in the entries, thus continous. 
