Topological Tensor Product is a Topological Ring Independent of the Choice of Basis Let $A, B$ be commutative rings containing a field $k$, with $B$ a finite dimensional $k$-module, $w_1, ... , w_N$ a basis.  If $w_iw_j = \sum\limits_{n=1}^N c_{ijn}w_n$, then we can define $C$ to be a free $A$-module on the objects $\bar{w_1}, ... , \bar{w_N}$, and make $C$ into a ring by defining $\bar{w_i} \bar{w_j}$ to be $\sum\limits_{n=1}^N c_{ijn} \bar{w_n}$.  So $C$ is an algebra over $k$, and in fact $C$ is a tensor product of $A$ and $B$ over $k$.  By the universal mapping property $C$ is defined up to a $k$-algebra isomorphism, independent of the choice of basis for $B/k$.  
If $A$ has some topology with respect to which addition and multiplication are continuous, then $C$ can be given the product topology by the correspondence $a_1 \bar{w_1} + \cdots + a_N\bar{w_N} \mapsto (a_1, ... , a_N) \in \prod\limits_{i=1}^n A$.  
This is all described in Cassels' section of Algebraic Number Theory.  He also claims that:
(i) This topology is independent of the choice of basis $w_1, ... , w_N$.
(ii) Addition is continuous in $C$ with respect to this topology.
(iii) Multiplication is also continuous.
I am currently not very familiar with topological hand-waving.  The only assertion that seems remotely obvious is (ii); since $A$ is continuous with respect to addition, so is $A \times \cdots \times A$ with respect to the product topology, hence so is $C$.  Can anyone explain to me (without too much hand-waving) why (i) and (iii) are also true?  
 A: Let me explain this topology in my own words.
Let $A,B$ be commutative $k$-algebras. Assume that $N:=\dim_k(B)$ is finite, so that there is an isomorphism of vector spaces $B \cong k^N$. Hence, there is an isomorphism of vector spaces $A \otimes_k B \cong A^N$. If $A$ is a topological vector space, it follows that also $A^n$ and thereby $A \otimes_k B$ is a topological vector space. We want to show that the topology doesn't depend on the chosen isomorphism $B \cong k^N$.
Any other isomorphism differs by an automorphism of the vector space $k^N$. Hence, it suffices to prove the following: If $k^N \to k^N$ is an automorphism, then the induced automorphism $A^N \to A^N$ of vector spaces is a homeomorphism. By symmetry, it suffices to show that it is continuous. More generally, we show that if $k^N \to k^M$ is any linear map, then the induced linear map $A^N \to A^M$ is continuous. We may assume that $M$ has only one element. Thus, we have to prove that for elements $\lambda_1,\dotsc,\lambda_n \in k$ the map $A^N \to A$, $(a_i) \mapsto \sum_i \lambda_i a_i$ is continuous. This is a sum of continuous maps of the form $A^N \to A \to A$, where $A^N \to A$ is some projection and $A \to A$ is given by $a \mapsto \lambda a$ for some $\lambda \in k$. This is continuous because $A$ is a topological vector space.
Now let $A$ be a topological commutative $k$-algebra. Then $A \otimes_k B$ is a commutative topological $k$-algebra: All we are left to prove is that the multiplication is continuous. Choose a basis $w_1,\dotsc,w_N$ of $B$ and write $w_i w_j = \sum_{n=1}^{N} c_{ijn} w_n$ with $c_{ijn} \in k$. Under the isomorphism of topological vector spaces $A \otimes_k B \cong A^N$, the multiplication of $A \otimes_k B$ becomes the $A$-bilinear map
$$A^N \times A^N \to A^N, ~ e_i \otimes e_j \mapsto \sum_{n} c_{ijn} e_n .$$
More generally, $((a_i)_i,(a'_j)_j)$ is mapped to $(\sum_{i,j} c_{ijn} a_i a'_j )_n$. This is continuous, which can be easily derived from the assumption that $A$ is a topological $k$-algebra. The argument is similar to the one in the second paragraph. Please let me know if you need more details.
