TITLE: Show that $\mathbb{Z}[G\times H]\cong \mathbb{Z}[G]\otimes_\mathbb Z \mathbb{Z}[H]$
QUESTION [7 upvotes]: This is exercise 5.5 of Rotman's 'Introduction to Homological Algebra'.
Let $G$ and $H$ be groups. We must show that $$\mathbb{Z}[G\times H]\cong \mathbb{Z}[G]\otimes_\mathbb Z \mathbb{Z}[H]$$ as rings.
This is my first attempt at it.
Define a map $\mathbb{Z}[G]\times \mathbb{Z}[H]\longrightarrow \mathbb{Z}[G\times H]$ given by $(\sum_{x\in G}m_xx,\sum_{y\in H}n_yy)\mapsto \sum_{(x,y)\in G\times H}m_xn_y(x,y)$.
This is $\mathbb{Z}$-bilinear, so it induces a $\mathbb{Z}$-linear map $\phi:\mathbb{Z}[G]\otimes_{\mathbb{Z}}\mathbb{Z}[H]\longrightarrow \mathbb{Z}[G\times H]$ s.t. $x\otimes y \mapsto (x,y)$. Its inverse is given by $\psi:(x,y)\mapsto x\otimes y$ and extended linearly to all of $\mathbb{Z}[G\times H]$.
So this shows that the two sets in question are isomorphic as abelian groups. If we define multiplication in $\mathbb{Z}[G]\otimes_{\mathbb{Z}} \mathbb{Z}[H]$ by $$(\sum_i a_ix_i\otimes y_i)(\sum_j b_jx^{\prime}_j\otimes y^{\prime}_j)=\sum_{i,j}a_ib_jx_ix^{\prime}_j\otimes y_iy^{\prime}_j$$ then $\mathbb{Z}[G]\otimes_\mathbb{Z} \mathbb{Z}[H]$ becomes a $\mathbb{Z}$-algebra and $\phi$ becomes an $\mathbb{Z}$-algebra homomorphism (I think). 
I'd like to know if the above argument is correct.
Is there a more elegant way of showing this?  
Many thanks for your help.

REPLY [2 votes]: A fancy way to prove this is that the functor ${\rm Groups}\to \rm Rings$ that sends a group $G$ to the algebra $\Bbb Z[G]$ is left adjoint to the units functor ${\rm Rings}\to \rm Groups$ that sends a ring $R$ to the groups of units $R^\times$. Indeed, there is a natural bijection between ring morphisms $\Bbb Z[G]\to A$ and group morphisms $G\to A^\times$. 
The direct product $G\times H$ of groups has a universal property that maps from  $G\times H$ to a group $K$ are the same as maps $G\to K$ and $H\to K$ with commuting images. Similarly the integral tensor product $R\otimes S$ of rings has a universal property that maps from $R\otimes S$ to a ring $T$ are the same as maps $R\to T$ and $S\to T$ again with commuting images. 
You can then use the above three universal properties to conclude.