Tensor product of commutative rings I need help with this question: 
Suppose that A, B, C are commutative rings with unit. Is it true that $A\otimes_\mathbb{Z}(B\times C)$ is isomorphic as rings with $(A\otimes_\mathbb{Z} B)\times(A\otimes_\mathbb{Z} C)$?
If you have a reference or a counterexample that would be great. 
Note: Here $\otimes$ = tensor product or rings and $\times$=direct product of rings.  
P.S. On the web I found a positive answer, but I don't see how the universal properties of tensor product and direct product can be used. 
Thanks.
 A: First, I think that writing an explicit isomorphism is pretty straightforward and you don't need to appeal to universal mapping properties. 
There is an obvious set map
$$
 \phi\colon A\otimes_{\mathbb Z}(B\times C)\to (A\otimes_{\mathbb Z}B)\times (A\otimes_{\mathbb Z}C), \quad \sum a_i\otimes(b_i,c_i)\mapsto (\sum a_i\otimes b_i, \sum a_i\otimes c_i).
$$
Then show that it is injective, surjective and a ring morphism.
However, you could show this by using the universal mapping property of the coproduct (=tensor product) or the product (=direct product). If you want to show $(A\otimes_{\mathbb Z}B)\times (A\otimes_{\mathbb Z}C)$ is isomorphic to $A\otimes_{\mathbb Z}(B\times C)$ it is enough to show it has the universal mapping property of the coproduct of $A$ and $B\times C$, i.e., you show that there are ring morphisms $s_1\colon A\to (A\otimes_{\mathbb Z} B)\times(A\otimes_{\mathbb Z} C)$ and $s_2\colon B\times C\to (A\otimes_{\mathbb Z} B)\times(A\otimes_{\mathbb Z} C)$ such that every other pair $f\colon A\to T$ and $g\colon B\times C\to T$ induces a unique factorization $h\colon (A\otimes_{\mathbb Z} B)\times(A\otimes_{\mathbb Z} C)\to T$. 
You could also do this the other way around by showing $A\otimes_{\mathbb Z}(B\times C)$ satisfies the universal mapping property of the product of $A\otimes_{\mathbb Z}B$ and $A\otimes_{\mathbb Z}C$. 
I don't think this is easier than explicitly giving an isomorphism by elements. But you can also try to find such maps as an exercise in this line of thinking. 
A: It is true for modules and for $\mathbf Z$-algebras: tensor product commutes with direct sums and finite products.
