Showing there's no ring isomorphism using tensor products

Question

I'm currently learning about tensor products, though I struggle a lot with it. In an exercise, it is stated that one can use the tensor product to show that there's no isomorphism between certain rings. Here are two examples:

Let $R:=\mathbb{Z}[X,Y]$. Show the following using a suitable tensor product functor:

$1.\quad R \oplus R/(X)\ncong_R R\oplus R/(X)\oplus R/(X)$

$2.\quad R/(X)\oplus R/(X)\oplus R/(X+1,Y) \ncong_R R/(X)\oplus R/(X+1,Y)\oplus R/(X+1,Y)$

My thoughts are that if I were to use a tensor product functor on $1.$ or $2.$, I could make abelian groups out of the above. Maybe that's easier to show? However, I don't know how to find a suitable tensor product functor for that nor do I understand how to go on once I have one, since I have problems working with tensor products. I also know that the tensor product is compatible with colimits, but this would only give us certain isomorphisms instead of showing there's none.

Can someone help me here? Thanks in advance.

Answer 1

Yes, taking tensor product makes problem easier.

For the first problem, assume that two modules given in our problem are isomorphic and take a tensor product by $R/(X)$. We can see that $R\otimes_R R/(X)\cong R/(X)$ and $R/(X)\otimes_R R/(X)\cong R/(X)$. As $\otimes$ is distributive under direct sum, we can see that $$(R/(X))^2\cong (R/(X))^3.$$ Then deduce a contradiction by the proposition of @AreaMan's answer.

For the second problem, take a tensor product by $R/(X+1,Y)$ and show that $R/(X)\otimes_R R/(X+1,Y)\cong 0$.

Answer 2

Here is an important example of this method, as a kind of hint:

Let $R$ be a commutative ring with unity. Let $m \not = n$ be natural numbers. Then $R^n \not \cong R^m$.

Pf: Pick a maximal ideal $\mathscr{m} \subset R$. Tensor the equation, $R^n \cong R^m$ with $R / \mathscr{m}$. Since tensor products distribute over sums, $R^n \cong R^m$ implies $(R / \mathscr{m})^m \cong (R / \mathscr{m})^n$. However, now $R / \mathscr{m}$ is a field, and we have concluded that a vector space over $R / \mathscr{m}$ of dimension $m$ is isomorphic to one of dimension $n$. This is a contradiction.