Example of a ring $R$ such that $R\otimes_R R\not\simeq R$

Question

I want to show the following statement:

If a ring $R$ is commutative and $I,J\triangleleft R$, then $$ R/I\otimes_R R/J\simeq R/(I+J). $$

I can easily show this, assuming that $1\in R$. What can be a counterexample if this assumption is removed?

Answer 1

I claim that $I=J=0$ already yields a family of counterexamples to $R\otimes_R R \cong R$. We must have $R\cdot R \neq R$ for this to work (I think this condition might also be sufficient for the natural map to not be an isomorphism, but I haven't worked out all the details).

$R=n\mathbb{Z}$ is such a rng: $n\mathbb{Z} \otimes_{n\mathbb{Z}} n\mathbb{Z} \cong n^2\mathbb{Z}\not\cong n\mathbb{Z}$.

For a more dramatic example, where we do not even have an isomorphism as abelian groups, let $R$ be any nonzero ring with trivial multiplication. Then $R\otimes_R R = 0 \not\cong R$.

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I wrote some general notes before getting to the above examples. I thought I should put the examples first, but the exposition is probably helpful to anyone who wants to understand this more deeply:

In general, $R/(I+J)$ is isomorphic to the pushout of $R/I \gets R \to R/J$ in the category of rngs (rings without identity), because a map $R\to S$ vanishes on $I$ and $J$ exactly when it vanishes on $I+J$.

What fails is that the pushout is no longer described by the tensor product. To take the simplest example: every bilinear map $\beta:A\times 0\to S$ is zero, because $\beta(a,0) = \beta(a,0\cdot 0) = 0\cdot \beta(a,0) = 0$. So $A\otimes_0 0 = 0$, breaking the expected rule of $A\otimes_R R \cong A$. But the coproduct of $A$ and $0$ is still $A$.

In general, I believe the correct coproduct of two $R$-algebras $A$ and $B$ is a quotient of $A\oplus B \oplus (A\otimes_R B)$, where we identify the image of $R$ in $A$ with the image of $R$ in $B$, and the product $ab$ with the tensor $a\otimes b$. When $R$ has an identity, it's a nice exercise to check that this is isomorphic to the usual tensor product. Some condition like $RA=A$, $RB=B$ is probably also sufficient.

The moral of all of this appears to be that an $R$-module $M$ behaves a lot more like a module over a unital ring when $RM=M$.