Ideals and Tensor Products

Question

I'm reading Osbourne's Basic Homological Algebra, and on page 18 he has this situation where we've got a ring $R$ and a right-ideal $I$, and some left $R$-module $B$. He says $I\otimes B$ is not a subgroup of $R\otimes B$, but he doesn't give an example, just states "the free abelian groups and equivalence relations defining them are different". Well, sure, but the free abelian group on $I\times B$ is certainly a subgroup of the one on $R\times B$, and I cannot see how restricting the equivalence relations on $R\times B$ to just elements in $I\times B$ is going to make some new group that doesn't exist inside $R\otimes B$.

I've tried to make examples, but failed to find one which demonstrated this.

I've seen examples of tensor products which do this with more relaxed conditions, such as $\mathbb{Z}\otimes_\mathbb{Z} (\mathbb{Q}/\mathbb{Z})$ is not a subgroup of $\mathbb{Q}\otimes_\mathbb{Z} (\mathbb{Q}/\mathbb{Z})$ since the latter is 0 while the former is not, although obviously $\mathbb{Z}$ is a subgroup of $\mathbb{Q}$ - but it is not an ideal.

I would appreciate some examples, and some idea of what causes it to happen.

Answer 1

The problem is that the "obvious" inclusion mapping $I \otimes B \to R \otimes B$, given by tensoring the inclusion mapping with the identity, need not be injective. Recall that, by definition, a module is flat if tensoring with that module preserves the head of exact sequences; but that is the same as saying that tensoring an injective homomorphism with that module always yields and injective homomorphism. So one should look at non-flat modules $B$.

For example, take $\mathbb{Z}/2 \otimes \mathbb{Z}/2 \to \mathbb{Z}/4 \otimes \mathbb{Z}/2$. Both the domain and codomain are isomorphic to $\mathbb{Z}/2$, but the non-zero element $1 \otimes 1$ of the domain is mapped to $2 \otimes 1 = 1 \otimes 2 = 0$.

Answer 2

One example is the case $R = \Bbb{Z}_4$, $B = I = \{0, 2\}$ that Osbourne is talking about at that point in the book. In this example, you find that $I \otimes B \simeq R \otimes B \simeq \Bbb{Z}_2$, but the natural mapping from $I\otimes B$ to $R \times B$ is not an injection. In fact, $2 \otimes 2$ generates $I \otimes B$ but maps to zero in $R \otimes B$, because in $R \otimes B$, we have $2 \otimes 2 = 4(1 \otimes 1) = 0$, but $1 \otimes 1$ isn't in the free abelian group on $I \times B$, and in $I \otimes B$ we don't have this equation. ($R \otimes B$ is generated by $1 \otimes 2$.)