Tensor product of boolean rings

Question

Is there an example of a Boolean ring $B$ such that the $B$-bimodule map $\mu:B\otimes_\mathbb{Z}B\rightarrow B,$ defined by $\mu(x\otimes y)=xy$, does not split? (To split means that there is a $B$-bimodule map $x:B\rightarrow B\otimes_\mathbb{Z}B$ such that $\mu x=id$.)

Answer 1

This condition is one of the equivalent definitions of separability. It implies that $B$ is semisimple (here we need to use the fact that Boolean rings are naturally $\mathbb{F}_2$-algebras and that we can replace $\otimes_{\mathbb{Z}}$ with $\otimes_{\mathbb{F}_2}$), and Boolean rings are semisimple iff they're finite-dimensional. So any infinite-dimensional Boolean ring, such as $\prod_{i=1}^{\infty} \mathbb{F}_2$, is a counterexample.

Answer 2

This question has a very nice geometric interpretation. For a general commutative ring $B$, writing $X=\operatorname{Spec}(B)$, the map $\mu$ corresponds to the diagonal map $\Delta:X\to X\times X$ of schemes. Since $\mu$ is surjective, $\Delta$ is the inclusion of a closed subscheme, and $\mu$ splits as a $B\otimes B$-module map iff the kernel of $\mu$ is generated by an idempotent. Geometrically, this means that $\Delta(X)$ is a clopen subset of $X\times X$.

When $B$ is a Boolean ring, $X$ is a Stone space, and the underlying topological space of the scheme product $X\times X$ coincides with the ordinary topological product $X\times X$. So (via Stone duality) you are asking: for what Stone spaces $X$ is the diagonal clopen in $X\times X$? But for any topological space $X$, it is easy to see the diagonal in $X\times X$ is clopen iff $X$ is discrete. When $X$ is a Stone space, this means $X$ is finite, or equivalently that $B$ is finite.

Thus we conclude that for a Boolean ring $B$, $\mu$ splits iff $B$ is finite. In particular, $\mu$ will fail to split for any infinite Boolean ring.