Posted by Jason Polak on 26. November 2016 · Write a comment · Categories: commutative-algebra · Tags:

Let $R$ be a commutative ring and let $A$ be an $R$-algebra. We say that $A$ is separable if $A$ is projective as an $A\otimes_RA^{\rm op}$-module. There is a multiplication map $\mu:A\otimes_RA^{\rm op}\to A$ given by $a\otimes a’\mapsto aa’$, whose kernel we’ll call $J$.

It’s a fact that $A$ is separable if and only if there exists an idempotent $e\in A\otimes_RA^{\rm op}$ such that $\mu(e) = 1$ and $Je = 0$. The most familiar examples of separable algebras to many readers are probably the finite separable field extensions of a given field $R$. So let’s see an example of the separability idempotent!

For example, $\Q(\sqrt{2})$ is a separable $\Q$-algebra. One can then write down explicitly the above definitions. For example, $\Q(\sqrt{2})\otimes_\Q\Q(\sqrt{2})\cong \Q(\sqrt{2})\oplus \Q(\sqrt{2})$ via the map given on pure tensors by $(a,b)\mapsto (ab,a\sigma(b))$ where $\sigma:\Q(\sqrt{2})\to\Q(\sqrt{2})$ is the $\Q$-algebra isomorphism given by $\sqrt{2}\mapsto -\sqrt{2}$. The inverse of this map is given by
$$ (u,v)\mapsto \frac{u+v}{2}\otimes 1 + \frac{u-v}{2\sqrt{2}}\otimes\sqrt{2} $$
The multiplication map under this isomorphism translates to a map
$$\Q(\sqrt{2})\oplus\Q(\sqrt{2})\to Q(\sqrt{2})\\
(a,b)\mapsto a
$$
Therefore, the kernel of the multiplication map is $J = 0\oplus\Q(\sqrt{2})$ and the separability idempotent is just $e = (1,0)$. Of course, it’s naturally to want to observe this idempotent in its natural habitat of the tensor product $\Q(\sqrt{2})\otimes_Q\Q(\sqrt{2})$. Here it is folks:
$$
e = \frac{1}{2}\otimes 1 + \frac{1}{2\sqrt{2}}\otimes\sqrt{2}
$$
It’s easy to see that in general $J$ is the ideal generated by the set $\{ 1\otimes a – a\otimes 1 : a\in A\}$. Now just try verifying $Je = 0$ directly! Or not.

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