Separable algebra

Remarks

Equivalently, $A$ is a projective $A^e$-module. Equivalently, there is a separability idempotent: an element $e = \sum_i x_i \otimes y_i$ of $A^e$ with $\sum_i x_i y_i = 1$ and $\sum_i a x_i \otimes y_i = \sum_i x_i \otimes y_i a$ for all $a \in A$; take $e = s(1)$.

Example: $A = M_n(K)$ is separable over $K$. With matrix units $E_{ij}$, the element $e = \sum_{i} E_{i1} \otimes E_{1i}$ satisfies $\sum_i E_{i1} E_{1i} = \sum_i E_{ii} = 1$, and for all $k, l$ the sums $\sum_i E_{kl} E_{i1} \otimes E_{1i}$ and $\sum_i E_{i1} \otimes E_{1i} E_{kl}$ both equal $E_{k1} \otimes E_{1l}$.

Non-example: let $K = \F_p(t)$ and $A = K(t^{1/p})$, a purely inseparable field extension of degree $p$. Then $A \otimes_K A \cong A[x]/(x - t^{1/p})^p$ has nonzero nilpotent elements, so it is not semisimple, and $A$ is not separable over $K$.

When $K$ is a field, a finite-dimensional algebra $A$ is separable if and only if $A \otimes_K L$ is semisimple for every field extension $L$ of $K$. Equivalently, $A$ is semisimple and the center of each simple factor of $A$ is a separable field extension of $K$. A finite field extension $L/K$ is a separable algebra exactly when it is a separable extension.