Let $E/F$ be a finite extension of fields. Not too long ago we saw that the $E/F$ is separable if and only if for every extension $L/F$, the ring $E\otimes_F L$ has zero Jacobson radical.
The notion of separable algebra generalizes this concept to any $R$-algebra where $R$ is a commutative ring.
The best kind of separable algebra is the central separable $R$-algebra. These are the separable $R$-algebras whose center is $R$. You can always put yourself in this case because a separable algebra is a central separable algebra over its center.
Here's an example of a central separable algebra: if $R$ is a commutative ring then the $n\times n$ matrix ring $M_n(R)$ is a central separable $R$-algebra. What's an example of an algebra that is not separable?
To explain how to construct them, suppose $A$ is an $R$-algebra and $B\subseteq A$ is an $R$-subalgebra. Let's write $A^B = \{ a\in A : ab = ba \forall b\in B\}$. In words, $A^B$ is the set of elements of $A$ that commute with $B$. The notation is perhaps jarring but convenient. If $A$ happens to be central separable and $B$ is separable, then it's a theorem that $C = A^B$ is also separable and $A^C = B$.
This gives us a way to find algebras that are not separable, as subalgebras of matrix algebras. For example, a natural guess might be $T$, the upper triangular $n\times n$, $n > 1$ matrix algebra over a commutative ring $R$. Then $T$ is a subalgebra of $M$, the full $n\times n$ matrix ring.
We claim that $T$ is not a separable $R$-algebra. Indeed, it is easy to compute that $M^T = R$, where $R$ here is embedded into $M$ as the diagonal subalgebra. If $T$ were separable, then as we have remarked, we would have $M^R = T$. But this is false: $M^R = M$. This argument applies more generally: any proper subalgebra of $M$ whose centralizer in $M$ is just $R$ cannot be separable.