Let $k$ be a field, and $A$ a commutative $k$-algebra. I noticed two definitions of what it means for $A$ to be separable over $k$:
Definition 1 [Matsumura, Commutative Algebra, 1980, (27.D) + (27.G)]: $A$ is separable over $k$ if for all field extensions $L$ of $k$, $A \otimes_k L$ is reduced.
Definition 2 [https://ncatlab.org/nlab/show/separable+algebra ]: $A$ is separable over $k$ if for all field extensions $L$ of $k$, $A \otimes_k L$ has zero Jacobson radical.
Note that Definition 2 $\Rightarrow$ Definition 1 since the nilradical of a ring is contained in the Jacobson radical, but I don't see why Definition 1 $\Rightarrow$ Definition 2, unless some restrictions are imposed on $A$ (for e.g. if $A$ is of finite type over $k$, or if $A$ is generated as a $k$-algebra by a set $\{x_i\}_{i \in I}$ such that the cardinality of $I$ is less than the cardinality of $k$ [Stacks, Tag 00FU]).
Question: Are the two definitions equivalent? If not, is one definition better than the other for commutative rings?
The definitions are not equivalent on the class of all commutative $k$-algebras. Take $k = \mathbb{C}$ and $A = \mathbb{C}[[t]]$. Since $A$ is a local PID with maximal ideal $(t)$, its Jacobson radical is thus $(t) \neq (0)$. On the other hand, for any perfect field $k$, if $A$ and $B$ are two reduced, commutative $k$-algebras, then $A \otimes_k B$ is also reduced (e.g. see here for references to this standard-but-not-so-easy fact).
In my own experience, Definition 1 -- i.e., universally reduced -- is more standard than Definition 2 -- i.e., universally semiprimitive. However, it should be mentioned that in many contexts, the separable algebras one is interested in are finite-dimensional over $k$. (Indeed, sometimes this finite-dimensionality is taken as part of the definition.) As you already know, in that case the two definitions are equivalent.