Proposition 4

— Let $A$ be a separable algebra over a field $K$ and let $B$ be the total ring of fractions of $A$; then the $K$-algebra $B$ is separable.

Let $S$ be the set of cancellable elements of $A$. We have identified $A$ with a subring of $B$ (I, p. 113); further, every element of $S$ is invertible in $B$ and every element of $B$ has the form $a s^{-1}$ with $a \in A$ and $s \in S$. Let $L$ be an extension of $K$ and $x$ a nilpotent element of $L \otimes_K B$. Then $x$ may be written in the form

with $y_i \in L, a_i \in A, s_i \in S$ for $1 \leq i \leq n$. If we put $s = s_1 \ldots s_n$ then $x(1 \otimes s)$ belongs to the subring $L \otimes_K A$ of $L \otimes_K B$; since $x(1 \otimes s)$ is nilpotent and $A$ separable over $K$, we have $x(1 \otimes s) = 0$, and since $s$ is invertible in $B$ we find that $x = 0$. This proves the ring $L \otimes_K B$ to be reduced, whence the proposition.