Separability of sums and products; Wraith's definition of formally unramified morphisms

Question

Let $\Bbbk\to A$ be a finite commutative algebra over a field. For an element $a\in A$, TFAE.

1. The minimal polynomial $\mu_a\in \Bbbk[x]$ is separable.
2. $a\in A$ is a (necessarily simple) root of a separable polynomial in $\Bbbk[x]$.
3. $a\in A$ is a simple root of some polynomial in $\Bbbk[x]$.

Following Lombardi and Quitté, say an element satisfying the second condition is separable w.r.t the $\Bbbk$-algebra.

A well-known result in field theory states that separable elements form a subalgebra. Hence the elements of a field extension generated by separable elements are all separable.

For some reason, a paper by Wraith characterizes separable field extensions as those where each element of the large field is a rational expression over the small field in a simple root of a polynomial over the small field. This rather weak description is then used to define formally unramified algebras. Specifically, Wraith does not (explicitly) require each element of the large field to itself be a separable.

Question 1. Why does Wraith define formally unramified algebras in this way? For a general ring morphism, is it not true that the sum and product of separable elements are separable?

Question 2. Is Wraith's definition of formally unramified algebras equivalent to the standard one?

Question 3. What is a direct constructive proof that sums and products of separable elements are separable over a field extension? Over a $\Bbbk$-algebra?

Answer 1

The answer to question 1 is "no".

Here's a concrete example. Write $x\in\tfrac{\mathbb Z/4\mathbb Z[X]}{\langle f\rangle}$ for the equivalence class of $X$ in the quotient. Assume $f\in\mathbb Z/4\mathbb Z[X]$ is separable, e.g the monic quadratic $x^2+x+c$ (its discriminant $1-4c$ over $Z/4\mathbb Z$ equals one). Then by construction, $x\in\tfrac{\mathbb Z/4\mathbb Z[X]}{\langle f\rangle}$ is a root of a separable polynomial. I claim $2x$ is not the root of a separable monic.

Indeed let $g(2x)=0$. Then the linear monomial of $g$ has coefficient zero or two. If it has coefficient zero, then $x^2\mid g$ whence $g$ is not separable. If the coefficient is two, then $g=x(2+xh)$ for some polynomial $h$. But the discriminant is algebraic, so its invertibility persists $\mod 2$. Having ruled out both cases, we deduce $g$ is not separable, as desired.