# Is the zero polynomial considered separable?

For $$K$$ a field, a polynomial $$f \in K[X]$$ is called separable if it splits into distinct linear factors in a splitting field.

For $$f(X) = c$$ a constant polynomial, for $$c \neq 0$$, we can say $$f$$ is separable as it splits as $$c$$, with the splitting field $$K$$.

But what about when $$c = 0$$? On the one hand, we can write it as $$0$$ in the field $$K$$, but technically $$(X - \alpha)$$ is a factor for any $$\alpha$$.

Note it is not too hard to prove that if $$f \neq 0$$, then $$f$$ is separable if and only if $$\mathrm{gcd}(f, f') = 1$$.

Is separability/inseparability even defined for the zero polynomial? If it is, what does it count as?

This is because the only real use for the concept I know is to define what it means for an element $$\alpha \in L$$ of an extension $$L$$ of a field $$K$$ to be separable over $$K$$: it means that $$\alpha$$ is algebraic over $$K$$ and its minimal polynomial is separable. But the minimal polynomial is always nonzero, and in fact is always nonconstant; the only elements which could plausibly be said to have a minimal polynomial of zero are transcendental elements, which are not under discussion at all. So there is no point in talking about separability for constant polynomials, and in particular for the zero polynomial.
• It is also possible to give an alternative definition of separability that does not directly mention polynomials: we can say that an algebraic element $\alpha$ is separable over $K$ if $K[\alpha]$ is an etale algebra over $K$, meaning that $K[\alpha] \otimes_K \bar{K} \cong \bar{K}^n$. With this approach we don't need to talk about separability of polynomials if we don't want to. Commented Jun 11 at 22:41