# Proof of separability of polynomials without derivatives

Is there a known proof without differentiating that proves that all irreducible polynomials over $$\mathbb{Q}$$ are separable? (Or even better, for all fields of characteristic $$0$$.)

EDIT: As people seem to question this thread; I do know a proof with derivatives - my motivation for one without is simply curiosity. Multiple approaches are always nice.

• Could you tell what you don't like in the $f'$ proof : if $(x-\alpha)^n | f, n \ge 2$ then $n (x-\alpha)^{n-1} | f'$ and $gcd(f,f')$. If $char(K) | n$ then $n (x-\alpha)^{n-1}=0$ so that doesn't help, otherwise it means $f$ isn't irreducible since $\frac{f}{gcd(f,f')}\in K[x]$. – reuns Feb 16 at 19:46
• I‘m simply curious for other proofs, as I was not able to construct a proof without using derivatives having thought about it a little bit. Knowing multiple approaches to facts in mathematics helps enhancing the understanding - at least that‘s my opinion. Don‘t get me wrong, the standard proof is nice, I‘m just curious about other ones. – Kezer Feb 16 at 20:02
• I‘m not sure why this is marked as a duplicate - the linked question clearly does not answer my question, as it uses derivatives - unless I‘m overlooking something. – Kezer Feb 16 at 20:05
• Non separable polynomials means there is a finite extension $K$ where $(x-\alpha)^n = \sum_{m=0}^n {n \choose m} (-\alpha)^m x^{n-m} \in K[x]$ is irreducible. It is clear this is possible only if $char(K) \ne 0$ and ${n \choose m} = 0$ for $m \ne 0,n$ ie. $n = char(K)^l$ – reuns Feb 16 at 20:11
• I realize you're asking about derivative-free proofs. Does a derivation count as not a derivative. "Derivations" are a study in their own right. The space of derivations $\text{Der}_k (L, \overline{k})$ of $k$-linear derivations forms a vector space, and we can tell separability from the dimension of the vector space. – Dean Young Apr 25 at 3:54

Let $$L$$ be the splitting field of $$f \in F[x]$$. By elementary properties of field extensions, the automorphism group of $$L/F$$ acts transitively on the roots, so they have equal multiplicity. Suppose wlog that $$f$$ is monic. Then $$f$$ is some $$n$$th power of a $$g \in L[x]$$. By inspection of coefficients, $$g^n \in F[x]$$ implies $$g \in F[x]$$. Contradiction.

In the last step, it is used that $$n \neq 0$$ in F.

More details for that step: Let $$d$$ be the degree of $$g$$. We show by induction that the coefficient $$a_{d-i}$$ of $$x^{d-i}$$ in $$g$$ lies in $$F$$. For $$i = 0$$ this is clear. Suppose true for $$0, \ldots, i-1$$, and look at the coefficient of $$x^{nd-i}$$ in $$g(x)^n$$. It equals $$n a_{d-i}$$ $$+$$ a polynomial expression involving the $$a_{d-j}$$ for $$j < i$$. Because $$n \neq 0$$ and those $$a_{d-j} \in F$$ by assumption, we have $$a_{d-i} \in F$$.

• Can you elaborate on the last "induction" step? I can see other ways to make that argument but not an argument by induction on the degree of $g$. – Eric Wofsey Feb 17 at 16:20
• I went too fast there, the argument is indeed not as I said. – punctured dusk Feb 17 at 16:43
• This is very nice, thank you! It also gives another argument that roots always have equal multiplicity that I didn't know of. – Kezer Feb 17 at 17:17