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.

  • $\begingroup$ 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 $\frac{f}{gcd(f,f')}\in K[x]$. $\endgroup$ – reuns Feb 16 at 19:46
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    $\begingroup$ 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. $\endgroup$ – Kezer Feb 16 at 20:02
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    $\begingroup$ 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. $\endgroup$ – Kezer Feb 16 at 20:05
  • $\begingroup$ 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. $n = char(K)^l$ $\endgroup$ – reuns Feb 16 at 20:11
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    $\begingroup$ 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. $\endgroup$ – 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 $a_{d-i} \in F$.

  • $\begingroup$ 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 $g$. $\endgroup$ – Eric Wofsey Feb 17 at 16:20
  • $\begingroup$ I went too fast there, the argument is indeed not as I said. $\endgroup$ – punctured dusk Feb 17 at 16:43
  • $\begingroup$ This is very nice, thank you! It also gives another argument that roots always have equal multiplicity that I didn't know of. $\endgroup$ – Kezer Feb 17 at 17:17

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