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We say that a polynomial $f(x) ∈ F[x]$ is inseparable if it has a repeated root in some field extension. Otherwise we say that $f(x)$ is separable. Prove that $f(x)$ is separable $\iff\gcd(f, Df) = 1$. Note $Df$ denotes the formal algebraic derivative of $f$.

Instead of proving the bi-conditional as written, I have elected to prove the contrapositive, that is I will prove that $f(x)$ is not separable $\iff\gcd(f, Df)$ is not $1$.

Suppose $f(x) ∈ F[x]$ is inseparable and has degree $n$. Then by definition, $f(x)$ must have a repeated root in some field extension. It is well-known that if $f$ has a repeated root at some $x = a$, $Df$ must also have a root at $x=a$; this follows because we can write $f∈F[x]$ as a product of linear factors in some field extension of $F$ and then apply the product rule for derivatives to find an expression for $Df$ in terms of the linear factors of $f$, which verifies the result above. Hence, $f$ and $Df$ share a common factor, namely $(x-a)^m$, where $m$ is some integer power between $1$ and $n-1$ and thus, $\gcd(f, Df)$ is not equal to $1$.

Conversely, suppose $gcd(f, Df)$ is not equal to $1$. Then it follows that $f$ and $Df$ share a non-trivial common divisor, which without loss of generality, we will call $(x-a)^m$, where $m$ is some integer power between $1$ and $n - 1$. This implies that $Df$ and $f$ share a root at $x = a$ and since $f$ is a polynomial, then $\deg(Df)$ = $n-1 <\deg(f)$ = $n$. In case $m > 1$, it consequently follows that $f$ has a repeated root at $x = a$. Otherwise, if $m = 1$, then both $f$ and $Df$ share the common linear factor $x-a$. However, it is well-known that if $f$ and $Df$ share a root at $x = a$, then $f$ must contain at least one repeated root at $x = a$; this is a consequence that directly follows from the product rule for derivatives. Hence in either case, it follows that $f$ has a repeated root, and thus $f$ is inseparable, as desired.

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    $\begingroup$ Conversely if $f(x)$ is irreducible and non-separable then $Df(x) = 0$ so $f(x) =g(x^p) = h(x)^p,h^\phi(x) = g(x)$ where $p = char(F)$ and $\phi$ is the Frobenius applied to the coefficients, doing the same with $h(x)$ we find $f(x) = (x-a)^{p^n}=x^{p^n}-a^{p^n}$ for some $a$. $\endgroup$ – reuns Apr 20 at 12:01
  • $\begingroup$ @reuns For the converse, I'm a bit confused. If $f(x)$ is irreducible and inseparable, aren't we assuming what we are trying to prove (i.e. $f(x)$ is inseparable)? The only thing we can work with is the fact that $gcd(f, Df)$ does not equal $1$. $\endgroup$ – Sanjoy Kundu Apr 20 at 18:15
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    $\begingroup$ If $f$ is non-separable and $Df \ne 0$ then $gcd(f,Df)$ divides $f$ so it is non-irreducible. Thus $f$ separable and irreducible implies $gcd(f,Df) = f, Df = 0$. $\endgroup$ – reuns Apr 20 at 20:00
  • $\begingroup$ Ahh that makes more sense. Thank you so much! :) $\endgroup$ – Sanjoy Kundu Apr 20 at 20:01
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    $\begingroup$ (i meant non-seperable and irreducibe..) $\endgroup$ – reuns Apr 20 at 20:19
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Your proof of the converse statement is not quite complete; you state that

...it follows that $f$ and $Df$ share a non-trivial common divisor, which without loss of generality, we will call $(x-a)$.

But you do not explain why this non-trivial common divisor can be assumed to be linear.

Next you claim that

Hence it follows that $f$ has a repeated root, and thus $f$ is inseparable, as desired.

But it is not at all clear (to me, at least) why it follows that $f$ has a repeated root.


My suggestion to improve your proof, is to make explicit your claim that

It is well known that if $f$ has a repeated root at some $x=a$, $Df$ must also have a root at $x=a$.

To prove this, write $f\in F[x]$ as a product of linear factors in some field extension of $F$, and then express $Df$ in terms of the factors of $f$. This expression will make the equivalence you are trying to prove immediately clear.

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  • $\begingroup$ This is exactly what I felt was missing, much much appreciated. +1 $\endgroup$ – Sanjoy Kundu Apr 20 at 10:45
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    $\begingroup$ In stead of saying "It is well known that...", simply show that if $f$ splits as $$f=\prod_{i=1}^n(x-\alpha_i)^{m_i},$$ in some splitting field, with the $\alpha_i$ distinct and $m_i>0$, then by the product rule $$Df=\sum_{i=1}^n\left(m_i(x-\alpha_i)^{m_i-1}\prod_{\substack{j=1\\j\neq i}}^n(x-\alpha_j)^{m_j}\right).$$ This immediately shows that $Df(\alpha_i)=0$ if and only if $m_i>0$, which is pretty much equivalent to the claim you want to prove follows. $\endgroup$ – Servaes Apr 20 at 20:16
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    $\begingroup$ For example, the polynomial $f=x^4+2x^2+1$ is inseparable, and $Df=4x^3+4x$, but they do not have a common factor of the form $(x-a)^m$ over $\Bbb{Q}$. The $\gcd$ of two polynomials is not (without loss of generality) of the form $(x-a)^m$. $\endgroup$ – Servaes Apr 20 at 20:27
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    $\begingroup$ About the beating around the bush; you make two claims of the form "It is well-known that...". These two claims put together are:$$\text{"It is well known that $f$ has a repeated root at some $x=a$ if and only if $Df$ has a root at $a$."}$$But from this claim, the statement you want to prove follows immediately; by definition this claim says precisely that $f$ is inseparable if and only if $(x-a)$ divides $\gcd(f,Df)$ for some $a$ in some field extension. So the rest of what you wrote I would call 'beating around the bush', and the 'well-known' claim is precisely what does require proof. $\endgroup$ – Servaes Apr 20 at 20:36
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    $\begingroup$ Ahh I will try to incorporate your insight and remove the parts that are unnecessary. Thank you for being so patient with me, I really appreciate your guidance! $\endgroup$ – Sanjoy Kundu Apr 20 at 20:36
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Think in this way... If the polynomial is separable, it can be factored such that for each root $\alpha \in f(x)$, we can write on factor as $(x- \alpha)$.And any of its derivatives must strictly not contain any factor of he initial polynomial due to product rule of differentials(there will be a part of the factor which will be summed up with a different term which will strictly not be any multiple of $(x - \alpha)$ or $(x - \beta)$). So similar cases will occur for others and so there follows the result.

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