# Non-Separable Polynomials and their Derivatives

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.

• 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 $a$. – reuns Apr 20 at 12:01
• @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 $1$. – Sanjoy Kundu Apr 20 at 18:15
• 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 $gcd(f,Df) = f, Df = 0$. – reuns Apr 20 at 20:00
• Ahh that makes more sense. Thank you so much! :) – Sanjoy Kundu Apr 20 at 20:01
• (i meant non-seperable and irreducibe..) – reuns Apr 20 at 20:19

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.

• This is exactly what I felt was missing, much much appreciated. +1 – Sanjoy Kundu Apr 20 at 10:45
• 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. – Servaes Apr 20 at 20:16
• 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 $(x-a)^m$. – Servaes Apr 20 at 20:27
• 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. – Servaes Apr 20 at 20:36
• 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! – Sanjoy Kundu Apr 20 at 20:36

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.