Assume that $f$ is some monic polynomial in $K[X]$ for some field $K$. I would like to know if the following implication holds:

$\deg(\gcd(f,f'))>0 \quad \Longrightarrow \quad \exists a,b \in K \ \text{such that } \ aX+b \ \text{divides both} \ f,f'.$

In other want to know if $f, f'$ both have zeros if the polynomials are not coprime. Can you help me with this?

Research effort

First I tried to find a counterexample, but I failed. I denote $h=gcd(f,f')$. Now we can write: $f=gh$ for some polynomial $g$, and $f'=\bar{g}h$ for some polynomial $\bar{g}$. Moreover $f' = g'h+gh'$. I tried to exploit this identity:

$$g'h+gh' = \bar{g}h \quad \iff \quad (\bar{g}-g')h=gh'$$ Another observation:

$$\deg g + \deg h = \deg f = \deg f' +1 = \deg \bar{g} + \deg h +1$$ Now we see that $\deg g' = \deg \bar{g}$.

That's pretty much how far I came. Can you tell me if the statement is true, and why? Thank you

  • $\begingroup$ If I understand your question correctly, this (math.uconn.edu/~kconrad/blurbs/galoistheory/separable1.pdf) is what you want. In particular theorem 2.1 $\endgroup$ – LASV Dec 4 '13 at 21:48
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    $\begingroup$ Not true as written, consider $(X^2+1)^2$ in $\mathbb{R}[X]$. If you make $K$ algebraically closed, it's true (but trivial). Instead of monic, you might want irreducible for a more interesting problem. $\endgroup$ – Daniel Fischer Dec 4 '13 at 21:48

Suppose that $K$ is algebraically closed, so that only the linear polynomials are irreducible. For convenience, we can also assume that our polynomial is monic (we can pull out the lead coefficient).

If $f=(x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_n)$, then we have $f'=\displaystyle\sum_k(x-\alpha_1)..(x-\alpha_{k-1})(x-\alpha_{k+1})..(x-\alpha_n) =\sum_k\frac{f}{x-\alpha_k}$.
If $\alpha_i=\alpha_j$, then $(x-\alpha_i)\,|\,f'$ so multiple roots give common factors of $f$ and $f'$.

On the other hand, if all $\alpha_i$ are distinct, then $$f'(\alpha_i)=\prod_{k\ne i}(\alpha_i-\alpha_k)\ne 0,$$ so $(x-\alpha_i)$ does not divide $f'$.

Note also, that if $K$ has $p$ characteristic, then all polynomials $f$ of $x^p$ (i.e. $f(x)=g(x^p)$ for some polynomial $g$) satisfy $f'=0$.


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