3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 4
    $\begingroup$ 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
1
$\begingroup$

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 $f'=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.