3
$\begingroup$

Given any polynomial $f\in \mathbb C[x]$ of degree $n>0$, $f$ can be written in the form$$f=c(x-a_1)^{r_1}\cdots(x-a_l)^{r_l},$$where $a_1,...,a_l$ are distinct. Also, $f'$ is the product$$f'=(x-a_1)^{r_1-1}\cdots(x-a_l)^{r_l-1}H,$$where $H\in \mathbb C[x]$ is a polynomial vanishing at none of $a_1,...,a_l$.

I need to prove$$\gcd (f,f')=(x-a_1)^{r_1-1}\cdots(x-a_l)^{r_l-1},$$and I intend to to show that they divides each other. By the definition of $\gcd$, that RHS divides LHS is easy to show. But I fail to show that LHS divides RHS. Any suggestions?

$\endgroup$

1 Answer 1

2
$\begingroup$

Hint: $\gcd(f,f') = (x-a_1)^{r_1-1}\!\cdots(x-a_l)^{r_l-1}\gcd(c(x-a_1)\cdots(x-a_l),H)$. The latter gcd $= 1$ since $H(a_i)\ne 0\ \Rightarrow\ H$ coprime to all $x-a_i\ \Rightarrow\ H\:$ coprime to their product by Euclid's lemma (or by unique factorization).

The essence of the matter (differentiating reduces the power of each factor $x-a$ by precisely $1$) is clear if you view a polynomial as a (formal) power series. Shifting from $x= a$ to $x= 0$, we have

$\qquad\qquad\quad\ \ \ f\: =\ c_n\: x^n + \cdots\quad\ $ has order $\:n > 0\ \ $ (i.e. $\ x^n\ | f,\ \ x^{n+1}\nmid f$)

$\qquad\qquad\Rightarrow\ f' = n c_n x^{n-1}+ \cdots\ $ has order $\:n-1\ $ since $\ n,c_n\ne 0\ \Rightarrow\ nc_n \ne 0$

which makes clear they key role played by the nonzero characteristic of the coefficient ring. Indeed, over $\:\mathbb Z/p,\:$ for $f = x^p,\ f' = 0,\:$ so $\:\gcd(f,f') = f$.

$\endgroup$
2
  • $\begingroup$ Thanks. About $\gcd(f,f') = (x-a_1)^{r_1-1}\!\cdots(x-a_l)^{r_l-1}\gcd(c(x-a_1)\cdots(x-a_l),H)$, is there a general result that for any two polynomials $f,g$, $gcd(f,g)=$a common divisor $h \times gcd(f/h, g/h)$? $\endgroup$
    – Vladimir
    Commented Feb 19, 2012 at 20:56
  • $\begingroup$ @Jack Yes, gcds satisfy the distributive law $\:(ac,bc) = (a,b)c\:$. See here for a simple proof. You could eliminate this by simply appealing to unique factorization, noting that $x-a$ is prime in $\mathbb C[x]$, and products of primes always factor uniquely. $\endgroup$
    – Math Gems
    Commented Feb 19, 2012 at 21:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .