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, RHS|LHS is easy to show. But I fail to show LHS|RHS. Any suggestions?
How to show that $GCD(f,f')=(x-a_1)^{r_1-1}\cdots(x-a_l)^{r_l-1}$
1 Answers
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.
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0@Jack Yes, gcds satisfy the distributive law $\:(ac,bc) = (a,b)c\:$. [See here](http://math.stackexchange.com/a/62066/23500) 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. – 2012-02-19