UTM Ideals Varieties and Algorithm - Chapter 1 Section 5 Exercise 14

Problem:

Solution:

Part (a)

$f' = r(x - a)^{r-1}h + (x-a)^r h' = (x - a)^{r-1}(rh + (x-a) h')$

Therefore $h_1 = rh + (x-a) h'$, $h_1(a) = rh(a) + (a - a)h' = rh(a) \ne 0$.

Part (b)

Using product rule, we differentiate one of the term and keep the rest, and then we sum them up, so the derivative is

$f' = \sum\limits_{k = 1}^{n}\left(cr(x-a_k)^{r_k - 1}\prod\limits_{j=1, j \ne k}^{n}(x - a_j)^{r_j}\right)$

Therefore, for the final sum, we can always factor out $(x - a_1)^{r_1 - 1} \cdots (x - a_n)^{r_n - 1}$

$f' = \prod\limits_{j=1}^{n}(x - a_j)^{r_j - 1} \sum\limits_{k = 1}^{n}\left(cr\prod\limits_{j=1, j \ne k}^{n}(x - a_j)\right)$

Now after factoring out, the $k$ term is simply $cr(x - a_1) \cdots (x - a_n)$ (the product goes without $(x - a_k)$), so the $k$ term does not vanish for $a_k$, but all other terms does, That's why $H$ does not vanish for any $a_k$.

Part (c)

With part (b), we proved $(x - a_1)^{r_1 - 1} \cdots (x - a_n)^{r_n - 1}$ is a common factor. The only roots in $f$ is $\{ a_1 \cdots a_n \}$, so if the common factor is not greatest, we contradict the fact that $H$ does not vanish for any of those roots.