Exercise 5.3.17

Thus $f^{\prime }(x)={(x-\beta )}^{m-1}{h}_1(x)$, where $h_1(x)=\mathit{mh}(x)+(x-\beta )h^{\prime }(x)$, $h_1(\beta )=\mathit{mh}(\beta )$.

By induction, suppose that there exists $h_k\in F[x]$, for $k<m$, such that

Then

where $h_{k+1}(x)=(m-k)h_k(x)+(x-\beta )h_k^{\prime }(x)$, thus

and the induction is done. The property ist true up to rank $k=m$, which gives

Conclusion : if $f(x)={(x-\beta )}^{m}h(x)$, then $f^{(m)}(\beta )=m!h(\beta )$.

$\text{∙}$ Suppose that ${\mathrm{ord}}_f(\beta )=m$. Then $f(x)={(x-\beta )}^{m}h(x),h\in F[x]$, and $h(\beta )\ne 0$, otherwise $(x-\beta )\mid h$, and so ${(x-\beta )}^{m+1}\mid f$.

By part (a), for all integer $k,0\le k\le m-1$, $f^{(k)}(x)={(x-\beta )}^{m-k}{h}_k(x),h_k\in F[x]$, thus $f(\beta )=f^{\prime }(\beta )=\cdots =f^{(m-1)}(\beta )=0$.

Moreover, $f^{(m)}(\beta )=m!h(\beta )\ne 0$, since $h(\beta )\ne 0$, and since the characteristic is 0, so $m!\ne 0$ in $F$.

We have proved $f(\beta )=f^{\prime }(\beta )=\cdots =f^{(m-1)}(\beta )=0,f^{(m)}(\beta )\ne 0$.

$\text{∙}$ Conversely, suppose that

As $f(\beta )=0$, $x-\beta$ divides $f$. We take as induction hypothesis, for $k<m$, that ${(x-\beta )}^k\mid f(x)$.

Then $f(x)={(x-\beta )}^k{h}_k(x)$, and part (a) shows that $f^{(k)}(\beta )=k!h_k(\beta )=0$, since $k<m$. As the characteristic of $F$ is 0, $k!\ne 0$, thus $h_k(\beta )=0$, therefore $(x-\beta )\mid h_k(x)$, so ${(x-\beta )}^{k+1}\mid f$.

This induction proves that ${(x-\beta )}^m\mid f(x)$.

Using again part (a), $f(x)={(x-\beta )}^{m}h(x)$, gives $h(\beta )=\frac{f^{(m)}}{m!}\ne 0$, thus $(x-\beta )\nmid h_k(x)$, so ${(x-\beta )}^{m+1}\nmid f(x)$. Consequently ${\mathrm{ord}}_f(\beta )=m$.

Conclusion: if the characteristic of $f$ is 0,

Moreover $m!\ne 0$ is equivalent to $p>m$.

So we can state:

if the characteristic of $F$is $p$, and if $m<p$,