# The multiplicity of a root $r$ of a irreducible polynomial is a power of $p$ characteristic

$$f$$ is an irreducible polynomial over a field $$K$$ of characteristic $$p$$. $$F$$ is a splitting field of $$f$$ over $$K$$ and $$u_1$$ a root of $$f$$.

I have shown that $$f=[(x-u_1)\cdots (x-u_n)]^{[K(u_1):K]_s}$$ and $$n=[K(u_1):K]_i$$, where $$[K(u_1):K]_s$$ is separable degree and $$[K(u_1):K]_i$$ is inseparable degree.

Then I have trouble showing that $$u^{[K(u_1): K]_i}$$ is separable.

One proof points out that $$[K(u_1):K]_i = p^k = r$$ whence $$f=[(x-u_1)\cdots (x-u_n)]^{[K(u_1):K]_s} = (x^r - u_1^r) \cdots (x^r - u_n^r)$$ is of $$K[x]$$ therefore $$(x-u_1^r) \cdots (x-u_n^r)$$ is of $$K[x]$$ with $$u_1^r$$ $$\cdots$$ $$u_n^r$$ distinct so $$u_1^r$$ is irreducible.

Then I have a problem. Why is $$[K(u_1):K]_i$$ is a power of $$p$$.

At the beggining you mixed separable and inseparable degree. The main fact here is: If $$K$$ is of characteristic $$p$$, and $$f\in K[x]$$ is irreducible, then there exist $$k\geqslant 0$$ and an irreducible and separable polynomial $$f_s\in K[x]$$ such that $$f(x)=f_s(x^{p^k})$$. If $$\alpha$$ is a root of $$f$$, then the separable degree of $$K(\alpha)/K$$ is $$[K(\alpha):K]_s=\deg(f_s)$$ and the inseparable degree of $$K(\alpha)/K$$ is $$[K(\alpha):K]_i=p^k$$. Note that $$[K(\alpha):K]= \deg(f)= p^k\deg(f_s)$$, so it doesn't have to be the power of $$p$$.
Also, $$\beta= \alpha^{p^k}$$ is separable because its minimal polynomial is $$f_s(x)$$: $$f_s(\beta)= f_s(\alpha^{p^k})= f(\alpha)=0$$ and $$f_s(x)$$ is irreducible and separable.
Maybe you can look at the following example: Take $$p\neq 2$$ and $$K=\mathbb F_p(t)$$, where $$t$$ is transendental over $$\mathbb F_p$$. Consider polynomial $$f(x)= x^{2p}-t$$. It is irreducible over $$K$$, and for it we have $$f_s(x)=x^2-t$$ and $$k=1$$, and $$f(x)= f_s(x^p)$$. ($$f_s(x)$$ is separable because $$p\neq 2$$.) So, if $$\alpha$$ is one of the roots of $$f(x)$$, then $$[K(\alpha):K]=2p$$, $$[K(\alpha):K]_s=2$$ and $$[K(\alpha):K]_i=p$$.
You can investigate this situation more explicitly. Since $$\alpha^{2p}=t$$, we can write $$f(x)= x^{2p}-\alpha^{2p}= (x^p-\alpha^p)(x^p+\alpha^p)= (x-\alpha)^p(x+\alpha)^p$$. So $$f(x)$$ has two distinct roots: $$\alpha$$ and $$-\alpha$$ (thus separable degree is $$2$$), both of which are of multiplicity $$p$$ (thus inseparable degree is $$p$$).