Roots of a non-separable irreducible polynomial If we are working in a field of characteristic $p>0$, $K$, and we have $f$ irreducible and non-separable over $K$. If we have $L/K$ be the splitting field of $f$ and $r$ a root of $f$, is it true that $f$ splits into $(x-r)^d$. 
I get the impression that this is true but cannot figure out how I would go about proving it.
Thank you for any help.
 A: This is not always the case. Assume that $\alpha$ is a root of a separable irreducible polynomial $f(x)\in K[x]$ of degree $\ge2$, and further assume that $\alpha$ is not a $p^{th}$ power in $K[\alpha]$. Then the minimal polynomial of $r=\alpha^{1/p}$ over $K$ will
be $f(x^p)$. Its zeros are the $p^{th}$ roots of those of $f(x)$, so they are not all equal.
For a concrete example let $K=\mathbb{F}_2(t)$, $t$ transcendental, and let $f(x)=x^2+x+t$.
The zeros of $f(x)$ are $\alpha$ and $\alpha+1$. Techniques of valuations show that they both have a simple pole in the unique point lying above the infinite point of $K$, so they cannot be squares in $K[\alpha]$. Therefore the polynomial $p(x)=x^4+x^2+t$ is irreducible in $K[x]$ and it factors in its splitting field $K[\sqrt{\alpha}]$ as follows
$$
p(x)=(x^2-\alpha)(x^2-\alpha-1)=(x-\sqrt{\alpha})^2(x-\sqrt{\alpha}-1)^2.
$$
A: The conclusion holds iff $f$ is purely inseparable, in fact almost by definition.
For an explicit counterexample, let $k$ be a field of positive characteristic $p$, let $\ell$ be a prime different from $p$, take $K = k(t)$ and $f(x) = x^{\ell p} - t$.  
