Definition of root and purely inseparable I have a problem with the definition of root and purely inseparable. 
My book defines the purely inseparable iff 'the min($F,\alpha$) has only one distinct root'.
(1). How to define 'root' in Algebra? Does it mean '$\alpha$ is root of $f(x)$ in $K$, iff $f(\alpha)=0$'? Here$f(x)\in F[x]$ and $K$ in an extension of $F$. And I need to regard the $f(x)$ to be in $K[x]$. 
(2). When I write $(x-\alpha)|f(x)$ , is it automatically assumed that $f(x)$ is in $K[x]$? And is it equivalent to say $\alpha$ is root of $f(x)$?
(3). For the definition of purely inseparable, does the author means that there is no other element in $K$ that making min$(F,\alpha)$ to be $0$ in the sense of (1)? ($K$ is not assumed to be splitting field).
These question seems silly, but I want to make sure...

 A: Let $K$ be a field. Then for each element $\alpha\in K$ we have the evaluation map $K[x]\to K$ which is $K$-linear and sends $x\mapsto\alpha$. We write $f(\alpha)$ for the image of a polynomial $f\in K[x]$.
We say that $\alpha$ is a root of $f$ provided $f(\alpha)=0$. By the division algorithm, this is equivalent to saying that $x-\alpha$ divides $f(x)$ in $K[x]$.
If now $F\subset K$ is a subfield, then we can regard $F[x]$ as a subring of $K[x]$, and the evaluation map restricts to a map $F[x]\to K$, which is $F$-linear and sends $x\mapsto\alpha$. Thus for $f\in F[x]$ we have $\alpha\in K$ is a root of $f$ provided $f(\alpha)=0$, or equivalently $x-\alpha$ divides $f(x)$ in $K[x]$.
That answers (1) and (2).
(3). You are correct that the definition from the book is a bit ambiguous. An algebraic element $\alpha\in K$ is purely inseparable over the subfield $F$ provided the minimal polynomial $m:=\mathrm{min}(F,\alpha)$ has only one root in every field extension of $K$. In this case it follows that $m=(x-\alpha)^d\in K[x]$ for some $d$, so that $m$ actually splits over $K$, and conversely if $m=(x-\alpha)^d\in K[x]$, then $\alpha$ is purely inseparable over $F$.
In summary, $\alpha\in K$ is purely inseparable over $F$ provided the minimal polynomial $\mathrm{min}(F,\alpha)$ factors as $(x-\alpha)^d$ in $K[x]$.
This is different to saying that $m$ has only the one root $\alpha$ in $K$, since it may not split over $K$. For example, let $\alpha=\sqrt[3]2\in\mathbb R$, and take $F=\mathbb Q$ and $K=\mathbb Q(\alpha)$. Then the minimal polynomial $m=x^3-2$ has only one root in $K$ but $\alpha$ is separable over $\mathbb Q$.
