MHB The irreducible polynomial is not separable

[mathmari]

Hey!

Let $F$ be a field, $D=F[t]$, the polynomial ring of $t$, with coefficients from $F$ and $K=F(t)$ the field of rational functions of $t$.
(a) Show that $t\in D$ is a prime element of $D$.
(b) Show that the polynomial $x^n-t\in K[x]$ is irreducible.
(c) Let $\text{char} F=p$. Show that the polynomial $x^p-t$, even if it is irreducible in $K[t]$, it is not separable, and if $C$ is an algebraic closure of $K$, this polynomial has only one root in $C$. I have done the following:

(a) We assume that $t\in D$ is not prime. Then there are non-constant polynomials $f(t),g(t)\in D$ such that $t=f(t)g(t)$. Then $\deg (t)=\deg (f(t)\cdot g(t)$. Since $D$ is an integral domain, we have that $\deg (t)=\deg (f)+\deg (g)\Rightarrow 1\geq 1+1$, a contradiction.
Is this correct? (Wondering)

(b) From the generalized criterion of Eistenstein we have that the prime $t$ divides the term $t$ but not the coefficient of the highest degree term, so it $x^n-t$ is irreducible in $K[x]$.
Is this correct? (Wondering)

(c) Could you give me a hint how we could show that? Why does this hold? (Wondering)

 

[mathmari]

(c) Let $\text{char} F=p$. Show that the polynomial $x^p-t$, even if it is irreducible in $K[t]$, it is not separable, and if $C$ is an algebraic closure of $K$, this polynomial has only one root in $C$.

Let $c$ be the root of $x^p-t$ in $C$, i.e., $t=c^p$. We have that $x^p-t=x^p-c^p=(x-c)^p$, since $\text{char} F=p$.
Therefore, we have that $c$ is the only root in $C$ of multiplicity $p$, and so $x^p-t$ is not separable.

Is this correct? (Wondering)