I am going over my lecture notes in my Field theory class and I saw this following statement without a proof: if $\mathrm{char}(F)=p>0$ then if $g(x)\in F[x]$ is irreducible then $g(x)$ have multiple roots iff $g'(x)=0$.
I believe I can prove that if $g(x)$ have multiple roots then $g'(x)=0$ but I am not sure and I am unable to prove that the converse is also true.
My reasoning is as follows: $g(x)$ have multiple roots imply there is an extension $K/F$ and $\alpha\in K$ s.t $g(\alpha)=g'(\alpha)=0$ (since if the multiplicity of $\alpha$ in $g(x)$ is $m>1$ (since it is not a simple root) then the multiplicity of $\alpha$ in $g'(x)$ is at least $m-1\gt 0$). Since $g(x)$ is irreducible and we may assume WLOG that $g(x)$ is monic then it follows that the minimal polynomial of $\alpha$ over $F$ is $g(x)$ , but $g'(\alpha)=0$ and if $g'\neq0$ then $\deg(g')<\deg(g)$ and this is a contradiction.
Is my argument correct, and how can I prove the converse ? help is appreciated!