I have a question on this simple corollary from Abstract Algebra by Saracino:
Corollary 24.11 If $F$ is of characteristic $0$ and $f(X) \in F[X]$ is irreducible over $F$, then $f(X)$ has distinct roots.
Proof. If the leading coefficient of $f(X)$ is $c_n \neq 0$ then the leading coefficient of $f'(X)$ is $nc_n \neq 0$ (since $F$ is of characteristic $0$), so $f'(X)$ is not the zero polynomial. Thus $f(X)$ has distinct roots.
Why do we need the characteristic of $F$ to be $0$? I assume here that $n \neq 0$ because otherwise we'd have $f(X) = c_0$ for some $c_0 \in F$ (and in this book, the term "irreducible" is only defined for nonconstant polynomials). Then $n \neq 0$ and $c_n \neq 0$ imply that $nc_n \neq 0$, since $F$ is a field and hence does not have any nonzero zero-divisors. (The book does not explicitly state that $F$ is a field, but I assume it must be since "characteristic" in this book is defined only for fields.) So I'm not seeing how the characteristic of $F$ matters here.
Best Answer
If the characteristic of $F$ divides $n$, then, in $F$, $n=0$. So, $nc_n=0$ too.
For instance, if the characteristic of $F$ is $2$ and if $F(x)=x^4$, then $F'(x)=4x^3=0$.