Why do $f$ and $f'$ generate all of $K[X]$? I have been studying Marcus' Number Fields. I am stuck on a remark in Appendix 2, page 258.
He says: A monic irreducible polynomial $f$ of degree n over $K$ (a subfield of $\mathbb{C}$) splits into n monic linear factors over $\mathbb{C}$. We claim these factors are distinct; any repeated factor would be a factor of the derivative $f'$. [This is clear to me.]
This is my question:
He says: But this is impossible since $f$ and $f'$ generate all of $K[X]$ as an ideal.
How would I know this $a$ $priori$?
He finishes using this saying $1$ is a linear combination of $f$ and $f'$ with coefficients in $K[X]$ to show $f$ and $f'$ have no common factors, etc., and thus $f$ has distinct roots in $\mathbb{C}$
I have no problem with the conclusion either. My only question is how to know $f$ and $f'$ generate all of $K[X]$ as an ideal.
Thanks as always for you help.
 A: Marcus says that $f$ is irreducible, so $(f,f')=1$. If not, then $f\mid f'$ (since the only non-constant divisor of $f$ is itself) and since $\deg f'<\deg f$ we have a contradiction.
A: Consider $I = \langle f, f' \rangle = \langle g \rangle$ for some $g$, since $K[x]$ is a PID.  Since $g \mid f$ and $f$ is irreducible, $g$ is associate to (i.e., equal to up to multiplication by a unit) either $1$ or $f$.  
Since $g \mid f'$ and (because the characteristic is zero!) $0 \leq \operatorname{deg} f' < \operatorname{deg} f$, we have $\operatorname{deg} g \leq \operatorname{deg} f' < \operatorname{deg} f$, so $g$ cannot be associate to $f$. Thus $g$ must be associate to $1$.
A: Consider $I = (f ,f')$ the ideal generated by $f$ and $f'$. It can be written $I = (P)$ with $P \in K[X]$ monic (since $K[X]$ is principal). Clearly, $(f) \subset I$ so $P | f$. And because $f$ is irreducible, we get either $P = 1$ of $P = f$. But clearly $f' \notin (f)$, so $P = 1$.  
A: Hint: the gcd of $f$and  $f'$ is $1$ hence by Bezout, there exist $u,v\in K[X]$ such that
$$
uf+vf'=1.
$$
Note: if you know that the gcd remains the same in every extension (by Bezout and its converse, for instance), then you see directly that $gcd(f,f')=1$ in $\mathbb{C}[X]$. So $f$ can't have a repeated root. And you get a little shortcut.
