Doubt in proof of Proposition 4.5 from Patrick Morandi's *Field and Galois Theory* Relevant definitions and results


*

*Let $F$ be a field and $f(x) = a_0 + a_1 x + \dots + a_n x^n \in F[x]$. The formal derivative $f'(x)$ of $f(x)$ is defined by $f'(x) = a_1 + 2a_2 x + \dots + n a_n x^{n-1}$.

*Two polynomials $f,g \in F[x]$ are said to be relatively prime if $\gcd(f,g) = 1$.

*Lemma 3.1. Let $f(x) \in F[x]$ and $\alpha \in F$. Then $\alpha$ is a root of $f$ if and only if $x - \alpha$ divides $f$.

I am reading Patrick Morandi's Field and Galois Theory, and on page 41, he states and proves the following proposition.

Proposition 4.5. Let $f(x) \in F[x]$ be a nonconstant polynomial. Then $f$ has no repeated roots in a splitting field if and only if $\gcd(f,f') = 1$ in $F[x]$.

In the proof, he first proves that $f$ and $f'$ are relatively prime in $F[x]$ if and only if they are relatively prime in $K[x]$, where $K$ is any extension field of $F$. Then, he proceeds as follows:
Proof. (contd.) Suppose that $f$ and $f'$ are relatively prime in $F[x]$. In particular, let $K$ be a splitting field of $\{ f, f' \}$ over $F$. If $f$ and $f'$ have a common root $\alpha \in K$, then $x - \alpha$ divides both $f$ and $f'$ in $K[x]$. This would contradict the fact that $f$ and $f'$ are relatively prime in $K[x]$. Therefore, $f$ and $f'$ have no common roots.
Conversely, let $f$ and $f'$ have no common roots in a splitting field $K$ of $\{ f, f' \}$, and let $d(x)$ be the greatest common divisor in $K[x]$ of $f(x)$ and $f'(x)$. Then $d$ splits over $K$ since $f$ splits over $K$ and $d$ divides $f$. Any root of $d$ is then a common root of $f$ and $f'$ since $d$ also divides $f'$. Thus, $d(x)$ has no roots, so $d = 1$. Therefore, $f$ and $f'$ are relatively prime over $K$; hence, they are also relatively prime over $F$. $$\tag*{$\blacksquare$}$$ 

My question is that, from the proof it seems that the author is proving the following result:

$f$ and $f'$ have no common roots in a splitting field $\iff$ $\gcd(f,f') = 1$ in $F[x]$.

However, is it obvious that $f$ has no repeated roots in a splitting field $\iff f$ and $f'$ have no common roots in a splitting field? No such result has been proved earlier. So, is the proof incomplete, or am I missing something obvious here?
 A: Yes, the proof is still incomplete, but the gap is easily filled. We just need to know that some standard identities involving derivatives are still valid for formal derivatives: if $f(x), g(x) \in F[x]$ and $a \in F$, then


*

*$(af(x))' = af'(x)$;

*$(f(x)g(x))' = f'(x) g(x) + f(x) g'(x)$.



Lemma. Let $f(x) \in F[x]$ be a nonconstant polynomial. Then, $f$ has no repeated roots in a splitting field if and only if $f$ and $f'$ have no common roots in a splitting field.
Proof. Let $f$ have a repeated root $\alpha$ in a splitting field $K$. Then, $f(x) = (x - \alpha)^2 g(x)$ for some $g(x) \in K[x]$. So, $f'(x) = 2(x - \alpha) g(x) + (x-\alpha)^2 g'(x)$. It is clear that $x - \alpha$ also divides $f'(x)$. By Lemma 3.1, $\alpha$ is a root of $f'$. So, $f$ and $f'$ have a common root in $K$.
Conversely, let $f$ have no repeated root in a splitting field $K$. Then, there exist $a \in F$ and $\alpha_1,\dots,\alpha_n \in K$ with $\alpha_i \neq \alpha_j$ if $i \neq j$, such that $$f(x) = a \prod_{i=1}^n(x-\alpha_i).$$ Then, $$f'(x) = a\sum_{j=1}^n \prod_{\substack{i=1 \\ i \neq j}}^n (x - \alpha_i).$$ None of the $\alpha_j$'s are a root of $f'$ because $f'(\alpha_j) = a\prod_{i \neq j} (\alpha_j - \alpha_i) \neq 0$. So, $f$ and $f'$ have no common root in $K$. $$\tag*{$\blacksquare$}$$
