0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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

  1. $(af(x))' = af'(x)$;
  2. $(f(x)g(x))' = f'(x) g(x) + f(x) g'(x)$.

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 $K$. $$\tag*{$\blacksquare$}$$

$\endgroup$
  • $\begingroup$ Hey there! I am quite sure i have seen this in Herstein Chapter 5. A humble suggestion would be to supplement Morandi with Herstein so that you proceed smoothly. $\endgroup$ – crskhr Jun 19 '18 at 14:39
  • $\begingroup$ @crskhr I’ll take a look, thank you :) $\endgroup$ – Brahadeesh Jun 19 '18 at 14:39
  • $\begingroup$ Page 233, Lemma 5.5.2 pa :D $\endgroup$ – crskhr Jun 19 '18 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.