Repeated Roots and Being Relatively Prime w/Derivative

[Bashyboy]

Homework Statement

Let $f(x) = (x-a_1)...(x-a_n) \in k[x]$, where $k$ is a field. Show that $f(x)$ has no repeated roots (i.e., all the ai are distinct elements in $k$) if and only if $gcd(f,f')=1$, where $f'(x)$ is the derivative of $f$

Homework Equations

$(x-a)^2 |f(x)$ implies $(x-a)|f'(x)$

$(x-a)|f(x)$ and $(x-a)|f'(x)$ implies $(x-a)^2|f(x)$

The Attempt at a Solution

First note that $f(x) = (x-a_1)...(x-a_n)$ has a repeated roots if and only if $(x-a_k)^p$ is a factor of $f(x)$ for some $k \in \{1,...,n\}$ and $p \ge 2$.

Suppose that $f(x)$ has no repeated. Note that $f(x) = (x-a_1)...(x-a_n)$ is the prime factorization of $f(x)$. Now if were the case that $gcd(f,f') \neq 1$, then both $f$ and $f'$ would have a common prime factor. Since we know what $f(x)$'s prime factors look like, we know there is a $k$ such that $x-a_k$ divides $f'$. But the second theorem cited above implies that $(x-a_k)^2$ divides $f(x)$ and therefore it has a repeated root. Hence, $gcd(f,f')$ must be $1$.

Now suppose that $(f,f')=1$. If $f(x)$ had a repeated root, then $(x-a_k)^2$ would divide it, for some $k$. But the first theorem cited above would imply $(x-a)|f'(x)$, contradicting the fact that $(f,f')=1$.

How does this sound?

 

[fresh_42]

Sounds correct, although the whole work is done in the two statements you use. You could as well simply differentiate $f(x)=\prod(x-a_k)^{n_k}$ and directly see the result.