Exercise 5.3.12

Proof. By Proposition 5.3.2(c), $f\in F[x]$ is separable if and only if $f$ is nonconstant and $f\wedge f^{\prime }=1$.

If $f,f^{\prime }$ have a common root $\alpha$ in an extension $L$ of $F$, then $x-\alpha$ divides $f$ in $L[x]$, and also $f^{\prime }$, so divides their gcd in $L[x]$, and so $\gcd (f,f^{\prime })\ne 1$ (we know that the gcd is the same in $F[x]$ and in $L[x]$).

We have proved that if $f\wedge f^{\prime }=1$, then $f,f^{\prime }$ have no common root in any extension of $F$.

Conversely, if $f\wedge f^{\prime }\ne 1$, then $f,f^{\prime }$ have a common nonconstant factor $g\in F[x]$. Let $L$ an extension of $F$ such that $g$ has a root $\alpha \in F$. Then $\alpha \in L$ is a root of $f$ and $f^{\prime }$.

Conclusion: $f\in F[x]$ is separable if and only if $f$ is nonconstant and $f$ and $f^{\prime }$ have no common roots in any extension of $F$. □