$f\in F[x]$ is separable iff its image under a homomorphism is separable

Question

I am trying to prove that $f\in F[x]$ is a separable polynomial iff $\phi (f) \in K[x]$ is a separable polynomial, where $\phi:F\to K$ is a homomorphism of fields.

$f \in F[x]$ is separable iff $f$ and $Df$ are relatively prime, where $Df$ is the derivative of $f$.

$f \in F[x]$ is separable iff $(f,Df)$ is the unit ideal.

Here is what I have:

$f\in F[x]$ is separable

$\iff$ any common divisor $p\in F[x]$ of $f$ and $Df$ is a unit in $F[x]$.

$\iff \forall p\in F[x]$, such that $f=pg$ and $Df=ph$ for some $g,h\in F[x]$, $p$ is a unit in $F[x]$

$\iff \forall p\in F[x]$, such that $\phi (f)=\phi(p) \phi(g)$ and $\phi(Df)=\phi(p)\phi(h)$, $\phi(p)$ is a unit in $K[x]$

$\iff$ $\phi(f)$ and $\phi(Df)$ are relatively prime in $K[x]$.

$\iff$ $\phi(f)\in K[x]$ is separable.

However, I am not convinced by this as I am not sure if $p$ is common divisor of $f$ and $Df$ $\iff$ $\phi(p)$ is a common divisor of $\phi(f)$ and $\phi(Df)$.

What do you think?

Answer 1

See K. Conrad Corollary 2.7.

Forward direction: if we can write $a(x) f(x) + b(x) Df(x) = 1$, then we can apply $\phi$ to get a similar relation on $\phi(f)(x)$ and $\phi (Df)(x)$.

Backwards direction (by contrapositive): suppose that $a(x) | f(x)$ and $a(x) |(Df)(x)$ where $a(x)$ is non-constant. Then $\phi(a)(x)$ is nonconstant and divides both $\phi(f)(x)$ and $\phi(Df)(x)$.