Let $K$ be a number field and $\theta $ be an algebraic integer such that $ K = \mathbb Q(\theta )$. Let $ d $ be the discriminant of the basis $\{ 1, \theta, ..., \theta ^{ n -1 } \}$ of $K$ over $\mathbb Q$ and $g (t)$ be the minimal polynomial of $\theta$ over $\mathbb Q$. (I already shown that $ d \in \mathbb Z$ and $ g(t) \in \mathbb Z [t] $ ). The question is : if $p$ is a prime number not dividing $d$, show that the polynomial $ \bar g (t) \in \mathbb F_p [t] $ has no multiple roots in its splitting field.
I showed that $ d = \Pi _{ i < j } ( \theta_i - \theta_j ) ^2 $ where $ \{ \theta_1,..., \theta_n \}$ are roots of $g(t)$ in some splitting field. And I know that $ g ' ( \theta_i ) | d $ for $ i = 1,...,n $. But this does not quite help me as I don't have any idea how the roots of $ \bar g(t) $ in its splitting field related to $ \theta_1,...,\theta_n $. Any help is appreciated.
Add: After thinking about it longer, I found that it suffices to show $ disc (\bar g ) = \bar d$. And I believe this is true but I just can't come up with a proof. Can anyone help me for this?
The discriminant is proportional to the resultant of $g$ and $g'$ (in any field) (see for example wikipedia). If $p$ does not divide $d$, $d$ is nonzero in ${\mathbb F}_p$, so the resultant of $g$ and $g'$ is nonzero in ${\mathbb F}_p$, so $g$ has only simple roots in ${\mathbb F}_p$.