$R[X]/(f)$ separable $\iff \Delta(f) \in R^*$? Let $R$ be a commutative ring with $1$. Let $f \in R[X]$ be monic. I have to prove the following:
$$
\text{The discriminant of } f \text{ is invertible}
\quad \quad
\iff
\quad \quad
R[X] / (f) \text{ a is separable algebra over R}
$$
Could you please help me to do so?

The definitions I work with
For me the discriminant of a polynomial $f$ is understood to be
$$
\Delta(f) \ := \ \prod_{1 \leq i < j \leq n} (\alpha_i - \alpha_j)^2.  
$$
We call an algebra $B$ separable iff the following map is bijective:
$$
\Phi \ : \ B \rightarrow Hom(B,R) \ : \ 
x \ \longmapsto \ \left(y \mapsto Tr(xy)\right)
$$
where the trace of an element $b \in B$ is defined as the trace of the map
$ x \mapsto bx$.

My own thoughts
We tried to use another fact about separability of algebras
from this source:

Let $R$ be a commutative ring with $1$, B an $R$-algebra with basis 
  $\{w_1, w_2, \dots, w_n \}$, then 
  $$
\det(Tr(w_iw_j):1 \leq i,j \leq n ) \text{ is invertible}
\quad \quad
\iff
\quad \quad
\text{ B is a separable algebra over $R$}
$$
  Where $Tr(a)$ is a notation for the trace of the map
  $$
B \ \longrightarrow \ B \ : \ x \ \longmapsto \ ax
$$

Thus it would make sense to try to prove that $\Delta(f) = (Tr(w_iw_j):i,j)$, but we only managed to prove this for polynomials of degree two. In the general case both sides are very hard to calculate

Some computational effort
Lets write $f(X) = a_0 + a_1X + \dots + a_nX^n$. 
In our case $\{1,X,X^2,\dots, X^{n-1}\}$ is a basis for $A[X]/(f)$.
The determinant we have to calculate looks like if $n=3$:
$$
 \left| \begin{array}
        Tr(1) & Tr(X) & Tr(X^2) \\
        Tr(X) & Tr(X^2) & Tr(X^3) \\
        Tr(X^2) & Tr(X^3) & Tr(X^4) \\
        \end{array}
\right|
$$
and is similar for other $n \in \mathbb{N}$. Below I "sketched" what the matrix of a map $f(X) \mapsto X^j f(X)$ looks like.
$$ \left[
    \begin{array}{cccc|cccc}
      0& 0& \dots &0 \\
      \vdots& \vdots & & \vdots \\
      0& 0& \dots &0 \\
      \hline
      1 &0 &\dots &0 & \\
      0 &1 &\dots &0 && \\
      \vdots & \vdots &  \\
      0 &0 & \dots &1  &&&&  \\
    \end{array}
\right] $$
The empty parts are hard to spell out. At least we can see this way that for taking the trace we only need te consider the elements on the right.
This is still hard, though. The discriminant isn't easy to calculate either.

Could you please help me to solve this? Feel free to use any method you like.
 A: Here is a quick way to prove it, but quick only because a lot of things are considered well known.
A finite type extension $S/R$ of commutative rings is separable if and only if $\Omega_{S/R}$, the module of Kähler differentials, is trivial.
Let's show that this is equivalent to $f$ having an invertible discriminant.
For sake of clarity: I usually define $\Delta(f)$ as the resultant of $f$ and $f'$.
In the special case where $f$ is monic, it turns out that $\Delta(f):=\operatorname{res}(f,f')$ equals the norm of $f'\in R[x]/(f)$ over $R$, i.e., the determinant of the $R$-linear endomorphism of $R[x]/(f)$ given by multiplication with $f'$.
In particular, $f'$ is a unit in $R[x]/(f)$ if and only if $\Delta(f)$ is a unit in $R$.
Associated with the surjection $S:= R[x]\to S' := R[x]/(f)$ with kernel $(f)$ we get an exact sequence
$$(f)/(f)^2\to \Omega_{S/R}\otimes_{S} S'\to\Omega_{S'/R}\to 0.$$
We aim to show that $\Omega_{S'/R} = 0$.
As $S = R[x]$, it is well known that $\Omega_{S/R} = S\mathrm{d}x$, so $\Omega_{S/R}\otimes_{S} S' = S'\mathrm{d}x$.
For $\Omega_{S'/R} = 0$ to hold, the leftmost map, on $h+(f^2)\in (f)/(f^2)$ defined as $\mathrm{d}h\otimes 1$, has to be surjective.
Plugging in a general element $h = fg$ shows that this holds if and only if $f' $ is a unit in $S'$.
By the previous remarks this shows that $S'/R$ is separable if and only if $\Delta(f)$ is a unit in $R$, as claimed.
Addendum: A rather elementary proof, closer to the OPs definitions. I found it in the (german) book on algebra by Bosch.
First, allow me another definition of the discriminant as indicated by the OP.
It differs from $\operatorname{res}(f,f')$ in a sign.
To clearly distinguish them, I'll denote the new one by $\delta_f$ instead.
Let me use the following without proof (key word: fundamental theorem of symmetric polynomials).
For each $n\in\mathbb{N}$ there exists a kind of universal discriminant for polynomials of degree $n$ as follows.
Let $s_0,s_1,\dots,s_n\in \mathbb{Z}[t_1,\dots,t_n]$ be the elementary symmetric polynomials; more precisely, the unique polynomials such that $\prod_{i=1}^n(x-t_i)=\sum_{i=0}^n s_ix^{n-i}$ in $\mathbb{Z}[x,t_1,\dots, t_n]$.
They are algebraically independent and generate the sub-ring of symmetric polynomials, i.e., the map
$$\mathbb{Z}[\sigma_1,\dots,\sigma_n]\to\mathbb{Z}[t_1,\dots,t_n], \sigma_i\mapsto s_i,$$
is injective and its image consists precisely of the symmetric polynomials.
In particular, there is a unique polynomial $\Delta^n\in\mathbb{Z}[\sigma_1,\dots,\sigma_n]$ which is mapped to the symmetric polynomial $\prod_{i<j}(t_i-t_j)^2$.
Now, if $R$ is any commutative ring and $f = x^n + f_{1}x^{n-1}+\dots+f_{n-1}x+f_n\in R[x]$, consider the unique ring map $\chi_f\colon \mathbb{Z}[\sigma_1,\dots,\sigma_n]\to R$ defined by $\sigma_i\mapsto f_i$.
Then the discriminant of $f$ is defined as $\delta_f := \chi_f(\Delta^n)$; in other words, $\delta_f = \Delta^n(f_1,f_2,\dots  f_n)$.
Note that, whenever $R\subset R'$ is such that $f = \prod_{i=1}^n(x-\alpha_i)$ with $\alpha_i\in R'$, since $f_i = \sigma_i(\alpha_1,\dots,\alpha_n)$, we get
$\delta_f = \Delta^n(f_1,\ldots,f_n) = \prod_{i<j}(\alpha_i-\alpha_j)^2,$
as expected.
This construction shows that if $\varphi\colon R\to R'$ is any ring homomorphism and $f\in R[x]$ is any monic polynomial, then $\delta_{\varphi(f)} = \varphi(\delta_f)$. This will be important later.
Now to the actual problem.
Recall that if $f\in R[x]$ is monic of degree $n$, then $S:=R[x]/(f)$ is free as $R$-module of rank $n$, with basis $1,x,\dots x^{n-1}$.
This allows the definition of $S/R$ to be separable via non-degeneracy of the trace form, as the OP does.
If we identify $\mathrm{Hom}(S,R)$, as $R$-module, with $R^n$ via the basis dual to the one chosen above, we see that $S/R$ is separable if and only if the matrix
$(\mathop{Tr}(x^{i+j}))_{0\leq i,j\leq n-1}$
is invertible.
Since this matrix is invertible if and only if its determinant is invertible, we get the equivalence cited by the OP.
We will reduce the problem to a special case, so this determinant needs a name; given a monic $f\in R[x]$ of degree $n$, let
$$\mathrm{disc}_{R,f} := \det(\mathop{Tr}(x^{i+j}))_{0\leq i,j\leq n-1}\in R.$$
We aim to show that $\mathrm{disc}_{R,f} = \delta_f$.
Observe that if $\varphi\colon R\to R'$ is a ring homomorphism and $f\in R[x]$ is a monic polynomial, then so is $\varphi(f)\in R'[x]$ and $\mathrm{disc}_{R',\varphi(f)} = \varphi(\mathrm{disc}_{R,f})$, since the trace behaves accordingly.
In particular,


*

*if $\delta_f = \mathrm{disc}_{R,f}$, then $\delta_{\varphi(f)} = \mathrm{disc}_{R',\varphi(f)}$ and

*$\delta_f = \mathrm{disc}_{R,f}$ if and only if $\delta_{\varphi(f)} = \mathrm{disc}_{R',\varphi(f)}$ in case $\varphi$ is injective.


As a first reduction step, let $f = x^n + f_{1}x^{n-1}+\dots+f_{n-1}x+f_n\in R[x]$ and consider the map $A:=\mathbb{Z}[\sigma_1,\dots, \sigma_n]\to R$, $\sigma_i\mapsto f_i$.
Then we see that to prove the claim for $f$, it suffices to prove it for $\sum_{i=0}^n\sigma_ix^{n-i}\in A[x]$.
In the next step, we will use the inclusion $A\subset \mathrm{Frac}(A) = \mathbb{Q}(\sigma_1,\dots,\sigma_n) =: k$.
Since it is an inclusion, it suffices to prove the claim for $f:= \sum_{i=0}^n\sigma_ix^{n-i}\in k[X]$.
$f$ is irreducible.
Therefore, we have reduced the problem to the case of an algebraic field extension $L:=k[x]/(f)$ over $k$, where $f$ is irreducible of degree $n$.
Choose an algebraic closure $\overline{k}$ of $k$ and let $f = \prod_{i=1}^n(x-\alpha_i)$ with $\alpha_i\in \overline{k}$.
Here comes the final trick: in this special situation, we have $\mathrm{Tr}_{L/k}(x^j) = \sum_{\ell=1}^n\alpha^j_\ell$.
The case $j=1$ is easy to see as $-\mathrm{Tr}(x)$ is the degree $(n-1)$-coefficient of the minimal polynomial of $x$, which is $f$.
The case $j>1$ follows from the former case by basic eigenvalue-theory. 
Finally, observe that
$$\mathrm{disc}_{k,f} = \det\left(\sum_{\ell=1}^n\alpha_\ell^{i+j}\right)_{i,j} = \det(V^tV) = \det(V)^2 = \left(\prod_{i<j}^n(\alpha_i-\alpha_j)\right)^2=\delta_f,$$
where $V = (\alpha_\ell^j)_{\ell,j}$ (here $1\leq \ell\leq n$, but $0\leq j\leq n-1$) is the Vandermonde matrix.
