Hyperdiscriminants of a polynomial over a ring Let $f=\sum_ic_ix^i\in \mathbb K[x]$ for $\mathbb K$ algebraically closed, with roots $\alpha_1,\dots ,\alpha_n$ (possibly repeated). The theory of elementary symmetric polynomials ensures that if $f$ is monic, there is for each $1\leq i\leq n$ a unique polynomial $\Delta_i(f)\in \mathbb Z[x_1,\dots ,x_n]$ such that $$\Delta_i(f)(c_0,\dots ,c_{n-1})=\sigma_i(f^\prime(\alpha_1),\dots,f^\prime(\alpha_n)).$$
Let us call $\Delta_i(f)$ the $i^\text{th}$ hyperdiscriminant of $f$.
I have read that one can describe $\Delta_i(f)$ for $f\in R[x]$ with $R$ any commutative ring. That is, without making any use of (possibly non-present) roots of $f$ in $R$, but using only the coefficients. How to do this?
 A: The $\Delta_i$ you have defined does not seem to me to depend on $f$ at all, but just to be an element of $\mathbb{Z}[x_1,\dots,x_n]$ that one can write down as soon as one knows $n$ and $i$. Since $\mathbb{Z}[x_1,\dots,x_n]$ has a homomorphism to any given ring $R$ sending $x_1,\dots,x_n$ to any chosen $c_0,\dots,c_{n-1}\in R$, it makes sense to look at the image of $\Delta_i$ under this map. In this way, one gets a value of $\Delta_i$ (not a polynomial but an actual element of $R$) for any given polynomial $f\in R[x]$ with coefficients $c_0,\dots,c_n\in R$. 
I think it is only in this latter sense that the thing you're defining wants to depend on the polynomial. (I.e. as a polynomial, it is independent of the specific polynomial to which it is being applied, but when a polynomial is given, it then specifies a value in the ground ring by being evaluated on the coefficients of the given polynomial.)
What I am getting at, with respect to the question "how to define $\Delta_i$ over a general ring?", is that $\Delta_i$ is actually defined without reference to any specific ring at all! One can calculate $\Delta_i$ as soon as one has $i,n$. In order to have a setting in which to conduct the calculation, one can do all the work in the ring $A = \mathbb{Z}[\alpha_1,\dots,\alpha_n]$, where $\alpha_1,\dots,\alpha_n$ are regarded as formal indeterminates. Then, let $F = \prod(x-\alpha_i)\in A[x]$, and let $C_i = \sigma_{n-i}(\alpha_1,\dots,\alpha_n)\in A$, for $i=0,\dots,n-1$. This construction is the universal context for defining something that is supposed to relate to the roots of a monic polynomial, i.e. if $R$ is any ring with $f\in R[x]$ a polynomial that splits in $R$, then the $A$ I have just defined has a homomorphism to $R$ such that $f$ is $F$'s image in $R[x]$. The homomorphism is obtained by mapping the $\alpha_i$'s to the roots of $f$ in $R$, and then it will also map the $C_i$'s to the coefficients of $f$. But note that although $A$ and $F$ have this universal property having to do with roots of things, it was possible to define $A$ and $F$ from scratch, without assuming the existence of any dubious roots in any particular ring.
The whole calculation can now be performed over $A$, arriving at a polynomial $\Delta_i\in \mathbb{Z}[x_1,\dots,x_n]$. There is no definitional challenge anywhere in this construction. Furthermore, the fundamental theorem on symmetric polynomials gives us that the construction of $\Delta_i$ this way is unique. See below for an example.
Once one has the $\Delta_i$ constructed this way, one can post facto apply it to any given polynomial $f$ over any ring $R$, by evaluating it on the coefficients $c_i$ of $f$, obtaining an element $\Delta_i(c_0,\dots,c_{n-1})$ of $R$. (It is only this element that I am inclined to call $\Delta_i(f)$, since up til now, $\Delta_i$ didn't depend on $f$. I.e. I think it should be that $\Delta_i(f)$ is a synonym for $\Delta_i(c_0,\dots,c_{n-1})$, and $\Delta_i(f)(c_0,\dots,c_{n-1})$ is redundant notation.) At this point, one can ask, "What is $\Delta_i(f)$ telling us about $f$?"
I think it is only in answer to this last question that the matter of the existence of $f$'s roots enters. If $f$'s coefficients are in an algebraically closed field, then this field also contains roots $\alpha_1,\dots,\alpha_n$ (now specific elements of a field and no longer indeterminates -- apologies for the notational collision), and then one has the interpretation $\sigma_i(f'(\alpha_1),\dots,f'(\alpha_n))$ for $\Delta_i(f)$. More generally, if $f$'s coefficients are in any integral domain, then this domain has a fraction field, and this fraction field has an algebraic closure, and the latter contains roots $\alpha_1,\dots,\alpha_n$ for $f$, and again, one can interpret $\Delta_i(f)$ as $\sigma_i(f'(\alpha_1),\dots, f'(\alpha_n))$. But if the coefficients of $f$ live in a ring $R$ that is not an integral domain, then one can still calculate $\Delta_i(f)$ as an element of $R$. 
The only obstruction is to interpreting this element of $R$ as having an expression in terms of $f$'s roots.
Even here, there is no real obstruction: one can always find an overring $\tilde R$ of $R$ in which $f$ splits, by suitable adjunctions to $R$, and then $\Delta_i(f)$ can be interpreted as $\sigma_i(f'(\alpha_1),\dots,f'(\alpha_n))$, with the latter calculation happening in $\tilde R$. We know this because the identity $\Delta_i(F) = \sigma_i(F'(\alpha_1,\dots,\alpha_n))$ holds inside $A$, the "universal" ring -- in fact, we built $\Delta_i$ precisely to satisfy this identity in this universal context -- and because $f$ splits in $\tilde R$, there is a homomorphism that maps $F\in A[x]$ to $f\in R[x]$ and the roots of $F$ in $A$ to the roots of $f$ in $\tilde R$, and the identity rides this homomorphism right into $\tilde R$. If $\tilde R$ is not an integral domain, it is possible that $f$ may not split uniquely, but this is okay: $\Delta_i(f)$, which is uniquely defined by the above construction, must equal $\sigma_i(f'(\alpha_1),\dots,f'(\alpha_n))$, for any $\alpha_1,\dots,\alpha_n$ satisfying $f(x) = (x-\alpha_1)\dots (x-\alpha_n)$.
TL;DR: I guess it comes down to this: the apparent definitional difficulty springs from treating an $f$, contingent on a specific ground field $R$, as the starting point of the construction of $\Delta_i$. I am arguing that a contingent $f$ is not actually the starting point. $\Delta_i$ is expressing a formal relationship between the coefficients of any monic polynomial and its roots, so the true starting point is the universal monic polynomial
$$f = (x-\alpha_1)\dots(x-\alpha_n),$$
where $\alpha_1,\dots,\alpha_n$ are purely formal indeterminates. Thus the roots are already built into the setup in which $\Delta_i$ is calculated, and actually they are best seen not as roots of a preexisting $f$ but instead as indeterminates from which $f$ is constructed in such a way that they become its roots.
This is just like Vieta's formulas. The statement $\Delta_i(f) = \sigma_i(f'(\alpha_1),\dots,f'(\alpha_n))$ has the exact same philosophical status as the statement that the linear coefficient of $f$ is $-\sigma_1(\alpha_1,\dots,\alpha_n)$. One proves it by doing a purely formal calculation with $(x-\alpha_1)\dots(x-\alpha_n)$. Then the truth of it carries into any ring at all (domain or not) along the homomorphism that maps the formal indeterminates $\alpha_1,\dots,\alpha_n$ into that particular ring.
Example of how the calcuation goes:
Take the case that $n=3, i=1$. Let
$$F = (x-\alpha_1)(x-\alpha_2)(x-\alpha_3).$$
One can see this statement as taking place in some $\mathbb{K}[x]$, with $\mathbb{K}$ some algebraically closed field, but critical to what I'm saying is that one could equally well be working in $A[x]$, where $A$ is the ring $\mathbb{Z}[\alpha_1,\alpha_2,\alpha_3]$, and $\alpha_1,\alpha_2,\alpha_3$ are formal indeterminates. I.e. $A$ is the universal ring with three indeterminates.
Then one has $C_0 = -\sigma_3(\alpha)$, $C_1 = \sigma_2(\alpha)$, and $C_2 = -\sigma_1(\alpha)$, where $\sigma_i$ is the $i$th elementary symmetric polynomial, and $\alpha$ is shorthand for $\alpha_1,\alpha_2,\alpha_3$.
Now,
$$F' = (x-\alpha_2)(x-\alpha_3) + (x-\alpha_1)(x-\alpha_3) + (x-\alpha_1)(x-\alpha_2),$$
so that $F'(\alpha_1) = (\alpha_1 - \alpha_2)(\alpha_1-\alpha_3)$, etc. Again, note, all this makes sense working in $A[x]$, where $A$ is the universal ring defined above. Seeing it this way, $C_0,C_1,C_2$, and $F'(\alpha_1),F'(\alpha_2),F'(\alpha_3)$ all refer to specific elements of the ring $A$.
We want to find $\Delta_1$. This is the polynomial that is going to express $\sigma_1(F'(\alpha_1),F'(\alpha_2),F'(\alpha_3))$ in terms of $C_0,C_1,C_2$. We have
\begin{align}
\sigma_1(F'(\alpha_1),F'(\alpha_2),F'(\alpha_3)) &= (\alpha_1-\alpha_2)(\alpha_1 - \alpha_3) + (\alpha_2 - \alpha_1)(\alpha_2 - \alpha_3) + (\alpha_3 - \alpha_1)(\alpha_3 - \alpha_2)\\
&=\alpha_1^2 + \alpha_2^2 + \alpha_3^2 - \alpha_1\alpha_2 - \alpha_1\alpha_3 - \alpha_2\alpha_3\\
&=C_2^2 - 3C_1.
\end{align}
Thus, $\Delta_1(C_0,C_1,C_2) = C_2^2 - 3C_1$, and thus as an element of $\mathbb{Z}[x_1,x_2,x_3]$ we have
$$\Delta_1=x_3^2 - 3x_2.$$
NB: It's not a coincidence that $C_0$ and thus $x_1$ doesn't appear here: it is eliminated from the info in $f$ when one takes the derivative. So my notational aesthetics would rather regard $\Delta_i$ as an element of $\mathbb{Z}[x_1,\dots,x_{n-1}]$, with $x_i$ mapping to $C_i$, in which case I would write $\Delta_1$ in the present case as $x_2^2 - 3x_1$.
