Ring homomorphism takes discriminant to discriminant Let $R[x] \xrightarrow{\sigma} S[x]$ be a ring homomorphism where $R,S$ are integral domains of characteristic $0$. Is it true that for any monic polynomial $f(x) \in R[x],\sigma(disc(f(x)))=disc(\sigma(f(x)))$.
My approach was as following: If $\alpha_1,\ldots,\alpha_n$ are roots of $f(x)$ in some extension of $Frac(R)$,then $disc(f(x))=\prod_{1\leq i<j\leq n}(\alpha_i-\alpha_j)^2$.Somehow,if $\sigma$ could somehow be extended such that $\sigma(\alpha_i)$ makes sense,then they would be roots of $\sigma(f(x))$ and since there are $n$ such,they are all the roots and we would be done.
 A: Basically, yes, assuming your homomorphism is a "psychologically pleasing one." That is to say:  assuming $\sigma(x)=x$.
The discriminant is a constant times a polynomial in the coefficients of the polynomial, $f$, you can see that by comparing with the resultant formula. I.e. if $f(x)=a_nx^n+\ldots a_1x+a_0$ then the discriminant is $cp(a_n,\ldots, a_0)$ for some polynomial $p(x)$ and $c\in\text{Frac}(F)$ (given explicitly by the resultant formula). But then any ring homomorphism commutes with such a map, i.e. $\sigma(cp(a_n,\ldots, a_0))=\sigma(c)p(\sigma(a_n),\ldots,\sigma(a_0))$, so the image of the discriminant is the discriminant of the image, assuming your $\sigma$ sends $x$ to $x$.
If you don't impose that condition, note $R=\Bbb Z$, $S=\Bbb Z[i]$ and $\sigma(x)=i$, and $\sigma(n)=n, \forall n\in\Bbb Z$ is a counterexample, because you have $f(x)=x^2+1$ has discriminant $D=-4$, but the discriminant of $\sigma(f(x))$ is just the discriminant of the constant $0$, i.e. $0$.
A: I’m glad that Adam Hughes answered your question first, because in the answer I was preparing, I was completely unwarrantedly assuming that $\sigma(x)=x$.
I would have complained that you were using a very unhelpful definition of the discriminant, and then I looked into Wikipedia, whose article seems to me to be seriously deficient. From the definition you used, it’s not at all hard to show that the discriminant also is (up to $\pm1$) the much more handleable $\prod_\rho f'(\rho)$, the product being taken over the roots of $f$. From this, I think you can persuade yourself that the foregoing product is also the determinant of $f'(x)$ as an $R$-linear transformation (in the regular representation) of the $R$-free module $R[X]\big/\bigl(f(X)\bigr)$. This brings the whole mess down to the level of the coefficients of $f$, without ever mentioning the roots of $f$. The polynomial in the coefficients that you get is the same as the one gotten by the far messier process involving the dreaded resultant.
