# Discriminant ideal in $\mathbb{Z}$

Given a monic polynomial $f \in \mathbb{Z}[X]$, I would like to consider the ideal $$(f, f')_{\mathbb{Z}[X]} \cap \mathbb{Z}$$ in $\mathbb{Z}$. In particular: is it true that this is generated by the discriminant $\Delta(f)$?

I know at least that $\Delta(f)$ is contained in this ideal.

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How do you define $\Delta(f)$? There are many known definitions of the discriminant out there. –  Patrick Da Silva Nov 26 '12 at 10:52
In On resultants, Proc. Amer. Math. Soc. 89 (1983), 419-420, available at ams.org/journals/proc/1983-089-03/S0002-9939-1983-0715856-2 I proved that if $c$ generates your ideal then the resultant of $f$ and $f'$ divides $c^n$, where $n$ is the degree of $f$. As the discriminant can also be defined in terms of this resultant, you might find the paper useful. –  Gerry Myerson Nov 26 '12 at 11:53
@PatrickDaSilva I define it as the product over differences of zeros in $\mathbb{C}[X]$. –  Dog Nov 26 '12 at 13:17
@GerryMyerson Thanks for the information! –  Dog Nov 26 '12 at 13:18
@Patrick, given $a$ in $B$, the map $T_a:B\to B$ given by $T_a(b)=ab$ is linear. You can get a matrix for it by choosing a basis. The obvious choice is $1,x,x^2,\dots,x^{n-1}$, where $n$ is the degree of $f$; the matrix you get is called the right regular representation of $a$. A different choice of basis may yield a different matrix, but the determinant of the matrix is independent of the choice of basis. –  Gerry Myerson Nov 26 '12 at 21:54

Take $f(x)=x^2+1$ as an example. Then $f'(x)=2x$ and $2=2f(x)-xf'(x) \in (f,f')_{\Bbb Z[X]} \cap \Bbb Z$, but $\Delta(f)=-4$.