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I am reading Rational Points on Elliptic Curves by Silverman and Tate, and on page 48 authors make the following remark: (bold emphasis mine)

If $f(x)$ is any polynomial with leading coefficient $1$ in the ring $\mathbb{Z}[x]$ of polynomials with integer coefficients, then the discriminant of $f(x)$ will always be in the ideal of $\mathbb{Z}[x]$ generated by $f(x)$ and $f'(x)$. This follows from the general theory of discriminants, but for our particular polynomial $f(x) = x^3 + ax^2 + bx + c$, the quickest proof is just to write out an explicit formula:

$$D = \left\{(18b - 6a^2)x - (4a^3 - 15ab + 27c)\right\} f(x) +\left\{(2a^2- 6b)x^2+(2a^3 - 7ab+9c)x+(a^2b+3ac- 4b^2)\right\}f'(x)$$ We leave it to you to multiply this out and verify it is correct.

Could someone show how this statement follows from the general theory of discriminants? I would be delighted to learn :)

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up vote 5 down vote accepted

This follows from two basic facts that can be found in most books:

  1. The discriminant (up to sign!) of $f$ is the resultant of $f$ and $f'$.
  2. If $P$ and $Q$ are any two polynomials, there always exists two polynomials $A$ and $B$ such that $AP+BQ=\text{res}(P,Q)$.
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@YACP Of course, I edited accordingly. Thank you! – Alex Youcis Aug 10 '13 at 17:48

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