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Question: Why do we need to include the factor $a_{n}^{2n-2}$ in the discriminant of a polynomial?

Here is the definition of the discriminant ($\Delta$) in terms of the roots $r_1,r_2,...$:

$$ \Delta=a_{n}^{2n-2}\prod_{i<j}(r_i-r_j)^2 $$ for the polynomial $a_nx^n+a_{n-1}x^{n-1}+...+a_0$.

If the polynomial equation has non-real coefficients (in particular if $a_n$ is not real), then it is pointless to tell whether $\Delta>0$ and so in general I can't tell the number of real roots using this way for a polynomial equation with non-real coefficients.

However, for a polynomial equation with only real coefficients, even if we just consider $\prod_{i<j}(r_i-r_j)^2$, where $a_{n}^{2n-2}$ is not included, we still have:

$$\Delta >0, \text{if the number of complex roots} \equiv 0\mod 4$$ $$\Delta =0, \text{if there is a multiple root}$$ $$\Delta<0, \text{otherwise}$$

So, why isn't this the definition of discriminant? What difference does the factor $a_{n}^{2n-2}$ make in the non-real coefficients case?

Any help will be appreciated!

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1 Answer 1

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You can easily see that for a polynomial $f(x)=a_nx^n+…+a_0$ of degree $n$ $$ Res(f,f’)=(-1)^{\frac {n(n-1)}2}a_n\Delta$$ where $Res(f,f’)$ stays for the resultant of the polynomials $f$ and its derivative $f’$. Since $f$ and $f’$ have the same leading coefficient, i.e. $a_n$, then $Res(f,f’)\in a_n\Bbb Z[a_0,…,a_n]$ and thus $$\Delta\in \Bbb Z[a_0,…,a_n]$$ that’s the reason to have the exponent in the definition of the discriminant: in this way the discriminant of a polynomial $f$ is always a integer polynomial in the coefficients of $f$!

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  • $\begingroup$ So the factor $a_{n}^{2n-2}$ is included to ensure the coefficients of the discriminant (as a polynomial) must be integers? $\endgroup$
    – GHG
    Commented Aug 23, 2022 at 10:32
  • $\begingroup$ Yes, exactly @GHG $\endgroup$ Commented Aug 23, 2022 at 11:52

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