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I'm having trouble understanding a part of the proof for the final theorem of this article https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089506003272.

The discriminant of $g(x)=x^4+px^2+qx+r$ is

$-27p^4-108p^3q-162p^2q^2-108pq^3-27q^4+256r^3$

We can rewrite $f(x)=x^4+ax^3+bx^2+cx+d = (x+\frac{a}{4})^4+(b-\frac{3}{8}a^2)x^2+(c-\frac{a^3}{16})x+(d-\frac{a^4}{256})$.

If we let

$p= b-\frac{3}{8}a^2$, $q=c-\frac{a^3}{16}$ and $r= d-\frac{a^4}{256}$ then the discriminant of $f$ is

$2^{24} \cdot(-27p^4-108p^3q-162p^2q^2-108pq^3-27q^4+256r^3)$

I don't understand why this is true since $f$ does not have the same shape as $g$ because of $(x+\frac{a}{4})^4$.

Also, where did the constant $2^{24}$ come from? Does it depend on the linear transformation $x \mapsto x+\frac{a}{4}$? Does it have to do with the fact that the discriminant is a homogeneous polynomial of degree $2n-2=2(4)-2=6$ in the coefficients of $f$?

Tasmia
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  • The substitution $x\leftarrow x+\dfrac a4$ doesn't change the discriminant, so we are free to do it, and use the (simpler) formula for the discriminant of the depressed quartic. I assume that the factor $2^{24}$ is added just to clear the denominators in a minimal way. This may play a role in e.g. number theoretic applications. – Jyrki Lahtonen May 14 '17 at 08:57
  • Thanks for the comment. Yes, the substitution $x \rightarrow x+a/4$ does not change the discriminant. However, only one x is substituted, the first term in particular. The other x remain the same. Does that change the discriminant or not? – Tasmia May 14 '17 at 09:13
  • This means that you're saying 'the discriminant of $x^4+px^2+qx+r$ is the same as $(x+a/4)^4+px^2+qx+r$?' – Tasmia May 14 '17 at 09:17
  • Oh, I missed that. Something strange is going on. – Jyrki Lahtonen May 14 '17 at 10:20

1 Answers1

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The discriminant of $f$ is the same as the discriminant of $f(x-a/4)$. The discriminant is the product of differences of zeros of the polynomial, so doesn't change when these zeros are all translated by the same amount.

Dietmann does not assert that the formula with the $2^{24}$ factor is the discriminant. Rather that is a square integer and when you multiply it by the discriminant it cancels off all the powers of two in the denominators, yielding an expression (i) which is an integer if $a,\ldots,d$ are integers, and (ii) is a square iff the discriminant is a square.

Angina Seng
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  • Thanks for the answer. I think you meant 'f has the same discriminant as f(x+a/4)' instead of 'f(x+a/4)'? – Tasmia May 14 '17 at 09:07
  • Also, not all x substitute into x+a/4, only the first term. Doesn't that make a difference? – Tasmia May 14 '17 at 09:09
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    @Tasmia It looks as if Dietmann made a blunder. He should have replaced $x$ by $x+a/4$ throughout. Then one would get more complicated formulae for $q$ and $r$. I don't think that would affect the rest of his arguments. Perhaps you should write to him? – Angina Seng May 14 '17 at 09:52