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I am trying to calculate the discriminant of the polynomial $f = X^5 + pX^2 + q \in K[X]$ where $K$ is a field. I define the discriminant as $\text{disc } f = \prod_{i < j} (\alpha_i - \alpha_j)^2$ where the $\alpha_i$ are the roots of $f$.

Note that we can derive a useful formula for the disccriminant in terms of the derivative. In particular $$f(X) = \prod_{i=1}^n (X-\alpha_i) \Rightarrow f'(X) = \sum_j \prod_{j \neq i} (X - \alpha_i)$$ from which we see $$\text{disc }f = (-1)^{n(n-1)/2} \prod_{i=1}^n f'(\alpha_i).$$

Using this representation of the discriminant I calculated the discriminant of a simpler quintic polynomial $g = X^5 + q$. I got $3125q^4$.

I want to calculate the discriminant of $f = X^5 + pX^2 + q$ but I do not know how to do it because I do not know how to get explicit forms for the roots $\alpha_i$ of this polynomial (which seem to be a prerequisite to finding the discriminant?).

How can we proceed?

J. W. Tanner
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Featherball
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  • I can tell you the discriminant of $ax^2+bx+c$ without finding its roots – J. W. Tanner May 08 '24 at 20:02
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    Use Vieta's formulas. – Gonçalo May 08 '24 at 20:03
  • As something close to a heuristic, if $f$ and $f' = 5X^4 + 2pX$ have a common root, then either $X=0$ so $q=0$, or $X = (-2p/5)^{1/3}$ so $(-2p/5+p) (-2p/5)^{2/3} + q = 0 \implies q = -(3p/5) (2p/5)^{2/3} \implies q^3 = -108/3125 p^5 \implies 108 p^5 + 3125 q^3 = 0$. That suggests that whatever polynomial in $p$ and $q$ the discriminant is, it should probably be divisible by $q(108 p^5 + 3125 q^3)$ and generate the same radical ideal of $\mathbb{Q}[p,q]$. – Daniel Schepler May 09 '24 at 00:43

2 Answers2

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The discriminant of a polynomial $p$ with degree $n$ can be obtained by the (tedious!) expansion of the determinant of its Sylvester matrix Res$(p,p')$ (with size $(2n-1)\times(2n-1)$) as an example is given here.

Instead, of calculating a $9 \times 9$ determiant, it is better to use the software tools we all have at our disposal...

For this kind of questions, SAGE is a very good public domain software : just call https://sagecell.sagemath.org/ then copy-paste the program below in the Edit window :

 R.<p,q>=QQ[]
 x=polygen(R) # indeterminate
 y=x^5+p*x^2+q
 d=y.discriminant().factor()
 print(d)

Then press "Evaluate" ; you will get the following result :

$$q * (108*p^5 + 3125*q^3)$$

Edit : (as wisely advised by Viktor Vaughn) The person who has created the program can use the "Share" button allowing to give the user to possibility to find the program ready to be run like here.

Jean Marie
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It is possible to calculate the resultant of two polynomials via an algorithm very similar to the Euclidean algorithm for calculating the gcd of polynomials. In this case, applying that algorithm gives: \begin{align*} \operatorname{Discr}(X^5 + pX^2 + q) & = \operatorname{Res}(X^5 + pX^2 + q, 5 X^4 + 2pX) \\ & = 5^5 \operatorname{Res}(X^5 + pX^2 + q, X^4 + \frac{2p}{5} X) \\ & = 5^5 \operatorname{Res}(X^5 + pX^2 + q - X(X^4 + \frac{2p}{5}X), X^4 + \frac{2p}{5}X) \\ & = 5^5 \operatorname{Res}(\frac{3p}{5} X^2 + q, X^4 + \frac{2p}{5}X) \\ & = 5^5 \operatorname{Res}(X^4 + \frac{2p}{5}X, \frac{3p}{5} X^2 + q) \\ & = 5^5 (3p/5)^4 \operatorname{Res}(X^4 + \frac{2p}{5}X, X^2 + \frac{5q}{3p}) \\ & = 5 \cdot 3^4 p^4 \operatorname{Res}(X^4 + \frac{2p}{5} X - \left(X^2 - \frac{5q}{3p}\right) \left(X^2 + \frac{5q}{3p}\right), X^2 + \frac{5q}{3p}) \\ & = 5 \cdot 3^4 p^4 \operatorname{Res}(\frac{2p}{5} X + \frac{25q^2}{9p^2}, X^2 + \frac{5q}{3p}) \\ & = 5 \cdot 3^4 p^4 (2p/5)^2 \operatorname{Res}(X + \frac{125q^2}{18p^3}, X^2 + \frac{5q}{3p}) \\ & = \frac{2^2 3^4 p^6}{5} \left[\left(\frac{125q^2}{18p^3}\right)^2 + \frac{5q}{3p}\right] = q(108p^5+3125q^3). \end{align*} (Here we do the calculation in $\mathbb{Q}(p,q)[X]$, which by functoriality of the resultant must give the same result as the resultant in $\mathbb{Z}[p,q][X]$, and again by functoriality evaluating that polynomial must give the correct answer for the resultant over $\mathbb{Z}[X]$ or $\mathbb{Q}[X]$ for any particular values of $p$ and $q$. In particular, this explains why we get a correct answer even in the case $p=0$.)

(Also note that in a hand calculation, it would be reasonable to stop already at the sixth line, and calculate the resultant as $5\cdot 3^4 p^4 g((-5q/3p)^{1/2}) g(-(-5q/3p)^{1/2})$ where $g(X) = X^4 + \frac{2p}{5}X$ and $\pm(-5q/3p)^{1/2}$ are the roots of $X^2 + \frac{5q}{3p}$ -- where $g(X_0) g(-X_0) = X_0^8 - \frac{4p^2}{25} X_0^2$. On the other hand, an automated calculation could very well continue from the second-to-last line and essentially perform synthetic division of the quadratic polynomial by the linear polynomial.)

Daniel Schepler
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  • for a monic polynomial of degree $5$, the discriminant is the resultant of the polynomial and its derivative – J. W. Tanner May 09 '24 at 18:23
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    @J.W.Tanner So are you suggesting I add $\operatorname{Discr}(X^5+pX^2+q) = \operatorname{Res}(X^5+pX^2+q,5X^4+2pX) = \cdots$ at the head to make that explicit? That makes sense. – Daniel Schepler May 09 '24 at 18:25
  • Yes +1. Of note to OP, you calculated the discriminant without calculating the roots of the polynomial – J. W. Tanner May 09 '24 at 19:40