6
$\begingroup$

I'm trying to prove that the discriminant of $x^4 + px + q$ over $\mathbb{Q}$ is $-27p^4 + 256q^3$, where we define the discriminant to be

$$ \Delta_f = \prod_{i < j}(\alpha_i - \alpha_j)^2 $$

I am given the hint that

"It is a symmetric polynomial of degree $12$ hence a linear combination of $p^4$ and $q^3$."

By the Fundamental Theorem on Symmetric Polynomials, the discriminant is a polynomial expression of the elementary symmetric polynomials $s_1,\ldots,s_4$, and in this case $s_1 = s_2 = 0$, $s_3 = -p$, $s_4 = q$, so the discriminant is certainly a polynomial expression of $p, q$. To justify the next step, I have had to do some very unpleasant acrobatics, which seems very unlikely to be the intended solution. Can anybody give me a simpler way of seeing that indeed the discriminant is a linear combination of $p^4$ and $q^3$?

Edit: I'm looking for a solution that does use the hint, but is less long-winded than mine, since to me the hint implies that its second statement follows immediately from its first, and I suspect that I'm missing something.


My Attempt

Let $f \in K[t_1, \ldots, t_n]$ be a symmetric polynomial. We say that $f$ is $\textbf{homogeneous}$ if there is some $d$ such that for each nonzero term $\lambda t_1^{a_1}\ldots t_n^{a_n}$ of $f$, we have $a_1 + \ldots + a_n = d$. In that case, we refer to $d$ as the $\textbf{degree}$ of $f$.

Lemma 1: If $f, g$ are distinct, homogeneous symmetric polynomials with the same degree $d$, then $f - g$ is also homogeneous with degree $d$.

Proof: This is obvious since purity is a statement about degree of individual terms. $$\tag*{$\blacksquare$}$$

Lemma 2: Let $f\in K[t_1,\ldots, t_n]$ be a homogeneous symmetric polynomial. Then by the Fundamental Theorem of Symmetric Polynomials, there is a unique $g \in K[t_1, \ldots, t_n]$ such that $f(t_1, \ldots, t_n) = g(s_1, \ldots, s_n)$, where the $s_i$ are the elementary symmetric polynomials.

Then for each nonzero monomial term $\lambda t_1^{a_1}\ldots t_n^{a_n}$ of $g$, the degree of $s_1^{a_1}\ldots s_n^{a_n}$ is the same as the degree of $f$ (i.e. $a_1 + 2a_2 + \ldots + na_n = d$).

Proof: We proceed by induction on the degree of $g$ under the lexicographic ordering on monomials. If $g$ is a monomial, we are done, so assume not.

If $g$ has more than one term, then let $\lambda t_1^{a_1}\ldots t_n^{a_n}$ be the first term of $g$ under lexicographic ordering. Then $f(t_1, \ldots, t_n) - \lambda s_n^{a_n}s_{n-1}^{a_{n-1} - a_n}\ldots s_1^{a_1 - a_2}$ is homogeneous by Lemma 1, since $g$ has more than one term, so by induction we are done.

$$\tag*{$\blacksquare$}$$

From here the thing I want to prove follows quite easily, because the discriminant is homogeneous with degree $12$, and $s_3^as_4^b$ has degree $3a + 4b$, so it follows that $(a,b) \in (0,0), (4,0), (0, 3)$. Then from here we can determine the coefficients of the polynomial by making appropriate substitutions for $p$ and $q$.

$\endgroup$
10
  • 1
    $\begingroup$ I think most mathematicians would call "homogeneous (of degree $d$)" what you call "pure with total degree $d$" $\endgroup$ May 31, 2020 at 11:55
  • $\begingroup$ @EwanDelanoy good point - I have heard the term but it temporarily escaped me. Have edited. +1 $\endgroup$ May 31, 2020 at 11:58
  • $\begingroup$ If I understand well your question, you'd like basically to avoid using the fundamential theorem of symmetric polynomials? $\endgroup$ May 31, 2020 at 12:04
  • 2
    $\begingroup$ I think that your two lemmas are trivial and the only justification that you need is in the very last paragraph, which is the intended reasoning. Of course it's completely fine if you prove them too. $\endgroup$
    – mathmo
    May 31, 2020 at 13:49
  • 1
    $\begingroup$ @DavidPopović I completely agree with DavidPopovíc's assessment, and I think the only reason why the argument sounds like "very unpleasant acrobatics" to you (Qwertitops) is only because you're not familiar with it. Your proof is 100% correct, but you're still afraid to accept it. One suggestion I would make to convince yourself more is to show a slightly stronger property about the part that puzzles you. For example : take a polynomial $P$ in $p$ and $q$ not of the required form, say it contains a $pq$ term. So you know that $P$ will not be homog of degree $12$, but can y say more ? $\endgroup$ May 31, 2020 at 16:54

1 Answer 1

1
$\begingroup$

Let $$P(x) = x^4 + px + q$$

By definition $$\Delta = \prod_{i < j}(\alpha_i - \alpha_j)^2$$

and this value $\Delta$ is invariant with respect to permutations of the roots $\alpha$, therefore - (by Lemma 1) - it lies in the base field $\mathbb Q(p,q)$.

Since there are 4 roots, $\Delta$ is a product of 6 factors, squared. When multiplied out fully each term will be a product of 12 $\alpha$ values.

Note that $q = \alpha_1 \alpha_2 \alpha_3 \alpha_4$ and $p = -(\sum_{i,j,k} \alpha_i \alpha_j \alpha_k)$. To get terms with 12 values we can use $q^3$, $p^4$ and it is easy to verify that no combinations like $q p^2$ work (that gives 10 terms rather than 12).

To finally calculate the discriminant formula I don't know any shortcuts, it can be done with a CAS that supports symmetric polynomials.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .