Using symmetric polynomials to find the discriminant of $x^4 + px + q$ over $\mathbb{Q}$

Question

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.

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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 $q$.

Answer 1

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.