Brian Bi
$\DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\lcm}{lcm} \newcommand\divides{\mathbin |} \newcommand\ndivides{\mathbin \nmid} \newcommand\d{\mathrm{d}} \newcommand\p{\partial} \newcommand\C{\mathbb{C}} \newcommand\N{\mathbb{N}} \newcommand\Q{\mathbb{Q}} \newcommand\R{\mathbb{R}} \newcommand\Z{\mathbb{Z}} \newcommand\pref[1]{(\ref{#1})} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Gal}{Gal}$

## Section 16.2. The Discriminant

Exercise 16.2.1 Suppose the variables $$u_i$$ and $$u_j$$ are swapped in (16.2.1). Factors that don't contain $$u_i$$ or $$u_j$$ are unaffected. Factors of the form $$(u_i - u_k)^2$$ and $$(u_j - u_k)^2$$, where $$k \ne i, j$$ transform into each other. The factor $$(u_i - u_j)^2$$ becomes $$(u_j - u_i)^2$$, which is the same polynomial. Thus the discriminant remains unchanged under transpositions. But the transpositions generate all permutations.

Exercise 16.2.2

1. A real cubic either has three real roots or one real root.

If it has three real roots, the discriminant will obviously be nonnegative.

If it has one real root, then its roots will be $$r, z, \overline{z}$$ where $$r$$ is real and $$z$$ is nonreal. In that case, the discriminant will be $$(r - z)^2(r - \overline{z})^2 (z - \overline{z})^2$$. Now $$z - \overline{z}$$ is pure imaginary and nonzero, so $$(z - \overline{z})^2$$ is negative. The product of the other two factors can be rewritten as $$((r - z)(r - \overline{z}))^2 = (r^2 - r(z + \overline{z}) + z\overline{z})^2$$. But $$z + \overline{z}$$ and $$z\overline{z}$$ are both real, so $$(r - z)^2(r - \overline{z})^2$$ is positive (it cannot be zero since neither $$r - z$$ nor $$r - \overline{z}$$ can be zero). Therefore, the discriminant is the product of a negative real and a positive real, and is therefore negative.

2. A real quartic has zero, two, or four real roots. Since the discriminant is positive, all four roots are distinct.

Obviously, if there are four real roots, the discriminant is positive.

If there are two real roots, then the roots are of the form $$r_1, r_2, z, \overline{z}$$ where $$r_1, r_2$$ are real and $$z$$ is nonreal. In the discriminant, as in part (a), we find that $$(z - \overline{z})^2$$ is negative while $$(r_i - z)^2 (r_i - \overline{z})^2$$ is positive for $$i = 1, 2$$. The last factor, $$(r_1 - r_2)^2$$, is positive. So the discriminant in this case is negative.

If the quartic has no real roots, then it has roots of the form $$z_1, \overline{z_1}, z_2, \overline{z_2}$$. The two factors $$(z_i - \overline{z_i})^2$$ are each real and negative. The factors $$(z_1 - z_2)^2$$ and $$(\overline{z_1} - \overline{z_2})^2$$ are complex conjugates of each other so their product is real and positive. The same goes for $$(z_1 - \overline{z_2})^2$$ and $$(\overline{z_1} - z_2)^2$$. So the discriminant must be positive.

Thus, given the information that the discriminant is positive, we can conclude that the number of real roots of the quartic is either 0 or 4 (and that all roots are distinct).

Exercise 16.2.3

1. The roots of the transformed cubic will simply be shifted by $$-s_1/3$$ compared to the roots of the old cubic, so all factors in the discriminant remain unchanged, so the discriminant remains unchanged.
2. This is trivial so I have skipped it.

Exercise 16.2.4

1. The coefficient $$p$$ has weighted degree 2 and the coefficient $$q$$ has weighted degree 3, and the discriminant is homogeneous with degree 6. The only nonnegative integer combinations of 2 and 3 that yield 6 are $$3\cdot 2$$ and $$2 \cdot 3$$. Thus, $$\Delta = ap^3 + bq^2$$ for some constants $$a, b$$. To determine the constants, we will use the polynomials $$x^3 - x$$ and $$x^3 - 3x - 2$$. The former has roots $$-1, 0, 1$$ so its discriminant is 4. Therefore, $$a(-1)^3 = 4$$, and $$a = -4$$. The latter has a double root at $$x = 1$$, as we can determine easily using the multiple root theorem; so its discriminant is 0. Therefore $$a(-3)^3 + b(-2)^2 = 0$$, from which we obtain $$b = -27$$. Therefore, $$\Delta = -4p^3 - 27q^2$$.

2. The coefficient $$p$$ has weighted degree 3 and the coefficient $$q$$ has weighted degree 4, and the discriminant has degree 12. The only possible monomials in the discriminant are therefore $$p^4$$ and $$q^3$$, that is, $$\Delta = ap^4 + bq^3$$. We first consider the polynomial $$x^4 - x = x(x^3 - 1)$$. Its discriminant contains six factors, three of which occur in the discriminant of $$x^3 - 1$$, which is $$-27$$ according to part (a). The remaining three factors are $$(e^{2\pi ij/3})^2$$ for $$j = 0, 1, 2$$, that is, $$1, \omega^2, \omega^4$$, the product of which is 1. Therefore, the discriminant of $$x^4 - x$$ is $$-27$$, which implies $$a = -27$$. Next we consider the polynomial $$x^4 - 4x + 3$$, which has a double root at $$x = 1$$. Its discriminant is $$0 = -27(-4)^4 + b\cdot 3^3$$, giving $$b = 256$$. Thus, $$\Delta = -27p^4 + 256q^3$$.

3. The coefficient $$p$$ has weighted degree 4 and the coefficient $$q$$ has weighted degree 5, and the discriminant has degree 20. Therefore the discriminant equals $$ap^5 + bq^4$$ for some constants $$a, b$$. Part (b) implies that the discriminant of $$x^5 - x$$ is that of $$x^4 - 1$$ times $$(1 \cdot i \cdot (-1) \cdot -i)^2 = 1$$, that is, $$-256$$. Therefore $$a = 256$$. To find $$b$$, we use the polynomial $$x^5 - 5x + 4$$, which has a double root at $$x = 1$$. Its discriminant is $$0 = 256(-5)^5 + b \cdot 4^4$$, from which we obtain $$b = 3125$$. Therefore $$\Delta = 256p^5 + 3125q^4$$.

Exercise 16.2.5 Set $$u_4 = 0$$ in the discriminant to obtain $$u_1^2 u_2^2 u_3^2 D(u_1, u_2, u_3) = s_3^2 D(u_1, u_2, u_3)$$. According to (16.2.5), $$D(u_1, u_2, u_3) = -4s_1^3 s_3 + s_1^2 s_2^2 + 18s_1 s_2 s_3 - 4s_2^3 - 27s_3^2$$ where the symmetric polynomials are for the three variables $$u_1, u_2, u_3$$. Lemma 16.1.11 then implies that $$D(u_1, u_2, u_3, u_4) = s_3^2 (-4s_1^3 s_3 + s_1^2 s_2^2 + 18s_1 s_2 s_3 - 4s_2^3 - 27s_3^2) + hs_4$$ for some symmetric polynomial $$h$$.

Exercise 16.2.6 Shifting all roots by $$t$$ doesn't change the discriminant, so $$\frac{\d\Delta}{\d t} = 0$$. We also have \begin{align*} \frac{\d s_1}{\d t} &= \frac{\d}{\d t}(u_1 + t + u_2 + t + u_3 + t) = 3 \\ \frac{\d s_2}{\d t} &= \frac{\d}{\d t}((u_1 + t)(u_2 + t) + (u_2 + t) (u_3 + t) + (u_3 + t)(u_1 + t)) = u_1 + t + u_2 + t + u_2 + t + u_3 + t + u_3 + t + u_1 + t = 2s_1 \\ \frac{\d s_3}{\d t} &= \frac{\d}{\d t}((u_1 + t)(u_2 + t)(u_3 + t) = (u_2 + t)(u_3 + t) + (u_3 + t)(u_1 + t) + (u_1 + t)(u_2 + t) = s_2 \end{align*} According to (16.2.5), $$\Delta(s_1, s_2, s_3) = -4s_1^3 s_3 + s_1^2 s_2^2 + 18s_1 s_2 s_3 - 4s_2^3 - 27s_3^2$$. Therefore, \begin{align*} \frac{\p \Delta}{\p s_1} &= -12s_1^2 s_3 + 2s_1 s_2^2 + 18s_2 s_3 \\ \frac{\p \Delta}{\p s_2} &= 2s_1^2 s_2 + 18s_1 s_3 - 12s_2^2 \\ \frac{\p \Delta}{\p s_3} &= -4s_1^3 + 18s_1s_2 - 54s_3 \end{align*} so that \begin{align*} \frac{\d \Delta}{\d t} &= \frac{\p \Delta}{\p s_1}\frac{\d s_1}{\d t} + \frac{\p \Delta}{\p s_2}\frac{\d s_2}{\d t} + \frac{\p \Delta}{\p s_3}\frac{\d s_3}{\d t} \\ &= 3(-12s_1^2 s_3 + 2s_1 s_2^2 + 18s_2 s_3) + 2s_1(2s_1^2 s_2 + 18s_1 s_3 - 12s_2^2) + s_2(-4s_1^3 + 18s_1s_2 - 54s_3) \\ &= -36s_1^2s_3 + 6s_1s_2^2 + 54s_2 s_3 + 4s_1^3 s_2 + 36s_1^2 s_3 - 24s_1 s_2^2 - 4s_1^3 s_2 + 18s_1 s_2^2 - 54s_2 s_3 \\ &= 0 \end{align*} as expected.

Exercise 16.2.7

Suppose $$P \in \C[x_1, \ldots, x_n]$$ is both symmetric and alternating (meaning it vanishes whenever any two of the $$x$$'s are equal). Then $$D \divides P$$, where $$D$$ is the discriminant $$\prod_{i < j} (x_i - x_j)^2$$.

If $$n = 1$$ then $$D = 1$$ and there is nothing to prove. Otherwise, first perform polynomial division of $$P$$ by $$x_1 - x_2$$ in the variable $$x_1$$, giving $$P = (x_1 - x_2)Q + R$$ where $$Q, R$$ depend only on $$x_2, \ldots, x_n$$. Now $$(x_1 - x_2)Q$$ is alternating, so $$R$$ must be alternating, but since $$R$$ doesn't depend on $$x_1$$, $$R$$ must vanish identically. So $$P = (x_1 - x_2)Q$$. Now divide $$Q$$ by $$x_1 - x_2$$ in the variable $$x_1$$, giving $$P = (x_1 - x_2)^2 S + (x_1 - x_2)T$$ where both $$S$$ and $$T$$ depend on $$x_2, \ldots, x_n$$ alone. Since $$P$$ is invariant under exchange of $$x_1, x_2$$, perform this exchange to yield $$P = (x_1 - x_2)^2 S' - (x_1 - x_2)T'$$ where $$S'$$ and $$T'$$ are obtained by replacing each occurrence of $$x_1$$ by $$x_2$$ in $$S$$ and $$T$$, respectively. Adding the equations $$P = (x_1 - x_2)^2 S + (x_1 - x_2)T$$ and $$P = (x_1 - x_2)^2 S' - (x_1 - x_2)T'$$, we obtain $$2P = (x_1 - x_2)^2 (S + S') + (x_1 - x_2) (T - T')$$. But $$T - T'$$ is alternating, so $$x_1 - x_2 \mid T - T'$$. Therefore, in fact, $$(x_1 - x_2)^2 \mid 2P$$. The above reasoning holds for any pair of variables among $$x_1, \ldots, x_n$$, therefore $$(x_i - x_j)^2$$ divides $$P$$ for all $$i, j$$. But $$\C[x_1, \ldots, x_n]$$ is a UFD and the elements $$x_i - x_j$$ are irreducible, so the factorization of $$P$$ contains at least two copies of each $$x_i - x_j$$. Therefore $$D \divides P$$.

We now return to the problem. Observe that $$q$$ is obtained from $$p$$ by exchanging any pair of the $$u$$'s. Therefore, $$(p - q)^2$$ is symmetric and alternating, and the Lemma implies that $$D \divides (p - q)^2$$. But $$D$$ is homogeneous with degree $$n(n-1)$$, and so is $$(p - q)^2$$, therefore $$(p - q)^2 = cD$$ where $$c$$ is a constant. Finally, observe that in $$D$$, the coefficient of $$u_1^2 u_2^4 \ldots u_{n-1}^{2(n-1)}$$ is 1, and the same is true of $$(p - q)^2$$. Therefore $$(p - q)^2 = D$$.