Lecture 33 Notes: Symmetric Polynomials and the Discriminant

Lecture 33: Symmetric Polynomials and the Discriminant 33 33.1 Symmetric Polynomials and the Discriminant Symmetric Polynomials Last class, we began discussing symmetric polynomials and the discriminant. The goal is to develop some tools to understand the structure of the solutions to a polynomial — if not to compute them in radicals, then at least to see how this works when possible. Galois theory can be used for this as well, not just proving impossibility. Last time, we considered the symmetric polynomials Z[x1 , . . . , xn ] ⊃ Z[x1 , . . . , xn ]Sn = Rn , and stated the following fundamental theorem: Theorem 33.1 (n) (n) (n) We have Rn = Z[σ1 , σ2 , . . . , σn ], where (n) σ1 (n) σ2 = x1 + · · · + xn , = x1 x2 + · · · + xn−1 xn , .. . σn(n) = x1 · · · xn . In other words, every symmetric polynomial P ∈ Rn can be written in terms of the elementary symmetric functions. Example 33.2 We have x21 + · · · + x2n = σ12 − 2σ2 , x31 + · · · + x3n = σ13 − 3σ1 σ2 + 3σ3 . We previously saw this example when n = 3. But a feature is that the right-hand side doesn’t really depend on n — when we write the expression in terms of the symmetric polynomials, n sort of disappears. The reason is that we have an obvious homomorphism Z[x1 , . . . , xn ] → Z[x1 , . . . , xn−1 ] sending xn → 0 (so we essentially just kill one of the variables). Under this homomorphism, we have Rn → Rn−1 (since if the polynomial was invariant under permutations of n variables, it’s also invariant under permutations of the first n − 1, where we killed the last variable). But this homomorphism is compatible with the elementary symmetric functions — (n) (n−1) it’s clear that σi 7→ σi , since the homomorphism essentially just kills the monomials containing xn (and (n) σn 7→ 0). Meanwhile, the power sums have the same property — we have xd1 + · · · + xdn 7→ xd1 + · · · + xdn−1 . This means if we can prove an identity for large n, we can automatically deduce it for smaller n as well. On the other hand, we can also use degree considerations. For example, x31 + · · · + x3n is a homogeneous (n) polynomial of degree 3 in the xi ; while σi is a homogeneous polynomial of degree i. This means we can only (n) (n) (n) use σ1 , σ2 , and σ3 in the expression; so if we can find the identity for n = 3, we can automatically deduce it for all larger n as well. Putting these observations together, we see that a formula as in Example 33.2 for n + 1 implies one for n, and conversely, using degree considerations it’s enough to check it for small n. (In our example, n = 3 was enough; note that n = 2 is too small, because σ3 is mapped to 0.) We’ll now prove the theorem. Proof of Theorem 33.1. We use induction in the number of variables. (n) (n) We need to check that every symmetric polynomial can be expressed as a polynomial in σ1 , . . . , σn , and that (n) this expression is unique. In other words, we can consider the map φn : Z[t1 , . . . , tn ] → Rn , where ti 7→ σi . 147 Lecture 33: Symmetric Polynomials and the Discriminant Then we want to check that φn is an isomorphism, meaning that it’s one-to-one and onto. We’ll check these two parts separately. First we’ll check that φn is injective. Suppose there is some polynomial Q(t1 , . . . , tn ) which φn maps to 0 (to prove injectivity, it suffices to show there is no such polynomial). We can first pull out the last variable, by writing Q = tdn Q′ , where tn ∤ Q′ . Then φn (tdn Q′ ) = 0, so we have Y xi d · Q′ (σ1 , . . . , σn ) = 0. Since the first factor is nonzero, this means Q′ (σ1 , . . . , σn ) = 0. So this essentially means that we can assume that Q is not divisible by tn . But now we can use the homomorphism from earlier — let rn be the restriction map Rn → Rn−1 sending xn 7→ 0. Then we have rn (Q′ (σ1 , . . . , σn )) = 0 (because we assumed Q′ (σ1 , . . . , σn ) = 0). But we have      (n) (n−1) (n−1) rn Q′ σ1 , . . . , σn(n) = Q σ1 , . . . , σn−1 , where Q is not identically 0 (since we assumed Q′ is not divisible by tn , and rn just maps tn 7→ 0). But this contradicts the inductive assumption (because then φn−1 would map a nonzero polynomial Qn−1 to 0). (The base case of the induction is n = 1, which is trivial.) Now we’ll check that φn is surjective. Start with a polynomial P ∈ Rn ; we can assume P is homogeneous of degree d. We want to check that P = φn (Q) for some Q; we’ll use induction on d. The idea is again to reduce the number of variables. Let (n−1) rn (P ) = T (σ1 (n−1) , . . . , σn−1 ) (by the inductive assumption). Now consider the polynomial   (n) (n) P − T σ1 , . . . , σn−1 . We know that rn maps this polynomial to 0. But the kernel of rn is generated by xn , so then   (n) (n) xn | P − T σ1 , . . . , σn−1 . Since the RHS, it then follows that each xi divides it; but by unique factorization, this means   (n) (n) x1 · · · xn | P − T σ1 , . . . , σn−1 . So then we can write   (n) (n) P − T σ1 , . . . , σn−1 = σn · Q, where Q is a symmetric polynomial of smaller degree. But Q = S(σ1 , . . . , σn ) by the inductive assumption, so P = S(σ1 , . . . , σn ) + T (σ1 , . . . , σn ), as desired. This is one possible proof, emphasizing the inductive structure; but in applications, the proof won’t matter that much. 33.2 The Discriminant By Theorem 33.1, we can write Y i< j (xi − xj )2 = ∆n (σ1 , . . . , σ) for some polynomial ∆n (since the LHS is a symmetric polynomial). 148 Lecture 33: Symmetric Polynomials and the Discriminant Definition 33.3 The discriminant of a polynomial P (z) = z n + an−1 z n−1 + · · · + a0 is D = ∆n (−an−1 , an−2 , . . . , (−1)n a0 ). In particular, we see that D = 0 if and only if P has a multiple root. Example 33.4 We can calculate the discriminant explicitly when P has low degree — for P (x) = x2 + bx + c we have D = b2 − 4c, while for P (x) = x3 + px + q we have D = −4p3 − 27q 2 . Proof. We proved this last time, but we’ll outline the proof of the second statement again. Degree considerations give that D = ap3 + bq 2 for some a and b. Then we can take P (x) = x3 − x (which has discriminant 4) to get that a = −4, and P (x) = (x − 1)2 (x + 2) (which has discriminant 0) to get that b = −27. Student Question. How did we show that D = ap3 + bq 2 ? Answer. To simplify the formulas, we assumed that σ1 = 0, so we’re trying to compute ∆3 (0, p, −q). But ∆3 is homogeneous of degree 6, as a polynomial in the xi ; meanwhile σi is homogeneous of degree i. We only have σ2 (of degree 2) and σ3 (of degree 3), and the only way to make 6 from 2’s and 3’s is 2 + 2 + 2 and 3 + 3, so the only possible terms we can have are σ23 and σ32 . Student Question. Does this argument only work when σ1 = 0? Answer. Yes — in the general case, there is still a formula for ∆3 , but it’s longer. But the case σ1 = 0 is actually enough for practical purposes, since it’s possible to reduce any cubic to this form (by shifting the variable). We’ll now get to the role of the discriminant in Galois theory. We can also consider Y δn (x1 , . . . , xn ) = (xj − xi ), i< j so ∆n = δn2 . Note that δn is not symmetric — if we swap xi and xi+1 , then this swaps the sign of δn . This means
δ xσ(1) , . . . , xσ(n) = (−1)sgn(σ) δ(x1 , . . . , xn ), so they have the same sign if σ is even and opposite sign if σ is odd. (Here σ denotes an arbitrary permutation of {1, . . . , n}.) Now let E/F be a field extension, where E is the splitting field of P ∈ F [x]. Assume that P does not have multiple roots (but it is allowed to be reducible). Q By definition, this means E = F (α1 , . . . , αn ), where P (x) = (x − αi ). We know that G = Gal(E/F ) is a subgroup of Sn , since elements of G must permute the roots of P . Q Now let E/F be a field extension, where E is a splitting field of a polynomial P ∈ F [x]. Let P (x) = (x − αi ), and assume that P doesn’t have multiple roots (we don’t need to assume it’s irreducible). By definition, this means E = F (α1 , . . . , αn ). We know that G = Gal(E/F ) must permute the roots Q αi , and is therefore a subgroup of Sn (where we look at how it permutes those roots). Now if we let δ = i< j (αj − αi ), we know that δ ∈ E and δ 2 ∈ F . We can immediately see how G acts on δ — for any σ ∈ G, we know ( δ if σ is even σ(δ) = −δ if σ is odd. In particular, this means δ ∈ F if and only if all permutations σ ∈ G are even. (This is because F is exactly the fixed field of G, by the main theorem; so δ is fixed by all elements of G if and only if it’s in F .) So the conclusion is the following: 149 Lecture 33: Symmetric Polynomials and the Discriminant Proposition 33.5 We have Gal(P ) ⊂ An if and only if the discriminant ∆ of P is a square. 33.3 Cubic Polynomials We can now apply this to a concrete situation: suppose that n = 3, and we know P is irreducible. There are only two transitive subgroups of S3 , which are A3 and S3 . So these are the only options for Gal(P ), and we have a concrete way of distinguishing between these two cases — the Galois group is S3 when ∆ is not a square, and A3 when ∆ is a square. Example 33.6 Find the Galois group of P (x) = x3 − 3x − 1 over Q. Solution. The discriminant of P is D = 4 · 27 − 27 = 81, which is square; so the Galois group is A3 ∼ = Z/3Z. Example 33.7 Suppose F contains a cube root of unity ω; find the Galois group of P (x) = x3 − a (assuming P is irreducible). √ Solution. The discriminant is D = −27a2 . But since ω ∈ F , then −27 is a square (since ω = 1± 2 −3 , we have √ that −3 ∈ F ). So then Gal(P ) ∼ = Z/3Z. (This is a special case of the theorem we saw last time.) We’ve now seen an effective way to find the Galois group for cubic polynomials; we’ll finish by discussing how to actually solve them. Proposition 33.8 If F contains a primitive cube root of unity ω, and E/F is a Galois extension with Gal(E/F ) = Z/3, then E = F (α) for some α3 = a ∈ F . The proof we’ll give is constructive, and if we explicitly write out the construction, this leads to Cardano’s Formula for the solutions to a cubic (which was actually published in 1545). Proof. Let σ be a generator for Gal(E/F ). It suffices to find α ∈ E such that σ(α) = ωα or ω 2 α — then we have σ(α3 ) = α3 , and since σ generates the Galois group, this means all elements of the Galois group fix α3 , so α3 ∈ F . Meanwhile, the Galois group does not fix α itself; so the degree of α is 3. Since the degree of the extension is 3 as well, this means E = F (α). Pick some β ∈ E which is not in F , and let α1 = β + ωσ(β) + ω 2 σ 2 (β), α2 = β + ω 2 σ(β) + ωσ 2 (β). Then it’s clear that σ(α1 ) = ω 2 α1 and σ(α2 ) = ωα2 , so it suffices to check that one of α1 and α2 is nonzero. But otherwise, (β, σ(β), σ 2 (β)) would be a solution to the system of linear equations a + ωb + ω 2 c = a + ω 2 b + ωc = 0. Then orthogonality of characters for Z/3Z (from the representation theory of cyclic groups) shows that the only solution is a = b = c. But then σ(β) = β, which contradicts the fact that β ̸∈ F (since if σ fixed β, then the entire Galois group would fix β). So one of α1 and α2 must be nonzero. 150 Lecture 33: Symmetric Polynomials and the Discriminant Note 33.9 The same proof works with 3 replaced with any prime; and with a bit more work, it can be generalized to any n. This can be used to prove the converse of the theorem from last class — last class, we saw that any radical extension is solvable. But this shows that any extension with a solvable Galois group is radical. 151