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Casus Irreducibilis CASUS IRREDUCIBILIS D. KATZ In this note we provide a proof for my algebra class of the classical result known as Casus irreducibilis - Latin for irreducible case. This result states that if f(x) 2 Q[x] is an irreducible polynomial of degree three that has all of its roots in R, then none if its roots are contained in a real radical extension of Q. Note that a field L is a real radical extension of Q if L is a radical extension of Q and L ⊆ R. Of course every degree three polynomial is solvable by radicals, so that this result shows that we cannot solve an irreducible cubic using real radicals; instead, complex radicals are required. We will need two preliminary results. 2 Lemma A. Let f(x) F [x] be an irreducible polynomialQ of degree n with distinct roots α1; : : : ; αn, so that − 2 its splitting field K is Galois over F . Set ∆ := 1≤i<j≤n(αi αj). If ∆ F , then, when we identity Gal(K=F ) as a subgroup of S , we in fact have Gal(K=F ) ⊆ A . n Q n 2 σ − 2 σ Proof. For any σ Sn, set ∆ := 1≤i<j≤n(ασ(i) ασ(i)), so that if σ Gal(K=F ), σ(∆) = ∆ . If σ = (i; j) and j = i + k, it is not difficult to check that ∆σ = (−1)2k−1 (since there are 2k − 2 sign changes not coming interchanging αi and αj). It follows that if σ 2 Gal(K=F ), σ is an even permutation if and only if σ(∆) = ∆. Since ∆ 2 F , σ(∆) = ∆, for all σ 2 Gal(K=F ), it follows that Gal(K=F ) ⊆ An. □ Lemma B. Let F ⊆ R, a 2 F , p a prime and α 2 R such that αp = a. Then [F (α): F ] = 1 or p. Proof. Suppose α 62 F . Then α is a root of xp − a. Let f(x) 2 F [x] be the minimal polynomial of α over F . If ϵ is a primitive pth root of unity, then xp − a = (x − α)(x − ϵα) ··· (x − ϵp−1α). Since f(x) divides xp − a, the constant term of f(x) has the form (−1)dϵjαd, for some j. It follows that ϵj 2 R. Thus, ϵj = ±1. Thus, αd 2 F . Now, if d =6 p, then d and p are relatively prime, so we can write 1 = ad + bp for integers a and b. Thus, α = αadαbp 2 F , a contradiction. Thus, d = p, which yields [F (α): F ] = p, as required. □ Before stating and proving the theorem, we make the following observation. Suppose F ⊆ F (α) ⊆ R, are field extensions with αn = a 2 F , for some n. Thus, F (α) is a simple radical extension of F . If we can factor n = cd, then we can increase the number of terms in the original radical extension by writing d d d c c d F ⊆ F (α ) ⊆ F (α )(α) = F (α) ⊆ R, where (α ) 2 F and α 2 F (a ). We mayp continuing to do so, until all exponents requiredp are prime. For example, if we take the real number α := 12 2 we can write the radical extension Q ⊆ Q( 12 2) ⊆ R as q q rq p p p p p 3 p Q ⊆ Q( 2) ⊆ Q( 2; 2) ⊆ Q( 2; 2; 2) ⊆ R; so that the degree of each extension is prime. This clearly applies to any radical extension. Theorem (Casus irreducibilis). Let f(x) 2 Q[x] be an irreducible cubic polynomial with three real roots. Then no root of f(x) lies in a real radical extension of Q. Proof. Let α; β; γ 2 R denote the roots of f(x) and set ∆ := (α − β)(α − γ)(β − γ). Write K for the splitting field of f(x) over Q, and set D := ∆2, the discriminant of f(x). Note that every element of Gal(K=Q) fixes D, so that D 2 Q. Set E := Q(∆), allowing for the possibility that E = Q. Note that since ∆2 = D 2 Q, E is a radical extension of Q and [E : Q] ≤ 2. Thus, no root of f(x) is contained in E, since adjoining any root of f(x) to Q gives an extension of degree three. Therefore, f(x) is irreducible over E. It follows by Lemma A that Gal(K=E) ⊆ A3, and therefore jGal(K=E) = 3j, from which it follows that [K : E] = 3. Thus, K is Galois over E so that if we adjoin any root of f(x) to E, we have also adjoined all roots of f(x) to E. Now suppose a root of f(x), say α, is contained in a real radical extension of Q. Then α is contained in a real radical extension of E, so we have E = L0 ⊆ L1 ⊆ · · · ⊆ Lr ⊆ R; 1 ni 2 2 ≥ ≤ ≤ with each Li = Li−1(ai) and ai Li−1, for some ai Li, ni 1 and 1 i r. By the observation above, we may assume that each ni := pi is prime. By Lemma B, each [Li : Li−1] = pi or 1. Eliminating redundant terms, we may assume each [Li : Li−1] = pi. Moreover, without loss of generality, we may assume that Lr−1 does not contain α, but Lr contains α. Then f(x) is irreducible over Lr−1, and thus, [Lr−1(α): Lr−1] = 3. Since [Lr : Lr−1] = pr is prime, we must have pr = 3, and Lr = Lr−1(α). If we let K0 denote the splitting field of f(x) over Lr−1, then since ∆ 2 Lr−1, Lemma A gives that Gal(K0=Lr−1) ⊆ A3. Since [K0 : Lr−1] = jGal(K0=Lr−1)j ≥ 3, we have K0 = Lr. Thus, Lr is a Galois extension of Lr−1. Since pr 3 2 3 2 3 − 2 ar = ar Lr−1, ar = u, for some u L. Thus, x u Lr−1[x] has a root in Lr. Since Lr is a splitting 3 2 field over Lr−1, x − u has all of its roots in Lr. In other words, ar; ϵar; ϵ ar 2 Lr ⊆ R, where ϵ is a primitive third root of unity. But then ϵ 2 Lr, which is a contradiction. It follows that no root of f(x) can lie in a real radical extension of Q. □ Remarks. (i) Note that the casus irreducibilis theorem states that no root of an irreducible cubic in Q[x] can be contained in a real radical extension of Q. This is stronger than saying that the splitting field of an irreducible cubic is not contained in a real radical extension of Q. In fact, in his paper Solutions of polynomials by real radicals I.M. Isaacs proves the following notable reuslt: Suppose F ⊆ K is a Galois extension, with K ⊆ F [a] ⊆ R and an 2 F , for some n ≥ 1. Then [K : F ] ≤ 2. He then uses this fact to generalize casus irreduciblis by showing that if f(x) 2 Q[x] has all of its roots in R and has a root in a real radical extension of Q, then the degree of f(x) is 2n for some n. In particular, any polynomial in Q[x] that is solvable by real radicals must have degree 2n, for some n. Isaacs' paper is very readable, as it does not use any techniques we have not covered in this class. (ii) Recall that if f(x) 2 F [x] has degree less than or equal to three, then f(x) is irreducible over F if and only if f(x) does not have a root in F . In general this fails for polynomials of degree greater than three. However, for certain polynomials, irreducibility is determined by the availability of roots. One way to interpret Lemma B is as follows. If F has characteristic zero, and p is prime, then for any a 2 F , xp − a is irreducible if and only if it does not have a root in F . Certainly, if xp − a is irreducible over F , it does not have a root in F . Conversely, suppose xp − a does not have a root in F and α is a root of xp − a in an extension of F . Then [F (α): F ] =6 1, and thus equals p. Therefore, the minimal polynomial of α over F has degree p. But this polynomial divides xp − a, and thus equals xp − a, so that xp − a is irreducible over F . Department of mathematics, The University of Kansas, Kawrence KS 66045 2.
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