Insolvability of the Quintic (Fraleigh Section 56 Last one in the book!) [I m sticking to Freleigh on this one except that this will be a combination of part of the section and part of the exercises because I want to show an example of a polynomial over Q that isn t solvable.] Definition (Radical Extensions): An extension K of a field F is an extension of F by radicals if there are elements α1,, αr K and positive integers n1,, n1, such that K = F(α1,, αr), α1 n1 F, and αi ni F(α1,, αi- 1) for all 1 < i r. A polynomial f(x) F[x] is solvable by radicals over F if the splitting field E of f(x) over F is contained in an extension by radicals. Example: Suppose α is the real (positive) fourth root of 2. Then Q(α, i) is a radical extension since α 4 = 2 is in Q and i 2 = - 1 is in Q Q(α). What s the point? Well, if a polynomial, f(x), is solvable by radicals then all the roots of f(x), can be expressed using a finite sequence of the operations of addition, subtraction, multiplication, division, and taking nith roots starting with elements of the original field F. I talked about this when I first introduced Galois Theory via an extended example. Here is a copy/paste of what I said then: ******************************************************************************************* Time Out: How Do Radical Extensions Relate To Things Like The Quadratic Equation? Recall that the whole point of this exercise is to figure out why there is no analogue to the quadratic equation for 5 th degree polynomials. So, let s try to figure out why radical extensions are connected to solvability by radicals. First, suppose that you have a 2 nd degree irreducible polynomial, f(x), over Q and its splitting field is Q(α) where α 2 is in Q. Let s say c = α 2. Then α = c. Now, we know that all the roots of f(x) are elements of the splitting field Q(α). So they can be expressed in the form a + bα where a, b Q. Substituting, we see that any root is of the form a + b a, b, c Q. This is exactly the form that the roots of a quadratic are expressed in in the quadratic equation. c where For a slightly more complex example, let s consider a 3 rd degree irreducible polynomial, f(x), over Q with splitting field Q(α, β ) where α 3 is in Q, and β 2 is in Q(α). [This is the kind of situation you would expect with an irreducible 3 rd degree polynomial you could extend using a real root first, and then factor out a linear piece and then extend using a root of the 2 nd degree factor.]
Anyhow, any root of this polynomial will look like aα 2 + bα + c + dα 2 β + eαβ + fβ, because the basis for Q(α, β ) is {1, α, α 2, β, αβ, α 2 β }. Now, if we let s = α 3, then we can substitute to get any root in the form a( 3 s ) 2 3 3 + b s + c + d( s ) 2 3 β + e s β + fβ. Now if we let lα 2 3 + mα + n = l( s ) 2 3 + m s + n = β 2 3, then β = l( s) 2 3 +m s +n, so, we can express any root in the form: 3 a( s ) 2 3 3 + b s + c + d( s ) 2 3 l( s) 2 3 3 3 +m s +n + e s l( s) 2 3 +m s +n + f 3 l( s) 2 3 +m s +n. As long as the splitting field for the polynomial lives inside a radical extension, we will always be able to do this kind of process. And if it does not live in side a radical extension, we will not be able to do this. ******************************************************************************************* Example (Some quintics are solvable): Consider the polynomial x 5 1 over Q. This polynomial has 5 roots; 1 and four complex roots, each of which is what is called a primitive fifth root of unity. [A primitive nth root of unity is just an nth root of 1 that is a generator for the multiplicative group of nth roots of 1.] One such root in this case is cos(2π 5) + isin(2π 5). If we let α = cos(2π 5) + isin(2π 5), then I claim that Q(α) is the splitting field for x 5 1 over Q. The polynomial clearly splits over this field since by closure, α 2 = cos(4π 5) + isin(4π 5), α 3 = cos(6π 5) + isin(6π 5), and α 4 = cos(8π 5) + isin(8π 5) are all in Q(α). We know this is the smallest field over which this polynomial splits because it is the smallest field containing Q and α. Note that it is a 4 th degree extension since α is a root of the irreducible polynomial x5 1 x 1 = x4 + x 3 + x 2 + x +1. (A basis for this extension is given by {1, α, α 2, α 3 }). The Plan: I will first show that a polynomial f(x) in F[x] is solvable by radicals over F only if its splitting field E over F has a solvable Galois group. [This result goes both directions, but we only need this direction.] Then I will show that there is a polynomial of degree 5 with coefficients in Q such that the Galois group is S5 (which we have already seen is not solvable). Lemma (56.3): Let F be a field of characteristic 0, and let a F. If K is the splitting field of x n a over F, then Gal(K/F) is a solvable group. Proof: Suppose first that F contains all the nth roots of unity. It is easy to show that the nth roots of unity form a cyclic subgroup (of the multiplicative group of F) generated by ζ = cos(2π n) + isin(2π n). The full list of nth roots of unity is: 1, ζ, ζ 2,, ζ n- 1.
Now, if β F is a root of x n a F[x], then the complete list of zeros is: β, βζ, βζ 2,, βζ n- 1. Note that all of this roots are in F(β). Since K = F(β), an automorphism σ in Gal(K/F) is determined by the value σ(β). Now, if σ(β) = βζ i and τ(β) = βζ j where τ is in Gal(K/F), then (τσ)(β) = τ(σ(β)) = τ(βζ i ) = τ(β)τ(ζ i ) = τ(β)ζ i = βζ j ζ i. Similarly (στ)(β) = βζ i ζ j. So, τσ = στ, and Gal(K/F) is abelian and therefore solvable. Now, suppose that F does not contain a primitive nth root of unity. Let ζ be a generator of the cyclic group of nth roots of unity under multiplication in F. Let β again be a zero of x n a. Since β and βζ are both in the splitting field K of x n a, ζ = (ζβ)/β is in K. Let F = F(ζ), so we have F < F K. Now, F = F(ζ) is the splitting field of the separable polynomial x n - 1, and hence is Galois over F. Thus we know the Fundamental Theorem of Galois Theory (Part 3&4), that Gal(K/F ) is a normal subgroup of Gal(K/F). This means that {i} Gal(K/F ) Gal(K/F) is a subnormal series of groups (actually normal). And since,, F = F(ζ) an automorphism η in Gal(F /F) is determined by η(ζ), and we must have = η(ζ) = ζ i since all the roots of x n - 1 are powers of ζ. If, µ(ζ) = ζ j for µ in Gal(F /F), then (µη)(ζ) = µ(η(ζ)) = µ(ζ i ) = ζ ij. And similarly (ηµ)(ζ) = ζ ji. Thus Gal(F /F) is abelian. We also know (by the first part of the proof) that since F contains all the nth roots of unity, Gal(K/ F ) is abelian. The Fundamental Theorem of Galois Theory (Part 4) that Gal(K/F)/Gal(K/F ) is isomorphic to Gal(F /F) which is abelian. Thus {i} Gal(K/F ) Gal(K/F) is a subnormal series of groups with the property that all of the quotient groups are abelian. I claim that 1) any finite group has a composition series, and 2) every refinement of a subnormal series with abelian quotient groups will also have abelian quotient groups. These two things together mean that Gal(K/F) is solvable. So, QED/(Sub- proofs 1&2). ******************************************************************************************* Sub- proof 1: Let G be a finite group. If it is simple then we are done because {e} G is a composition series. Let H1 be the largest proper normal subgroup of G (if there is a tie, just pick one). Now pick the largest proper normal subgroup H1 of H2 (if such a thing doesn t exist you can stop). Keep doing this until you hit a simple subgroup. Reverse the numbering on these guys so the series matches how subnormal series usually look.
I claim that { e} H 0 H 1... H n 1 H n = G is a composition series. Suppose that quotient group Hk/Hk- 1 is not simple. Let N be a non- trivial normal subgroup of Hk/Hk- 1. Then consider the pre- image H of N under the canonical map (h hhk- 1). We know that it is a subgroup of Hk. I claim it is normal Hk in as well. Suppose g is in Hk and n is in H, then we know that (ghk- 1)(nHk- 1) (ghk- 1) - 1 = (gn g - 1 )H is in N because nhk- 1 is in N and N is normal. Thus gng - 1 is in H. And hence H is normal. But H contains Hk- 1 (Hk- 1 is the pre- image of the identity of N under the canonical map). This is a contradiction because Hk- 1 was selected to be the largest normal subgroup of Hk. So, sure enough { e} H 0 H 1... H n 1 H n = G is a composition series. Sub- proof 2: Suppose that e quotient groups Hk/Hk- 1 are abelian. Let e and consider the quotient group Ki/Ki- 1. There are four cases: { } H 0 H 1... H n 1 H n = G is subnormal and all the { } K 0 K 1... K m 1 K m = G be a refinement Case 1: Both Ki and Ki- 1 are in {Hi}. Then by assumption Ki/Ki- 1 is abelian. Case 2: Only Ki is in {Hi}, say Ki = Hl. Then we have H l 1 K i 1 H l = K i. Then by the Third Isomorphism Theorem, Ki/Ki- 1 is isomorphic to (Hl/Hl- 1)/(Ki- 1/Hl- 1). But we know by assumption that Hl/Hl- 1 is abelian, thus any quotient group of it will also be abelian. Therefore, Ki/Ki- 1 is abelian. Case 3: Only Ki- 1 is in {Hi}, say Ki - 1 = Hl. Then we have H l = K i 1 K i H l+1. Note that Ki/Ki- 1 is as subgroup of the abelian group Hl+1/Ki - 1 = Hl+1/Hl so it is also abelian. Case 4: Neither Ki nor Ki- 1 are in {Hi}. Then choose the largest l such that H l K i 1 Then we must have H l K i 1 K i H l+1 (otherwise, Ki = Hl+1). In this case, since Hl is normal in Hl+1 it is normal in Ki. Thus by the Third Isomorphism Theorem, we have that Ki/Ki- 1 is isomorphic to (Ki/Hl)/(Ki- 1/Hl). But Ki/Hl is a subgroup of the abelian group Hl+1/Hl and so it is abelian. This means that any of its quotient groups are abelian. Therefore Ki/Ki- 1 is abelian. So, in all cases, we found that Ki/Ki- 1 is abelian. The end! ****************************************************************************************** This next theorem will finish the first part of our job, showing that a polynomial f(x) in F[x] is solvable by radicals over F only if its splitting field E over F has a solvable Galois group. Theorem (56.4). Let F be a field of characteristic zero, and let F E K F, where E is a normal extension of F (since we are working over characteristic zero, this just means it is the splitting field of an irreducible polynomial) and K is an extension of F by radicals. Then Gal(E/F) is a solvable group.
Proof: We first show that K is contained in a finite normal extension L of F by radicals and that the group G(L/F) is solvable. Since K is an extension by radicals, K= F(α1,, αr) where a i n i F(α1,, αi- 1) for all 1 < i r and α1 n1 F. To form L, we first form the splitting field L1 of f1(x) = x n 1 α n 1 1 over F. Then L1 is a normal extension of F, and Lemma 56.3 shows that Gal(L1 /F) is a solvable group. Now, n a 2 2 L1 and we form the polynomial f2(x) = (x n 2 σ (α 2 )n 2. σ Gal(L 1 /F ) Since this polynomial is invariant under action by any σ in Gal(L1 /F), we see that f2(x) is in F[x]. We let L2 be the splitting field of f2(x) over L1. Then L2 is the splitting field over F also and is a normal extension of F by radicals. We can form L2 from L1 via repeated steps as in Lemma 56.3, passing to splitting field of (x n 2 σ (α 2 )n 2 at each step. By Lemma 56.3, we see that the Galois group over F of each new extension thus formed continues to be solvable. We continue this process of forming splitting fields over F in this manner: At stage i, we form the splitting field of the polynomial fi(x) = (x n i σ (α i ) n i over L1. We finally obtain a field L = Lr that is a normal extension of σ Gal(L i 1 /F ) F by radicals and we see that G(L/F) is a solvable group. We see from construction that K L. To conclude, note that by the Fundamental Theorem of Galois Theory (Part 4), we have Gal(E/F) Gal(L/F)/Gal(L/E). Thus Gal(E/F) is a quotient group and hence a homomorphic image of Gal(L/F). In this case Exercises 27-29 of Section 35 show that Gal(E/F) is solvable.
An Unsolvable Polynomial! In three easy steps I ll prove there is a polynomial of degree 5 in Q[x] that is not solvable by radicals over Q. Step 1: First I claim that if a subgroup H of S5 contains a cycle of length 5 and a transposition, then H = S5. Proof: Suppose that (abcde) is in H. Then so are (acebd), (adbec), and (aedcb). Suppose the transpositions is (ab), then we can conjugate by each 5 cycle to obtain: (abcde) (ab)(aedcb) = (bc) (acebd) (ab)(adbec) = (cd) (adbec) (ab)(acebd) = (de) (aedcb) (ab)(abcde) = (ae) Then (bc)(ab)(bc) = (ac), (bc)(cd)(bc) = (bd), (cd)(de)(cd) = (ce), (de)(ae) (de) = (ad), and (ce)(bc)(ce) = (be). This is all ten transpositions! And we know that any permutation can be written as a product of transpositions. Clearly this would have worked just as well if we had used any of the conjugates of (ab) we produced using the 5- cycles. So, let s check this with one of the transpositions we didn t generate this way. So, suppose that the transposition we have to start with is (ac). Again, I ll conjugate by my 5- cycles: (abcde) (ac)(aedcb) = (bd) (acebd) (ac)(adbec) = (ce) (adbec) (ac)(acebd) = (ad) (aedcb) (ac)(abcde) = (be) Then (bd)(ad)(bd) = (ab). And hey, we know that this one is good enough to get the rest! And since we can get (ac) from any of its conjugates, all of those will work as well. So, no matter what transposition we have we can generate all of S5. Done! Step 2: Show that if f(x) is an irreducible polynomial in Q[x] of degree 5 having exactly two complex and three real zeros in C, then the group of f(x) over Q is isomorphic to S5. Proof: If we extend first by adjoining one of the real zeros then we will get a 5 th degree extension that will not be a splitting field. The degree of the splitting field as an extension must then be a multiple of 5. This means that the order of the Galois group must be of order 5n for some n. [Also note that since the splitting field will only 5 roots, the Galois group can be identified with a subgroup of S5.] Then by the First Sylow theorem, the Galois group must contain an element of order 5 (which of course will correspond to a 5- cycle in S5
each element of the Galois group can be associated with a permutation of the roots). When we have adjoined all the real roots, we get an subfield, F, of the splitting field, over which we can factor f(x) into three linear factors and an irreducible quadratic factor (whose roots are the two complex roots). Then any automorphism of the splitting field that fixes F will have to send these two roots to each other and leave the rest alone. Thus we get a transposition in the Galois group. So, by the previous problem the Galois group of f(x) is isomorphic to S5. Step 3: I claim that the polynomial f(x) = 2x 5-5x 4 + 5 has is irreducible and has exactly two complex and three real zeros. Proof: The polynomial is irreducible by Eisenstein with p=5. Let s do some calculus! f (x) = 10x 4-20x 3. Let s see where the derivative is zero: 10x 4-20x 3 = 10x 3 (x- 2). So, the derivative is zero at x = 0 and x = 2. Let s check the concavity at those points. f (x) = 40x 3-60x 2. So f (2) = 320 240 >0 we can see (by the second derivative test) that we have a relative minimum at x = 2. And f(2) = 64 80 +5 = - 11 < 0. Let s try the first derivative test for x = 0. If we put in a small negative number into f (x) = 10x 4-20x 3 then we will for sure get a positive number. If we put in a small positive number, the negative part will overwhelm the positive part (the 4 th power of a really small number is way smaller than a 3 rd power of that number) so we ll get a negative first derivative. So by the first derivative test we have a relative max at x = 0. And f(0) = 5. We have no more relative extrema. This means that the function values are increasing from - to 0. This means we must have exactly one real root on the interval (-, 0). The function must then decrease from on the interval (0, 2) and cross the x- axis to get from y = 5 to y = - 11. So, there is exactly one real root on the interval (0, 2). Then the function must increase on the (2, ) starting from y = - 11, so there is on more real root on this interval. This gives us exactly three real roots total. The last two must of course be complex roots. Done! And here is a graph as a bonus: