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1 A. Miller M542 Galois Theory Spring 2000 For the material on Galois theory we will be assuming that the fields all have characteristic zero. When we get to solvability by radicals we will assume that all fields are subfields of the complex numbers C. 1 Review The following results are from the review sheet for the midterm. Theorem 1.1 Minimal polynomials are irreducible: If α is in some extension field of F and define: I = {f(x) F [x] : f(α) = 0} Then: (a) I is an ideal in F [x] (b) (assuming I is nontrivial) I = p(x) where p(x) is any polynomial of minimal positive degree in I and (c) p(x) is irreducible and so I is a maximal ideal. Theorem 1.2 If p(x) is an irreducible polynomial of degree n in F [x] and α a root of p in some extension field, then 1, α, α 2,..., α n 1 is a basis for F (α) as a vector space over F and hence [F (α) : F ] = n. Theorem 1.3 Suppose char(f )=0 and p(x) F [x] is irreducible, then p does not have any multiple roots in any extension of F. Theorem 1.4 Suppose char(f )=0 and [E : F ] <, then there exists α E such that E = F (α). Theorem 1.5 If p(x) F [x] is an irreducible polynomial and α and β are two roots of p, then F (α) F (β) with an isomorphism which fixes F and takes α to β. Theorem 1.6 If σ : F F is an isomorphism, f F [x] any polynomial, E F a splitting field of f over F, and E F a splitting field of σ(f) over F, then there exists an isomorphism ρ σ such that ρ : E E. 2 Galois Theory In this section we assume that all fields have characteristic 0. Definition 2.1 The field K F is a Galois extension of the field F iff K is the splitting field over F of some polynomial with coefficients in F. Proposition 2.2 Suppose F E K are fields and K is a Galois extension of F. Then K is a Galois extension of E. 1

2 Suppose F 3 is the splitting field of f F 1 [x]. This means that F 3 is the smallest field containing F 1 and all the roots of f. Then f F 2 [x] since F 1 F 2 and therefore F 3 is the smallest field containing F 2 and all the roots of f. Definition 2.3 For fields F E define aut(e F ) to be the set of all automorphisms σ of E which fix F, i.e., σ(a) = a for all a F. Proposition 2.4 aut(e F ) is a group. Furthermore, if F E K are fields, then aut(k E) is a subgroup of aut(k F ). If σ, τ aut(e F ), then clearly σ τ is an automorphism of E. Given a F we have that σ τ(a) = σ(τ(a)) = σ(a) = a and so σ τ fixes F and therefore σ τ aut(e F ). Similarly, σ(a) = a implies σ 1 (a) = a and so aut(e F ) is a group under composition. aut(k E) is a subgroup of aut(k F ) since the operation (composition) is the same and any automorphism which fixes E must also fix the smaller field F. Lemma 2.5 Suppose σ, ρ aut(f (α) F ). Then σ = ρ iff σ(α) = ρ(α). Similarly, if σ, ρ aut(f (α 1, α 2,..., α n ) F ) then σ = ρ iff σ(α k ) = ρ(α k ) for all k = 1, 2,..., n. The elements of F (α) have the form f(α) where f(x), g(x) F [x] are polynomials with g(α) g(α) 0. But then σ( f(α) g(α) ) = σ(f(α)) σ(g(α)) = f(σ(α)) g(σ(α)) this last is true because elements of F are fixed by σ. Since the same is true for τ, if σ(α) = τ(α), then σ( f(α) g(α) ) = f(σ(α)) g(σ(α)) = f(τ(α)) g(τ(α)) = τ(f(α) g(α) ) The similar result holds for F (α 1, α 2,..., α n ) since its elements have the form f(α 1, α 2,..., α n ) g(α 1, α 2,..., α n ) where f, g are polynomials in F [x 1, x 2,..., x n ]. Theorem 2.6 Suppose that K is the splitting field of a polynomial in F [x] of degree n. Then aut(k F ) is isomorphic to a subgroup of S n. 2

3 Let K = F (α 1, α 2,..., α n ) where the α i are all the roots of f F [x] and K is the splitting field of f. If σ aut(k F ), then σ(f) = f and so if α is a root of f then so is σ(α), i.e., f(α) = 0 implies f(σ(α)) = σ(f(α)) = σ(0) = 0. Hence σ must permute the roots α i of f. By Lemma 2.5 we see that elements of aut(k F ) are determined by this permutation. Hence the mapping: h : aut(k F ) S n given by h(σ) = δ where σ(α i ) = α δ(i) is a one-to-one homomorphism and therefore aut(k F ) is isomorphic to its image which is a subgroup of S n. Lemma 2.7 Suppose K is a Galois extension of F, K L, and σ : K L an embedding which fixes F. Then and σ(k) = K. Note that an embedding is a one-to-one homomorphism and σ(k) = {σ(a) : a K}. Since K = F [α 1, α 2,..., α n ] where the α i are the roots of some polynomial f F [x] and since σ fixes the coefficients of f it must be that for every i there is j such that σ(α i ) = α j. Hence σ(k) = K. Theorem 2.8 Suppose K is a Galois extension of F, then aut(k F ) = [K : F ] By Theorem 1.4 there exists α K such that K = F [α]. By Theorem 1.1 there exists an irreducible polynomial p(x) F [x] such that p(α) = 0. Claim. K is the splitting field of p. Proof of Claim: Let L be any extension field in which p splits and suppose β is any root of p in L. Then there exists an isomorphism σ : F (α) F (β) which fixes F and sends α to β (Theorem 1.5). But then by Theorem 1.6 the map σ extends to an automorphism ρ : L L. Since ρ fixes F and by Lemma 2.7 ρ(k) = K. Since ρ(α) = β this means β is in K. Since β was an arbitrary root of p this means that p splits in K. End proof of Claim. We have that by Theorem 1.2 that [K : F ] = n where n is the degree of p and since irreducible polynomial have distinct roots (Theorem 1.3), the roots of p are {α 1, α 2,..., α n }. By Theorem 1.5 for every i there exists σ i : F (α) F (α i ). But automorphisms of F (α) which fix F are determined by the values on α (Lemma 2.5) we have that aut(k F ) = {σ 1, σ 2,..., σ n } Hence the theorem is proved. 3

4 Lemma 2.9 Suppose F E, E K, K is a Galois extension of F and σ : E E and isomorphism which fixes F. Then there exists ρ aut(k F ) which extends σ. This is just a special case of Theorem 1.6. If K is the splitting field of f F [x] over F, then σ(f) = f since σ fixes F and K = K is the splitting field of f over both E and E. Theorem 2.10 Suppose K and E are Galois extensions of F and F E K. Then aut(k E) is a normal subgroup of aut(k F ) and Define the map aut(k F ) aut(k E) aut(e F ) h : aut(k F ) aut(e F ) by h(ρ) = ρ E i.e., we restrict ρ to E. By Lemma 2.7 we have that ρ(e) = E and so this restriction is in aut(e F ). It is easy to check that h is a group homomorphism. Claim. The kernel of h is aut(k E). This is clear because h(ρ) =identity iff ρ E fixes E iff ρ aut(k E). Claim. h is onto. This follows from Lemma 2.9, since if we take E = E in that Lemma then for every σ aut(e F ) there exists ρ aut(k) such that ρ σ. But this means h(ρ) = ρ E = σ. Hence h is onto. The homomorphism theorem of group theory says the range of h is isomorphic to the quotient group of its domain by its kernel (which is a normal subgroup) and so the Theorem is proved. 3 More Galois Theory In this section we develop some more of the basics of Galois Theory. It will not be needed for the section on solvability by radicals. Theorem 3.1 Suppose K is a Galois extension of F and p(x) F [x] is an irreducible polynomial. Then either all the roots of p are in K or none of them are. Suppose K is the splitting field of f(x) over F. Let L be the splitting field of p(x) over K. It follows that L is the splitting field of f(x)p(x) over F and so it is a Galois extension of F. 4

5 Suppose α K and p(α) = 0. We need to show all the roots of p are in K. Let β L be any root of p. We have that there exists σ : F (α) F (β) which fixes F and maps α to β. By Lemma 2.9 that there exists ρ : L L which extends σ. But by Lemma 2.7 we have that ρ(k) = K. But this implies that ρ(α) = β K. Hence K contains all the roots of p. Example 3.2 (a) There exists fields F 1 F 2 F 3 such that F 3 is a Galois extension of F 1 but F 2 is not a Galois extension of F 1. (b) There exists fields F 1 F 2 F 3 such that F 3 is a Galois extension of F 2 and F 2 is a Galois extension of F 1, but F 3 is not a Galois extension of F 1. Hint: (a) Q Q(β) Q(β, αβ, α 2 β) where α = e 2π 3 i and β = 3 2 (b) Q Q( 2) Q( 1 + 2) Definition 3.3 For H aut(k F ) define fix(h) = {a K : σ H σ(a) = a} Proposition 3.4 Suppose F K are fields and H aut(k F ), then E = fix(h) is a field such that F E K. Since σ aut(k F ) implies it fixes each element of F, this implies that F E. E K by definition. To see that E is a subfield of K. Suppose x, y E. Then for every σ H we know that σ(x) = x and σ(y) = y. Since σ is a field automorphism we have that σ(x + y) = σ(x) + σ(y) = x + y and σ(xy) = σ(x)σ(y) = xy and so x + y E and xy E. Similarly x y and x/y are in E, so its a subfield. Lemma 3.5 Suppose K is a Galois extension of F, then F = fix(aut(k F )). F fix(aut(k F )) is trivial, since by definition every σ aut(k F ) fixes F. To see the other way that fix(aut(k F ) F, what we need to show is that for every α K with α / F there exists ρ aut(k F ) such that ρ(α) α. So fix such an α and let p(x) F [x] be the minimal (irreducible) polynomial such that p(α) = 0. We know that since irreducible polynomials have distinct roots (Theorem 1.3) that p has a root β α. Since p splits in K (Lemma 3.1) there is such a β K. Let σ : F (α) F (β) be an isomorphism fixing F and taking α to β (Theorem 1.5). Using Lemma 2.9 with E = F (α) and E = F (β) we get that there exists ρ aut(k F ) extending σ. Since ρ(α) = β α we are done. 5

6 Lemma 3.6 Suppose K is a Galois extension of F and G a subgroup of aut(k F ) and fix(g) = F, then G = aut(k F ). As in the proof of Theorem 2.8 let K = F [α], p(x) F [x] the minimal polynomial of α, and p(x) = (x α 1 )(x α 2 ) (x α n ) where α i K are the roots of p. In that proof we showed that aut(k F ) = n. Suppose for contradiction that G = m < n. By reordering the α i let Consider the polynomial {α 1, α 2,..., α m } = {σ(α) : σ G} q(x) = (x α 1 )(x α 2 ) (x α m ) Note that each σ G permutes the set {α 1, α 2,..., α m } and hence σ(q(x)) = (x σ(α 1 ))(x σ(α 2 )) (x σ(α m )) = (x α 1 )(x α 2 ) (x α m ) = q(x) It follows if we write q(x) = x m + b m 1 x m b 0 that for each σ G that σ(b k ) = b k for each k. But this implies (since fix(g) = F ) that q(x) F [x]. This is a contradiction since q divides p but p is supposed to be irreducible in F [x]. Theorem 3.7 Suppose K is a Galois extension of F, then there is a one-to-one correspondence between the subgroups of aut(k F ), G = {H : H aut(k F ) is a subgroup} and the set of intermediate fields between F and K, and F = {E : F E K and E is a field} This correspondence is given by the two maps which are inverses of each other. φ : G F defined by φ(h) = fix(h) ψ : F G defined by φ(e) = aut(k E) Suppose E F. Then K is a Galois extension of E and so by Lemma 3.5 we have that E = fix(aut(k E)). But this means E = φ(ψ(e). Suppose G G. Let E = fix(g). By Lemma 3.6 we have that G = aut(k E). But this just means that G = ψ(φ(g). Hence the two maps are inverses of each other and the Theorem is proved. 6

7 4 Tartaglia s method for solving cubics Step 1. Given x 3 + ax 2 + bx + c = 0 substitute x + α for x and pick α so as to eliminate the coefficient of x 2. Step 2. Given x 3 + px + q = 0 substitute x + β x the two most complicated coefficients. for x and choose β so as to eliminate The resulting equation will be: x 3 + β3 x 3 + q = 0 which is quadractic in x 3. If u is a solution of it, then u + β u original equation. + α is a solution of the Definition 4.1 A polynomial f(x) Q[x] is solvable by radicals iff its roots are in the smallest subfield S C which is closed under taking all complex roots, i.e., if a S and n N then all n complex roots of a are in S, i.e., x n a splits in S. It is also true that every f(x) Q[x] of degree 4 is solvable by radicals. Hint: To solve the quartic below factor it into quadratics: x 4 + px 2 + qx + r = (x 2 + ax + b)(x 2 ax + c) and then show that a 2 is the solution of a cubic: c + b a 2 = p (c b)a = q bc = r (c + b) 2 (c b) 2 = 4bc = 4r (p + a 2 ) 2 q 2 /a 2 = 4r 5 Solvability by radicals In this section we assume all our fields are subfields of the complex numbers C. Definition 5.1 For G a finite group define G is a solvable group by induction on G. G is solvable iff either G is abelian or there exists a normal subgroup H G such that both H and G/H are solvable. Definition 5.2 Given fields F E C we say that E is a radical Galois extension of F iff there exists n N and a F such that E is the splitting field of the polynomial x n a over F. 7

8 Theorem 5.3 Suppose that E is a radical Galois extension of F, then aut(e F ) is a solvable group. Suppose E is the splitting field of x n a over F. Let β C be any complex number such that β n = a. Let α = e 2πi n. Then we have that (Note that αβ β = α so α E.) E = F (β, βα, βα 2,..., βα n 1 ) = F (α, β) F F (α) F (α, β) = E the field F (α) is a Galois extension of F since it is the splitting field of x n 1, so we have (by Theorem 2.8) that H = aut(f (α, β) F (α)) is a normal subgroup of G = aut(f (α, β) F ) and aut(f (α) F ) is isomorphic to their quotient G/H. Claim. G/H aut(f (α) F ) is abelian. Suppose σ, τ aut(f (α) F ), then for some i we have σ(α) = α i and for some j τ(α) = α j. But then σ(τ(α)) = σ(α j ) = (α j ) i = α ji = α ij = τ(σ(α)) Hence (by Lemma 2.5) we have that στ = τσ. Claim. H = aut(f (α, β) F (α)) is abelian. Suppose σ, τ aut(f (α, β) F (α)), then for some then for some i we have σ(β) = α i β and for some j we have τ(β) = βα j. By definition both fix α. But then σ(τ(β)) = σ(α j β) = (α j )σ(β) = α j+i β = α i+j β = τ(σ(β)) Hence we have that στ = τσ. Since H and G/H are abelian they are solvable and so by definition G is solvable. Lemma 5.4 Suppose that G is solvable group and G is a homomorphic image of G, then G is solvable. Hence quotient groups of solvable groups are solvable. Let h : G G be an onto homomorphism. We prove the lemma by induction on G. Suppose G is abelian. Then G is abelian since h(x)h(y) = h(xy) = h(yx) = h(y)h(x) Suppose there exists H G such that H and G/H are solvable. Let H = h(h). Then H is a normal subgroup of G because given y H and x G we have h(x)h(y)h(x) 1 = h(xyx 1 ) H 8

9 By induction H is solvable. It is also true that G /H is the homomorphic image of G/H since we can define k : G/H G /H by k(ah) = h(a)h This is well-defined because ah = bh implies b 1 a H implies h(b 1 a) h(h) = H which implies h(a)h = h(b)h. It is easy to check that k is an onto homomorphism. Hence by induction G /H is solvable. By definition G is solvable. Theorem 5.5 Suppose that F 1 F 2 F 3 F m is a sequence of radical Galois extensions, i.e., F k+1 is a radical Galois extension of F k for each k = 1, 2,..., m 1. Suppose that K is a Galois extension of F 1 such that K F m. Then aut(k F 1 ) is a solvable group. This is proved by induction on m m = 2: In this case we have that F 1 K F 2. We have that aut(k F 1 ) is isomorphic to a quotient of aut(f 2 F 1 ) (by Theorem 2.8) and we have that aut(f 2 F 1 ) is a solvable group by Theorem 5.3. Hence by Lemma 5.4 we have that aut(k F 1 ) is solvable. m > 2: In this case suppose that F 2 is splitting field of x n a F 1 [x] over F 1 and K is the splitting field of f(x) F 1 [x]. Let L be the splitting field of (x n a)f(x) over F 1. Then we have F 1 F 2 L F m and F 1 K L By induction on m we have that aut(l F 2 ) is a solvable group. We also know that L is a Galois extension of F 1 and by Theorem 2.8 we that H = aut(l F 2 ) is a normal subgroup of G = aut(l F 1 ) with quotient group isomorphic to G/H aut(f 2 F 1 ). Since both H and G/H are solvable, we have that G is solvable. Finally since F 1 K L we know (by Theorem 2.8) that aut(k F 1 ) is isomorphic to a quotient of aut(l F 1 ) and hence by Lemma 5.4 is solvable. Corollary 5.6 Suppose f Q[x] is a polynomial which is solvable by radicals. Then if K is the splitting field of f over Q, then aut(k Q) is a solvable group. Theorem 5.7 The group A 5 is simple. Recall: A 5 is the subgroup of S 5 consisting of those permutations which can be written as the product of an even number of transpositions. A group is simple if has no nontrivial normal subgroups. Let H A 5. Case 1. H contains a 3-cycle. First note that the 3-cycles generate A n. This is because: (12)(23) = (123) and (123)(234) = (12)(34). These equations show that for 9

10 any pair of transpositions if they overlap their product is a 3-cycle and if they don t overlap, then they can written as a product of 3-cycles. Hence the 3-cycles generate A n. Next note that since H is a normal subgroup, if it contains one 3-cycle then it contains them all: Suppose not. If a H and b / H are 3-cycles (and n-cycles have order n), then we see that 3 divides H and since bh has order 3 in A 5 we see that 3 divides H A 5. But by Lagrange Theorem A H 5 = A 5 H we would then have that 9 divides H A 5 = = 60. Hence H contains all 3-cycles and so H = A 5. Case 2. H contains a 5-cycle. By a similar argument to Case 1, we have that H contains all 5-cycles. But (12345)(54312) = (132) is a 3-cycle in H and so we are done by Case 1. Case 3. H contains a product of two disjoint transposition. For example, suppose (12)(34) = α H. Let β = (345) A 5. Then by normality of H, βαβ 1 H and (using that disjoint cycles commute): βαβ 1 = (345)(12)(34)(543) = (12)(345)(34)(543) = (12)(35) But the product of α and βαβ 1 is in H and and so we are done by Case 1. α(βαβ 1 ) = (12)(34)(12)(35) = (34)(35) = (543) Every permutation in S 5 can be written as a product of disjoint cycles, e.g., (12) (123) (1234) (12345) (12)(34) (123)(45) The cycle structures of the elements of A 5 are exactly covered by the 3 cases. Corollary 5.8 The group S 5 is not solvable. A 5 is not solvable since it is not abelian and has no nontrivial normal subgroups. A subgroup of solvable group is solvable (exercise) and hence S 5 is not solvable. Lemma 5.9 Suppose G is a subgroup of S 5 which contains a transposition and a 5- cycle. Then G = S 5. By automorphing G around we can assume without loss of generality that (12345) and (1i) are elements of G. Again by symmetry it really reduces to cases (1i) = (12) or (1i) = (13). In the first case by using the 5-cycle it is clear that any adjacent 10

11 transposition (i i + 1) can be obtained, i.e., rotate around until i on top, switch the top two, then rotate back. But it is easy to see that adjacent transposition generate S n : visual a stack of plates numbered 1 thru n. Any plate can be brought to the top by adjacent switches of plates. Then any plate below the top can then be brought to the second position, etc. Hence any permutation of the plates can be obtained by repeatedly switching adjacent plates. In the other case, (1i) = (13), by using the 5-cycle we can obtain (35). But (13)(35)(13) = (15) and (15) is an adjacent transposition. Example 5.10 There is a polynomial f Q[x] of degree 5 whose splitting field K has aut(k Q) isomorphic to S 5, i.e., the Galois group of f is S 5. Suppose f(x) is irreducible and has exactly three real roots. Then we claim that its Galois group G is S 5. Since the nonreal roots of a real polynomial must occur in complex conjugate pairs, the roots of f must be α 1, α 2, α 3, β, β. Conjugation is an automorphism of C which fixes the α i and swaps β and β. Hence G contains a transposition. Also since [Q(α 1 ) : Q] = 5 we know that 5 divides [K : Q] where K is the splitting field of f. But since G = [K : Q] (Theorem 2.8) we have that 5 divides the order of G. By Cauchy s Theorem G has an element of order 5. The only elements of S 5 of order 5 are 5-cycles, hence by Lemma 5.9 it must be that G S 5. There are many examples of such polynomials. Let f(x) = x 5 80x + 5 Then f is irreducible by Eisenstein s criterion with prime 5. f (x) = 5x 4 80 has zeros at 2 and 2, f( 2) > 0 and f(2) < 0. But since f is positive in the interval (, 2), its increasing there and so has exactly one real root in this interval. Similarly f is negative in ( 2, 2) so f is decreasing and so has exactly one real root in this interval. Finally f has exactly one real root in the interval (2, ). Corollary 5.11 (Abel) Fifth degree polynomials are not solvable by radicals. 6 Solvable by real radicals, constructible polygons Definition 6.1 A polynomial f(x) Q[x] is solvable by real radicals iff its roots are in the smallest subfield S R which is closed under taking real roots, i.e., if a S, a > 0 and n N then n a S. Lemma 6.2 Suppose F C is a subfield, p a prime, and a F. Then f(x) = x p a is reducible in F iff it has a root in F. 11

12 Suppose f(x) = x p a = g(x)h(x) is reducible in F. We must find a root of it in F. So let β C be any root of f(x) and let α = e 2πi p be the usual primitive pth root of unity. Then we know that the zeros of f are: β, αβ, α 2 β,..., α p 1 β and f factors in C by p 1 f(x) = x p a = (x α i β) Now suppose that the degree of g(x) is k with 1 k < p. If we factored g in C we would see that it contains k of the factors of f, i.e., there exists a set A {0, 1,..., p 1} of size k such that g(x) = i A(x α i β) Suppose b F is the constant term of g. We will show how to use b and a to construct a root of f. First note that b = ( 1) k α l β k for some l. Let c = α l β k and notice that c p = (β p ) k = a k since α p = 1. Now since p is prime and 1 k < p we have that there exists s, t Z with sk + tp = 1. It follows that Hence (c s a t ) F is a root of f. i=0 a 1 = a sk+tp = ((a k ) s )((a t ) p ) = ((c p ) s )((a t ) p ) = (c s a t ) p Theorem 6.3 Suppose f(x) Q[x] is an irreducible cubic with three real roots. Then f(x) is not solvable by real radicals. For contradiction suppose K S is the splitting field of f(x). Claim. There exists a subfield F 0 such that Q F 0 K and [K : F 0 ] = 3. Note that the Galois group of f is a subgroup of the group S 3 of permutations of its three roots. Since f is irreducible we have that [Q(α) : Q] = 3 where α is any of the three roots of f. So if K = Q(α) we can take F 0 = Q. Otherwise the Galois group of f is S 3. Let H S 3 be any (the) subgroup of S 3 with H = 3. Then the fixed field: F 0 = fix(h) = {x K : σ H σ(x) = x} has the property desired, i.e., [K : F 0 ] = aut(k F 0 ) = H = 3 by Theorems 2.8 and 3.7. This proves the Claim. Note: f is irreducible over F 0. This is because [F 0 : Q] is either 1 or 2 but f is an irreducible cubic, so if α any root of f we have [Q(α) : Q] = 3. Note: If F 0 F then either f is irreducible over F or f splits in F. This is because [K : F 0 ] = 3 and hence for any root α of f we have that F 0 [α] = K. Hence if one root exists in F then all roots are in F. 12

13 Note: S is the smallest subfield of R closed under taking real roots of prime degree. This is because, for example, the sixth root is the square root of the cube root, etc. Therefore, it follows that if K S there must be some subfield F S with F 0 F S, a F and p prime such that f is irreducible over F but f is reducible over F ( p a). By the Lemma we have that x p a is irreducible over F and so [F ( p a) : F ] = p. We also know that for any root α of f that [F (α) : F ] = 3 and since F F (α) [F [ p a] we have that 3 divides p. Since p was prime we must have that p = 3 and therefore F (α) = F ( 3 a). Since adding one root of f adds all roots of f we know that F ( 3 a) is the splitting field of f over F. But this contradicts Theorem 3.1 which says that for a Galois extension an irreducible which has a root splits. But the irreducible polynomial x 3 a has only one root in F ( 3 a), the other two roots are not real. Theorem 6.4 (Gauss) Let p be a prime. Then the regular p-gon is constructible with straight edge and compass iff p is a Fermat prime, i.e., p = 2 2n + 1 for some n. See Chapter

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