GALOIS THEORY AT WORK: CONCRETE EXAMPLES KEITH CONRAD 1. Examples p p Example 1.1. The field extension Q( 2; 3)=Q is Galois of degree 4, so its Galoisp group phas order 4. The elementsp of the Galoisp groupp are determinedp by their values on 2 and 3. The Q-conjugates of 2 and 3 are ± 2 and ± 3, so we get at most four possible automorphisms in the Galois group. Seep Table1.p Since the Galois group has order 4, these 4 possible assignments of values to σ( 2) and σ( 3) all really exist. p p σ(p 2) σ(p 3) p2 p3 p2 −p 3 −p2 p3 − 2 − 3 Table 1. p p Each nonidentity automorphismp p in Table1 has order 2. Since Gal( Qp( 2p; 3)=Q) con- tains 3 elements of order 2, Q( 2; 3) has 3 subfields Ki suchp that [Q( 2p; 3) : Ki] = 2, orp equivalently [Ki : Q] = 4=2 = 2. Two such fields are Q( 2) and Q( 3). A third is Q( 6) and that completes the list. Here is a diagram of all the subfields. p p Q( 2; 3) p p p Q( 2) Q( 3) Q( 6) Q p Inp Table1, the subgroup fixing Q( 2) is the first and secondp row, the subgroup fixing Q( 3) isp the firstp and thirdp p row, and the subgroup fixing Q( 6) is the first and fourth row (since (− 2)(− 3) = p2 3).p p p The effect of Gal(pQ( 2;p 3)=Q) on 2 + 3 is given in Table2. The 4 valuesp p are allp different,p since 2 and 3 are linearlyp independentp over Q. Therefore Q( 2; 3) = Q( 2 + 3). The minimal polynomial of 2 + 3 over Q must be p p p p p p p p (X − ( 2 + 3))(X − (− 2 + 3))(X − ( 2 − 3))(X − (− 2 − 3)) = X4 − 10X2 + 1: In particular, X4 − 10X2 + 1 is irreducible in Q[X] since it's a minimal polynomial over Q. 1 2 KEITH CONRAD p p p p σ(p 2) σ(p 3) σ(p 2 +p 3) p2 p3 p2 + p3 p2 −p 3 p2 − p3 −p2 p3 −p2 + p3 − 2 − 3 − 2 − 3 Table 2. By similar reasoning if a field F does not have characteristicp p 2 and a and b are nonsquares in F such that ab is not a square either, then [F ( a; b): F ] = 4 and all the fields between p p F and F ( a; b) are as in the following diagram. p p F ( a; b) p p p F ( a) F ( b) F ( ab) F p p p p Furthermore, F ( a; b) = F ( a + b). The argument is identical to the special case above. p p p 4 4 4 Example 1.2. The extension Q( 2)=Q is not Galois,p but Q( 2) lies in Q( 2; i), which 4 is Galoisp over Q. We will use Galois theory for Q( 2; i)=Q to find the intermediate fields 4 in Q( 2)=Q. p The Galois group of Q( 4 2; i)=Q equals hr; si, where p p p p r( 4 2) = i 4 2; r(i) = i and s( 4 2) = 4 2; s(i) = −i: p (Viewing elements of Q( 4 2; i) as complex numbers, s acts on them like complex conjuga- tion.) The group hr; si is isomorphic to D4, where r corresponds to a 90 degree rotation of the squarep and s corresponds to a reflectionp across a diagonal. What is the subgroup H of Gal(Q( 4 2; i)=Q) corresponding to Q( 4 2)? p Q( 4 2; i) f1g 2 2 p (1.1) Q( 4 2) H 4 4 Q D4 p 4 Since s isp a nontrivialp element of the Galois group that fixes Q( 2), s 2 H. The size 4 4 of Hp is [Q( 2; i): Q( 2)] = 2, so H p= f1; sg = hsi. By the Galois correspondence for Q( 4 2; i)=Q, fields strictly between Q( 4 2) and Q correspond to subgroups of the Galois group strictly between hsi and hr; si. From the known subgroup structure of D4, the only 2 subgroup lyingp strictly between hspi and hr; si is hr ; si. Therefore only one field lies strictly between Q( 4 2) and Q. Since Q( 2) is such a field it is the only one. GALOIS THEORY AT WORK: CONCRETE EXAMPLES 3 Remark 1.3. While Galois theory provides the most systematic method to find intermedi-p 4 ate fields, it may be possiblep to argue in other ways. Forp example, suppose Q ⊂ F ⊂ Q( 2) with [F : Q] = 2. Then 4 2 has degree 2 over F . Since 4 2 is a root of X4 − 2, its minimal 4 polynomial overp F has to be a quadratic factor of X − 2.p There are three monic quadraticp 4 2 4 factors with 2 as a root, butp only one of them, X − 2, has coefficientsp in Q(p2) (let 2 4 alone in R). Therefore X p− 2 must be the minimal polynomial of 2 over F , so 2 2 F . Since [F : Q] = 2, F = Q( 2) by counting degrees. p 4 2πi=8 Example 1.4. Let's explore Q( 2; ζ8), where ζ8 = ep is a root of unity of order 8, 4 4 whose minimal polynomialp over Q is X + 1. Both Q( 2) and Q(ζ8) have degree 4 over 2 4 4 4 Q. Since ζ8 = i, Q( 2; ζ8) is a splitting field over Q of (X − 2)(X + 1) and therefore is Galois over Q. What is its Galois group? We have the following field diagram. p 4 Q( 2; ζ8) ≤4 ≤4 p 4 Q( 2) Q(ζ8) 4 4 Q p 4 Thus [Q( 2; ζ8): Q] is at mostp 16. We will see the degree is not 16: there are some hidden 4 algebraic relations betweenp 2 and ζ8. 4 Any σ 2 Gal(Q( 2; ζ8)=Q) is determined by its values p p a × 4 b 4 (1.2) σ(ζ8) = ζ8 (a 2 (Z=8Z) ) and σ( 2) = i 2 (b 2 Z=4Z): There are 4 choices each for a and b. Taking independent choices of a and b, there are at most 16 automorphisms in thep Galois group. But the choices of a and b can not be made 4 independently because ζ8 and 2 are linked to each other: π p p 2 (1.3) ζ + ζ−1 = e2πi=8 + e−2πi=8 = 2 cos = 2 = 4 2 : 8 8 4 p p 4 This says 2 belongs top both Q(ζ8p) and Q( 2). Here is a field diagram that emphasizes 4 the common subfield Q( 2) in Q( 2) and Q(ζ8). This subfield is the source of (1.3). p 4 Q( 2; ζ8) ≤4 ≤4 p 4 Q( 2) Q(ζ8) 2 2 2 p Q( 2) Q(i) 2 2 Q p p p −1 2 4 Rewriting ζ8 + ζ8 = 2 as ζ8 − 2ζ8p+ 1 = 0, ζ8 has degree at mostp 2 over Q( 2). 4 4 Since ζ8 is not real, it isn't inside Q( 2), so it has degree 2 over Q( 2). Therefore 4 KEITH CONRAD p 4 [Q( 2; ζ8): Q] = 2 · 4 = 8 and the degrees marked as \≤ 4" in the diagram both equal 2. We rewrite the field diagram below. p 4 Q( 2; ζ8) 2 2 p 4 Q( 2) Q(ζ8) 2 2 2 p Q( 2) Q(i) 2 2 Q p p 4 4 Returning to the Galois group, (1.3) tells us thep effect of σ 2 Gal(Q( 2; ζ8)=Q) on 2 4 2 −1 partially determines it on ζ8, and conversely: (σ( 2)) = σ(ζ8) + σ(ζ8) , which in the notation of (1.2) is the same as ζa + ζ−a (1.4) (−1)b = 8 p 8 : 2 p p 2πi=8 a −a Since ζ8 = e p = (1 + i)= 2, a calculation shows ζ8 + ζ8 = 2 if a ≡ 1; 7 mod 8 and a −a 7 −1 5 −3 ζ8 + ζ8 = − 2 if a ≡ 3; 5 mod 8 (note ζ8 = ζ8 and ζ8 = ζ8 ). Thus if a ≡ 1; 7 mod 8 b b we have (−1) = 1, so b ≡ 0; 2 mod 4, whilep if a ≡ 3; 5 mod 8 we have (−1) = −1, so 4 3 b ≡ 1; 3 mod 4. For example, σ can't both fix 2 (b = 0) and send ζ8 to ζ8 (a = 3) because (1.4) would not hold. p 4 The simplest way to understandp Q( 2; ζ8) is to use a different set of generators. Since 2πi=8 πi=4 ζ8 = e = e = (1 + i)= 2, p4 p4 Q( 2; ζ8) = Q( 2; i); and from the second representation we knowp its Galois group over Q is isomorphic to D4 4 with independent choices of wherep to send 2 (to any fourth root of 2) and i (to any 4 square root of −1) rather than 2 and ζ8. A different choice of field generators can make it easier to see what thep Galois group looks like. We also see immediately from the second 4 representation that [Q( 2; ζ8): Q] = 8. p 8 8 Example 1.5. Let's considerp the splittingp field of X − 2 over Q, which is Q( 2; ζ8), 2πi=8 8 wherep ζ8 = e = (1 + i)= 2.