MATH 101A: ALGEBRA I, PART D: GALOIS THEORY 295. Finite fieldsOn the last day I asked several questions about F28and we tried toanswer them.5.1. Galois group. The first question was: What is the Galois groupof F28/F2? Since F28is a vector space of dimension 8 over F2, its degreeis 8. So, Gal(F28/F2) has 8 elements. Which group is it?To answer this we looked for intermediate fields. If Fqis containedin F28then 28= qnfor some n ≥ 1. So, q = 1, 2, 22, 24, 28. So, thereare only two intermediate fields: F4, F16. This means the Galois grouphas exactly 2 nontrivial proper subgroups. So, it must be cyclic.Gal(F28/F2)∼=Z/8.In fact the Galois group of any finite field is cyclic.Theorem 5.1. The Galois group of Fpn/Fpis a cyclic group of ordern generated by the Frobeniusφ(x) = xp.Proof. The Frob enius is a homomorphism φ : K → K for any fieldK or characteristic p. Its kernel is trivial since xp= 0 implies x = 0.Therefore, φ is an automorphism for any finite field. So, it is an elementof the Galois group. The fixed field of this element is the set of all rootsofXp− XBut the p elements of the prime field Fpare roots of this polynomial.SoFp= F!φ"pn.By the Galois correspondence, this implies that %φ& = Gal(Fpn/Fp). !Corollary 5.2. Gal(Fpnm/Fpn) is the cyclic group generated by φn.5.2. the field F4. has only 4 elements: F4= {0, 1, α , α + 1}.Proposition 5.3. The irreducible polynomial of α isX2+ X + 1.30 MATH 101A: ALGEBRA I, PART D: GALOIS THEORYProof. 0, 1 are not roots of this polynomial. So, the polynomial isirreducible and defines a degree 2 extension of F2.In class I gave another proof. There are only 4 polynomials of degree2: the one above andX2+ X, X2+ 1, X2.But these polynomial are not irreducible:X2+ X = X(X + 1), X2+ 1 = (X + 1)2So, X2+ X + 1 is the unique degree 2 irreducible polynomial over F2.So, it must be the irreducible polynomial of α. !If we were to represent elements of F4in binary notation we wouldwrite the elements as: (0, 0), (0, 1), (1, 0), (1, 1) where(0, 0) = 0, (0, 1) = 1, (1, 0) = α, (1, 1) = α + 1.Addition is coordinate-wise and multiplication is given by the irre-ducible polynomial.5.3. the field F16. We know that F16= F4(β) for any element β ofF16which is not in F4. We want the irreducible polynomial of β overF2since this will let us write elements of F16as strings of four 0’s and1’s and tell us how to multiply them.The irreducible polynomial of β over F4is quadratic. One possibilityisf(X) = X2+ X + α.The four elements of F4are not roots of this polynomial. So, it isirreducible. If β is a root then the irreducible polynomial of β over F2isff = (X2+ X + α)(X2+ X + α)where conjugation α is given by the element of the Galois group, φ.So, α = φ(α) = α2= α + 1 andirr(β, F2) = (X2+X+α )(X2+X+α+1) = (X2+X+α )2+(X2+X+α )= X4+ X2+ α + 1 + X2+ X + α = X4+ X + 1.This polynomial has 4 roots. Since F16has 16 − 4 = 12 generators andeach degree 4 irreducible polynomial has 4 roots, there must be twomore irreducible polynomials.Theorem 5.4. There are exactly three irreducible degree 4 polynomialsover F2:(1) f1(X) = X4+ X + 1(2) f2(X) = X4+ X3+ 1(3) f3(X) = X4+ X3+ X2+ X + 1.MATH 101A: ALGEBRA I, PART D: GALOIS THEORY 31Proof. There are only 8 polynomial of degree 4 with nonzero constantterm. The other 5 are reducible since:X4+ 1, X4+ X2+ X + 1, X4+ X3+ X + 1, X4+ X3+ X2+ 1have an even number of terms and thus have X = 1 as a root andX4+ X2+ 1 = (X2+ X + 1)2.I also pointed out that f2is the reverse of f1and is thus irreducible(the roots of f2are the inverses of the roots of f1). The roots of f3are5th roots of unity, so they are not elements of F4. !Here is an example of how multiplication is done in F16using theirreducible polynomial X4+ X + 1.(1101)(0101) = 1101 + 11, 0100 = 11, 1001 = 1, 1111 = 1100where the last two reductions use the irreducible polynomial which is1,0011:11, 1001 = 11, 1001 + 10, 0110 = 1, 1111 = 1, 1111 + 1, 0011 = 1100.The commas are just to make it easier to read the numbers.5.4. the field F256. We didn’t get very far with this. The usual irre-ducible polynomial isg(X) = X8+ X7+ X2+ X + 1 = 1, 1000, 01115.4.1. number of irreducible polynomials. The number of irreduciblepolynomials of degree 8 is256 − 168=2408= 30.The number of polynomials of degree 8 with nonzero constant term is27= 128. Half of these have an even number of terms making X = 1a root. This leaves 64. There are 25= 32 polynomials which have αas a root and half of them have an odd number of terms. This leaves64 − 16 = 48. We can multiply any two of the three irreducible degree4 polynomials. There are 6 ways to do that. This leaves 42 left. Thereare two irreducible polynomials of degree 3, namely,X3+ X + 1, X3+ X2+ 1.And there are 6 irreducible polynomials of degree 5. (Three are 8polynomials of degree 5 with nonzero constant term and an odd numb erof terms. Two of them factor as X2+X +1 times one of the two degree3 irreducibles.) This makes 2 · 6 = 12 products leaving 42 − 12 = 30irreducible polynomials of degree 8.32 MATH 101A: ALGEBRA I, PART D: GALOIS THEORY5.4.2. irreducible polynomial over intermediate fields. Let γ be a rootof the polynomial g(X). Then what is the irreducible polynomial of γover F4? over F16?Since φ4generates the Galois group of F28/F16, the polynomial of γover F16is(X − γ)(X − φ4(γ)) = (X − γ)(X − γ16) = X2+ (γ + γ16)X + γ17.Using a computer, I calculated this in binary notation:γ17= 1101, 1110, γ16= 0110, 1111.So,irr(γ, F16) = X2+ 0110, 1101X + 1101, 1110.Some further computer calculations show thatα = 1010, 1010, β = 1101, 1110satisfy the equations:α2+ α = 1, β2+ β = α, β4+ β = 1.So, we can identify them with the generators of F4and F16that wechose earlier. In this notation we have:irr(γ, F16) = X2+ (α + 1)(β + 1)X + β.Applying the generator φ2of Gal(F16/F4) we getirr(γ4, F16) = X2+ (α + 1)βX + β + 1The product of these isirr(γ, F4) = X4+ αX3+ αX2+ αX + α.If we multiply this with the conjugate:irr(γ2, F4) = X4+ αX3+ αX2+ αX + αwe get back the original irreducible polynomialirr(γ, F2) = X8+ X7+ X2+ X + 1.5.4.3. the order of γ. One last point: The calculationγ17= βimplies that γ is a generator of the cyclic groupF×256∼=Z/255∼=Z/17 ⊕ Z/5 ⊕ Z/3since β has order 15. (The elements of order 3 lie in F4and the elementsof order 5 are roots of the irreducible polynomial f3(X). β is a root