GALOIS THEORY - TUTORIAL 9 MISJA F.A. STEINMETZ 1. Finite Fields Suppose that p is a prime number. Recall that Fp := Z=pZ is a field. Moreover, if K is a finite field of characteristic p then K is a finite extension of Fp. Writing [K : Fp] = r; it follows that #K = pr: Conversely, we may ask: does there exist a finite field of size pr for any r? The answer is yes. Non-Example 1 The ring Z=prZ is not a field for any r > 1: If this is not the way to get fields of size pr; how do we get them? For q = pr, if K has size q, then the multiplicative group K× has size q − 1: This means that αq−1 = 1 for any α 2 K×: If we include 0 we thus see that all the elements of K satisfy the equation Xq − X: This suggests the following definition. q Definition. Let Fq be the splitting field of X − X over Fp: Remark. Since splitting fields are only well defined up to isomorphism, this definition only defines Fq up to isomorphism. r Theorem. Suppose p is prime, q = p and Fq is as above. Then (i) #Fq = q; (ii) any finite field of size q is isomorphic to Fq; (iii) The extension Fq=Fp is Galois and the Galois group Gal(Fq=Fp) is cyclic of order r gen- p erated by φ : Fq ! Fq defined by φ : x 7! x : Example 2 ∼ If f(X) 2 Fp[X] is an irreducible polynomial of degree r, then Fp[X]=(f(X)) = Fpr : Date: 19 March 2018. 1 2 MISJA F.A. STEINMETZ Classwork 1 4 Factorise X − X 2 F2[X] and, hence, write an addition and multiplication table for F4 (where you can write α for one of the roots of the polynomial that are not in F2). Classwork 2 4 Let L = F2(α) where α is a root of the irreducible polynomial f(X) = X + X + 1 2 F2[X]: Show that Gal(L=F2) contains exactly one subgroup H of order 2 and find β 2 L such that H L = F2(β): GALOIS THEORY - TUTORIAL 9 3 We have looked at finite extensions K=Fp; but what about finite extensions K=Fq? Recall that this simply means that there is a homomorphism Fq ! K: Classwork 3 Prove that there exists a homomorphism Fpr ! Fps if and only if s is divisible by r: Proposition. If L=K is an extension of finite fields with #K = q; then L is Galois over K, and Gal(L=K) is cyclic, generated by the element φ : L ! L defined by φ(α) = αq: Proposition. Let q = pr, where p is prime and r ≥ 1: Then the polynomial f(X) = Xq − X 2 Fp[X] is the product of all monic irreducible polynomials in Fp[X] of degree dividing r: Classwork 4 Prove that if p; r are primes then the number of monic irreducible polynomials of degree r r in Fp[X] is (p − p)=r: 4 MISJA F.A. STEINMETZ Classwork 5 Find the number of irreducible monic polynomials in F2[X] of degree 6. Strand 5.12, King's College London E-mail address: [email protected] URL: https://www.nms.kcl.ac.uk/misja.steinmetz/.