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Math 250A, November 19 Lecture. Fall 2015

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Math 250A, November 19 Lecture. Fall 2015 Math 250A, November 19 Lecture. Fall 2015 Today: we will classify all finite fields k - these are the fields with finitely many elements - and understand their Galois groups. Classification problems In mathematics we frequently want to classify all objects up to some notion of equivalence (usually isomorphism in some category). Question: what are some examples of classification problems that you know complete answers to? (eg. finite dimensional vector spaces (over some field), finitely generated abelian groups, finitely generated modules over PID.) We will see that finite fields can be classified by the set f(p, n) j p prime, n 2 Z≥1g. More precisely: n Theorem. Let p be prime, n 2 Z≥1. Then, there exists a finite field k with p elements. Moreover, any finite field has pn elements, for some p, n, and finite fields with the same number of elements are isomorphic. The proof is quite straightforward: n `every finite field k is a splitting field for X p − X , for some p, n.' Examples: Here are some examples of finite fields p def ∼ 2 Fp = Z=pZ, Fp[X ]=(f ) for f irreducible, Z[ 2]=(3) = F3[X ]=(X + 1). We will prove an additional result: Theorem. Every finite field k is isomorphic to Fp[X ]=(f ), for p prime, f irreducible. Prime subfield: For any ring R there is a unique homomorphisms of rings ι : Z ! R ; 1 7! 1R . If R = k is a finite field then this map must have nontrivial kernel. Hence, we see that Z= ker ι is a domain, so that ker ι is a prime ideal. Thus, there is a prime p 2 Z such that ker ι = (p) ⊂ Z. Hence, there is a unique subfield in k isomorphic to Fp: we call this the prime subfield. Using multiplication in k, we can consider k as a Fp-vector space. Since k is finite it must be ∼ n a finite dimensional Fp-vector space. This shows that k = Fp as an Fp-vector space, so that jkj = pn. To summarise: n any finite field has p elements, for p prime, n 2 Z≥1. Fix k a finite field. Denote k∗ def= k − f0g; it is a finite abelian group via multiplication in k. In fact: Claim: k∗ is a cyclic group. Proof: Let N = maxfo(x) j x 2 k∗g, where o(x) is the order of x 2 k∗. Since N j jk∗j = pn−1, we must have that N ≤ pn − 1. Applying the fundemantal theorem of finite abelian groups we see that k∗ is isomorphic to Z=m1Z × · · · Z=mr Z, where mi divides mi+1. 1 ∗ In particular, we must have that N = mr and every element in k has order dividing N. Thus, xN = 1 2 k∗, for every x 2 k∗, so that the polynomial X N − 1 2 k[X ] admits at least pn − 1 distinct roots. Therefore, we have pn − 1 ≤ N ≤ pn − 1 so that there must exist an element ξ 2 k∗ of order pn − 1. The result follows. We call a generator of k∗ a primitive root. Fix ξ 2∗ a primitive root. Then, we obtain a ring homomorphism evξ : Fp[X ] ! k ; m 7! m(ξ). k k ∗ Since fevξ(X ) = ξ j k 2 Z≥0g = k we see that evξ is surjective. Hence, Theorem. Any finite field is isomorphic to Fp[X ]=(f ), for f irreducible. Remark: in general, it is very difficult to determine primitive roots in k∗. From what I've read online(!), there is no known algorithm to determine primitive roots, other than a brute-force approach. As such, it is at least as difficult to find a realisation of the above Theorem: to obtain the polynomial f above, we would need to find a primitive root ξ. Then, f is the minimal polynomial of ξ in Fp[X ]. Example: the following fields all have nine elements 2 2 2 F3[X ]=(X + 1), F3[X ]=(X + X + 2), F3[X ]=(X + 2x + 2). In the first example, it can be verified that X is not a generator of the nonzero elements. However, X is a generator of the nonzero elements in the second and third examples. Also, we have yet to explicitly see why the above fields should all be isomorphic. Splitting fields: Recall that a splitting field of a polynomial f 2 E[X ], where E is a field, is an extension F ⊃ E such that f splits into linear factors in F [X ], and so that there exists no proper intermediate field E ⊂ K ⊂ F with this property. Splitting fields have the property that they are (essentially) unique: if F , F 0 are splitting fields of f 2 E[X ], then there exists an isomorphism α : F ! F 0 such that α(u) = u, for all u 2 E. n pn Claim: any finite field k, such that jkj = p , is a splitting field of X − X 2 Fp[X ]. n Proof: Any finite field k with jkj = p is an extension of Fp. Moreover, we've seen that any nonzero u 2 k satisfies upn−1 = 1 so that upn = u, for all u 2 k. Hence, the polynomial X pn − X splits into linear factors in k[X ]. Since X pn − X is seperable (use the derivative test), it admits pn distinct roots. Thus, k is a splitting field for X pn − X . Now, since any two fields of the same cardinality are splitting fields of the same polynomial, they must be isomorphic. Theorem. Any two fields of the same size are isomorphic. pn Furthermore, for any prime p, n 2 Z≥1, we consider F , a splitting field for X − X . Then, the subset fu 2 F j upn = ug is a subfield of F (use binomial theorem). Thus, it must be equal to F . In particular, jF j = pn, again using the X pn − X is seperable. Hence, n Theorem. For any prime p, n 2 Z≥1, there exists a field with p elements. We have now completed the classification of finite fields. Combining the above results we can obtain a bit more. Suppose that f 2 Fp[X ] is irreducible, n deg f = n. Then, k = Fp[X ]=(f ) is a finite field with p elements so that it is a splitting field of X pn − X . In fact, we have seen that upn = u, for every u 2 k. In particular, if u = X 2 k is pn n the coset containing X , then we have X − X = 0 2 k so that f divides X p − X .. This must be true for every irreducible f 2 Fp[X ] so that we have the following 2 pn Theorem. Let f 2 Fp[X ] be irreducible, deg f = n. Then, f divides X − X. Example: Let p = 2. Then, we find X 2 − X = X (X − 1) X 4 − X = X (X − 1)(X 2 + X + 1) X 8 − X = X (X − 1)(X 3 + X + 1)(X 3 + X 2 + 1). In particular, we see that there are exactly two distinct degree 3 irreducible polynomials with coefficients in F2. Subfield lattice: We now provide the subfield structure of a finite field k with pn elements. Let d be a divisor of n. Then, there is exactly one subfield in k of order pd . Moreover, every subfield in k must have order pd , for some divisor d of n. Hence, the subfield lattice for k is isomorphic to the divisor lattice of n. Question: do you know any other mathematical object whose subobject lattice is isomorphic to the divisor lattice of n? (A: a cyclic group of order n) The Galois group of k: Let k be a finite field with pn elements. Then, any automorphism of k must necessarily fix the prime subfield Fp. Thus, we have that Aut(k) = Gal(k=Fp). pn Since k is the splitting field of X − X , it is a normal extension of Fp, and as this polynomial is seperable it is also a seperable extension. Hence, Fp ⊂ k is a Galois extension so that jGal(k=Fp)j = [k : Fp] = n. Observe that the Frobenius morphism ζ : x 7! xp is an automorphism of k. We show that ζ r is a generator of Gal(k=Fp): let r be the order of ζ. Then, ζ = idk so that, for every u 2 k, u = ζr (u) = ζ(u)r = upr . Hence, we must have r ≥ n, which gives that r = n; that is, ζ is a generator of Gal(k=Fp). The appearance of the divisor lattice as the subgroup lattice of Z=nZ and the subfield lattice of k, can be seen as a manifestation of the Galois correspondence for k. 3.
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