# Notes on Field Extensions

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1 Notes on Field Extensions Ryan C. Reich 16 June Definitions Throughout, F K is a finite field extension. We fix once and for all an algebraic closure M for both and an embedding of F in M. When necessary, we write K = F (α 1,..., α n ), and K 0 = F, K i = F (α 1,..., α i ), q i the minimal polynomial of α i over F i 1, Q i that over F. 1.1 Definition. Aut(K/F ) denotes the group of automorphisms of K which fix F (pointwise!). Emb(K/F ) denotes the set of embeddings of K into M respecting the chosen embedding of F. 1.2 Definition. For an element α K with minimal polynomial q, we say q and α are separable if q has distinct roots, and we say K is separable if this holds for all α; conversely, we say they are purely inseparable if q has only one root. We say K is splitting if each q splits in K. 1.3 Definition. By deg(k/f ) we mean the dimension of K as an F -vector space. We denote K s /F the set of elements of K whose minimal polynomials over F have distinct roots; by 2.12 this is a subfield, and deg(k s /F ) = deg s (K/F ) and deg(k/k s ) = deg i (K/F ) by definition. 1.4 Definition. If K = F (α) for some α with minimal polynomial q(x) F [x], then by 2.2, q(x) = r(x pd ), where p = char F (or 1 if char F = 0) and r is separable; in this case we also denote deg s (K/F ) = deg(r), deg i (K/F ) = p d. 2 Theorems 2.1 Lemma. q(x) F [x] is separable if and only if gcd(q, q ) = 1, where q is the formal derivative of q. Proof. Passing to M, we may factor q: and by the product rule, q (x) = n q(x) = (x r i ) mi j=1 m j (x r j ) mj 1 i j (x r i ) mi. gcd(q, q ) is independent of the field since K[x] is a PID, hence is the product of the common factors of q and q over M. q and q have a common factor in M if and only if they have a common root, since they both split, and therefore if and only if q vanishes at some r j. Given the above representation, this holds if and only if m j > Lemma. If char F = 0 then K s = K. If char F = p > 0, then for any irreducible q(x) K[x], there is some d 0 and polynomial r(x) K[x] such that q(x) = r(x pd ), and r is separable and irreducible. 1

2 Proof. By formal differentiation, q (x) has positive degree unless each exponent is a multiple of p; in characteristic zero this never occurs. If this is not the case, since q is irreducible, it can have no factor in common with q and therefore has distinct roots by 2.1. If p > 0, let d be the largest integer such that each exponent of q is a multiple of p d, and define r by the above equation. Then by construction, r has at least one exponent which is not a multiple of p, and therefore has distinct roots. 2.3 Corollary. In the statement of 2.2, q and r have the same number of roots. Proof. α is a root of q if and only if α pd is a root of r; i.e. the roots of q are the roots of x pd β, where β is a root of r. But if α is one such root, then (x α) pd = x pd α pd = x pd β since char K = p, and therefore α is the only root of x pd β. 2.4 Lemma. The correspondence which to each g Emb(K/F ) assigns the n-tuple (g(α 1 ),..., g(α n )) of elements of M is a bijection from Emb(K/F ) to the set of tuples of β i M, such that β i is a root of q i over K(β 1,..., β i 1 ). Proof. First take K = F (α) = F [x]/(q), in which case the maps g : K M over F are identified with the elements β M such that q(β) = 0 (where g(α) = β). Now, considering the tower K = K n /K n 1 /... /K 0 = F, each extension of which is primitive, and a given embedding g, we define recursively g 1 Emb(K 1 /F ) by restriction and subsequent g i by identifying K i 1 with its image and restricting g to K i. By the above paragraph each g i corresponds to the image β i = g i (α i ), each of which is a root of q i. Conversely, given such a set of roots of the q i, we define g recursively by this formula. 2.5 Corollary. Emb(K/F ) = n deg s(q i ). Proof. This follows immediately by induction from 2.4 by Lemma. For any f Emb(K/F ), the map Aut(K/F ) Emb(K/F ) given by σ f σ is injective. Proof. This is immediate from the injectivity of f. 2.7 Corollary. Aut(K/F ) is finite. Proof. By 2.6, Aut(K/F ) injects into Emb(K/F ), which by 2.5 is finite. 2.8 Proposition. The inequality is an equality if and only if the q i all split in K. Aut(K/F ) Emb(K/F ) Proof. The inequality follows from 2.6 and from 2.7. Since both sets are finite, equality holds if and only if the injection of 2.6 is surjective (for fixed f Emb(K/F )). If surjectivity holds, let β 1,..., β n be arbitrary roots of q 1,..., q n in the sense of 2.4, and extract an embedding g : K M with g(α i ) = β i. Since the correspondence f f σ (σ Aut(K/F )) is a bijection, there is some σ such that g = f σ, and therefore f and g have the same image. Therefore the image of K in M is canonical, and contains β 1,..., β n for any choice thereof. If the q i all split, let g Emb(K/F ) be arbitrary, so the g(α i ) are roots of q i in M as in 2.4. But the q i have all their roots in K, hence in the image f(k), so f and g again have the same image, and f 1 g Aut(K/F ). Thus g = f (f 1 g) shows that the map of 2.6 is surjective. 2

3 2.9 Corollary. Define D(K/F ) = Then the chain of equalities and inequalities deg s (K i /K i 1 ). Aut(K/F ) Emb(K/F ) = D(K/F ) deg(k/f ) holds; the first inequality is an equality if and only if each q i splits in K, and the second if and only if each q i is separable. Proof. The statements concerning the first inequality are just 2.8; the interior equality is just 2.5; the latter inequality is obvious from the multiplicativity of the degrees of field extensions; and the deduction for equality follows from the definition of deg s Corollary. The q i respectively split and are separable in K if and only if the Q i do and are. Proof. The ordering of the α i is irrelevant, so we may take each i = 1 in turn. Then Q 1 = q 1 and if either of the equalities in 2.9 holds then so does the corresponding statement here. Conversely, clearly each q i divides Q i, so splitting or separability for the latter implies that for the former Corollary. Let α K have minimal polynomial q; if the Q i are respectively split, separable, and purely inseparable over F then q is as well. Proof. We may take α as the first element of an alternative generating set for K/F. The numerical statement of 2.9 does not depend on the particular generating set, hence the conditions given hold of the set containing α if and only if they hold of the canonical set α 1,..., α n. For purely inseparable, if the Q i all have only one root then Emb(K/F ) = 1 by 2.9, and taking α as the first element of a generating set as above shows that q must have only one root as well for this to hold Corollary. K s is a field and deg(k s /F ) = D(K/F ). Proof. Assume char F = p > 0, for otherwise K s = K. Using 2.2, write each Q i = R i (x pd i ), and let β i = α pd i i. Then the β i have R i as minimal polynomials and the α i satisfy s i = x pd i β i over K = F (β 1,..., β n ). Therefore the α i have minimal polynomials over K dividing the s i and hence those polynomials have but one distinct root. By 2.11, the elements of K are separable, and those of K purely inseparable over K. In particular, since these minimal polynomials divide those over F, none of these elements is separable, so K = K s. The numerical statement follows by computation: deg(k/k ) = p di = 2.13 Theorem. The following inequality holds: deg(k i /K i 1 ) deg s (K i /K i 1 ) = deg(k/f ) D(K/F ). Aut(K/F ) Emb(K/F ) = deg s (K/F ) deg(k/f ). Equality holds on the left if and only if K/F is splitting; it holds on the right if and only if K/F is separable. Proof. The numerical statement combines 2.9 and The deductions combine 2.10 and

4 3 Definitions Throughout, we will denote as before K/F a finite field extension, and G = Aut(K/F ), H a subgroup of G. L/F is a subextension of K/F. 3.1 Definition. When K/F is separable and splitting, we say it is Galois and write G = Gal(K/F ), the Galois group of K over F. 3.2 Definition. The fixed field of H is the field K H of elements fixed by the action of H on K. Conversely, G L is the fixing subgroup of L, the subgroup of G whose elements fix L. 4 Theorems 4.1 Lemma. A polynomial q(x) K[x] which splits in K lies in K H [x] if and only if its roots are permuted by the action of H. In this case, the sets of roots of the irreducible factors of q over K H are the orbits of the action of H on the roots of q (counting multiplicity). Proof. Since H acts by automorphisms, we have σq(x) = q(σx) as a functional equation on K, so σ permutes the roots of q. Conversely, since the coefficients of σ are the elementary symmetric polynomials in its roots, H permuting the roots implies that it fixes the coefficients. Clearly q is the product of the polynomials q i whose roots are the orbits of the action of H on the roots of q, counting multiplicities, so it suffices to show that these polynomials are defined over K H and are irreducible. Since H acts on the roots of the q i by construction, the former is satisfied. If some q i factored over K H, its factors would admit an action of H on their roots by the previous paragraph. The roots of q i are distinct by construction, so its factors do not share roots; hence the action on the roots of q i would not be transitive, a contradiction. 4.2 Corollary. Let q(x) K[x]; if it is irreducible, then H acts transitively on its roots; conversely, if q is separable and H acts transitively on its roots, then q(x) K H [x] is irreducible. Proof. Immediate from Lemma. If K/F is Galois, so is K/L, and Gal(K/L) = G L.. Proof. K/F Galois means that the minimal polynomial over F of every element of K is separable and splits in K; the minimal polynomials over L = K H divide those over F, and therefore this is true of K/L as well; hence K/L is likewise a Galois extension. Gal(K/L) = Aut(K/L) consists of those automorphisms σ of K which fix L; since F L we have a fortiori that σ fixes F, hence Gal(K/L) G and consists of the subgroup which fixes L; i.e. G L. 4.4 Corollary. If K/F and L/F are Galois, then the action of G on elements of L defines a surjection of G onto Gal(L/F ). Thus G L is normal in G and Gal(L/F ) = G/G L. Conversely, if N G is normal, then K N /F is Galois. Proof. L/F is splitting, so by 4.1 the elements of G act as endomorphisms (hence automorphisms) of L/F, and the kernel of this action is G L. By 4.3, we have G L = Gal(K/L), so G L = Gal(K/L) = [K : L] = [K : F ]/[L : F ], or rearranging and using that K/F is Galois, we get G / G L = [L : F ] = Gal(L/F ). Thus the map G Gal(L/F ) is surjective and thus the induced map G/G L Gal(L/F ) is an isomorphism. Conversely, let N be normal and take α K N. For any conjugate β of α, we have β = g(α) for some g G; let n N. Then n(β) = (ng)(α) = g(g 1 ng)(α) = g(α) = β, since g 1 ng N by normality of N. Thus β K N, so K N is splitting, i.e., Galois. 4.5 Proposition. If K/F is Galois and H = G L, then K H = L. 4

5 Proof. By 4.3, K/L and K/K H are both Galois. By definition, Gal(K/L) = G L = H; since H fixes K H we certainly have H < Gal(K/K H ), but since L K H we have a fortiori that Gal(K/K H ) < Gal(K/L) = H, so Gal(K/K H ) = H as well. It follows from 2.13 that deg(k/l) = H = deg(k/k H ), so that K H = L. 4.6 Lemma. If K is a finite field, then K is cyclic. Proof. K is then a finite extension of F p for p = char K, hence has order p n, n = deg(k/f p ). Thus α pn = α for all α K, since K = p n 1. It follows that every element of K is a root of q n (x) = x pn x. For any d < n, the elements of order at most p d 1 satisfy q d (x), which has p d roots. It follows that there are at least p n (p 1) > 0 elements of order exactly p n 1, so K is cyclic. 4.7 Corollary. If K is a finite field, then Gal(K/F ) is cyclic, generated by the Frobenius automorphism. Proof. First take F = F p. Then the map f i (α) = α pi is an endomorphism, injective since K is a field, and surjective since it is finite, hence an automorphism. Since every α satisfies α pn = α, f n = 1, but by 4.6, f n 1 is nontrivial (applied to the generator). Since n = deg(k/f ), f = f 1 generates Gal(K/F ). If F is now arbitrary, by 4.5 we have Gal(K/F ) = Gal(K/F p ) F, and every subgroup of a cyclic group is cyclic. 4.8 Corollary. If K is finite, K/F is primitive. Proof. No element of G fixes the generator α of K, so it cannot lie in any proper subfield. F (α) = K. Therefore 4.9 Proposition. If F is infinite and K/F has only finitely many subextensions, then it is primitive. Proof. We proceed by induction on the number of generators of K/F. If K = F (α) we are done. If not, K = F (α 1,..., α n ) = F (α 1,..., α n 1 )(α n ) = F (β, α n ) by induction, so we may assume n = 2. There are infinitely many subfields F (α 1 + tα 2 ), with t F, hence two of them are equal, say for t 1 and t 2. Thus, α 1 + t 2 α 2 F (α 1 + t 1 α 2 ). Then (t 2 t 1 )α 2 F (α 1 + t 1 α 2 ), hence α 2 lies in this field, hence α 1 does. Therefore K = F (α 1 + t 1 α 2 ) Corollary. If K/F is separable, it is primitive, and the generator may be taken to be a linear combination of any finite set of generators of K/F. Proof. We may embed K/F in a Galois extension M/F by adjoining all the conjugates of its generators. Subextensions of K/F are as well subextensions of K /F and by 4.5 the map H (K ) H is a surjection from the subgroups of G to the subextensions of K /F, which are hence finite in number. By 4.8 we may assume F is infinite. The result now follows from Corollary. If K/F is Galois and H G, then if L = K H, we have H = G L. Proof. Let α be a primitive element for K/L. The polynomial h H (x h(α)) is fixed by H, and therefore has coefficients in L, so α has H conjugate roots over L. But since α is primitive, we have K = L(α), so the minimal polynomial of α has degree deg(k/l), which is the same as the number of its roots. Thus H = deg(k/l). Since H G L and G L = deg(k/l), we have equality Theorem. The correspondences H K H, L G L define inclusion-reversing inverse maps between the set of subgroups of G and the set of subextensions of K/F, such that normal subgroups and Galois subfields correspond. Proof. This combines 4.5, 4.11, and

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