Math 201C Assignment 6

Math 201C Assignment 6 Adam (Esteban) Navas May 18, 2010 Exercise (Additional) Let F be a finite field with F = p n. Show that F is a splitting field of x pn x over Z p. Every nonzero u F satisifies u pn 1 = 1 F, so that u is a root of the polynomial x pn 1 1 F F [x]. But this implies that u is a root of x(x pn 1 1) = x pn x Z p [x]. Hence there are p n distinct roots of x pn x (including 0), so that x pn x splits over F. u F was arbitrary, thus F is the splitting field of x pn x over Z p. Exercise V.4.1 Suppose f K[x] splits as f = (x u 1 ) n1 (x u n ) n k (u i distinct; n i 1). Let v 0,..., v k be the coefficients of the polynomial g = (x u 1 ) (x u n ) and let E = K(v 0,..., v k ). Then (a) F is a splitting field of g over E. (b) F is Galois over E. (c) Aut E F = Aut E F. (a) E(u 0,..., u k ) F by construction of E, and since E(u 0,..., u k ) is the splitting field of g over K, E is the minimal such field in which g splits. The reverse containment therefore follows, so that F = E(u 1,..., u k ). Thus F is the splitting field of g over K. (b) This result follows from Exercise V.3.13(iii), because the irreducible factors of g are separable (the roots of g are distinct), and because F is a splitting field of g over E. (c) Aut E F Aut K F because K E, so if σ fixes elements of E, it must fix elements of K. Aut E F Aut K F because σ Aut E F only acts on u 1,..., u k, which implies that σ fixes all elements in Aut K F. Thus, Aut K F = Aut E F. Exercise V.4.2 Suppose K is a subfield of R (so that F may be taken to be a subfield of C) and that f is irreducible of degree 3. Let D be the discriminant of f. Then (a) D > 0 if and only if f has three real roots. (b) D < 0 if and only if f has precisely one real root. 1

(a) By Exercise III.6.16, every polynomial of odd degree in R[x] has a real root, u 1. If f factors into (x u 1 ) and an irreducible degree 2 polynomial, then the remaining roots of f must be conjugate to each other because they are elements of C. ( = ) Suppose D > 0 and f has only 1 real root. Let u 2 = a + bi and u 3 = u 2. Then = (u 1 u 2 )(u 1 u 3 )(u 2 u 3 ) = [(u 1 a) bi][(u 1 a) + bi][2bi] = [(u 1 a) 2 + b 2 ](2bi) = D = 2 < 0. This contradicts D > 0, hence f must have 3 real roots. ( = ) If f has 3 real roots, then (u 1 u 2 ), (u 1 u 3 ), (u 2 u 3 ) R, which implies = (u 1 u 2 )(u 1 u 3 )(u 2 u 3 ) R. (b) This part is true by taking the contrapositive of the forward and backward implications of part (a), and because the roots of f in C R are complex conjugates. Exercise V.4.5 If chark 2 and f K[x] is a cubic whose discriminant is a square in K, then f is either irreducible or factors completely ink. D is the square of an element in K, so by Corollary V.4.6, the Galois group of f must consist only of even permutations. Suppose that f factors into (x u 1 )g(x), where g(x) is a degree 2 irreducible polynomial in K[x]. g must be separable because char K 2, which implies that the Galois group of g is isomorphic to Z 2, by Corollary V.4.3. Therefore there is an automorphism σ that maps u 2 to u 3, and maps u 3 to u 2. But this means σ( ) = σ((u 1 u 2 )(u 1 u 3 )(u 2 u 3 )) = (u 1 u 3 )(u 1 u 2 )(u 3 u 2 ) = (u 1 u 2 )(u 1 u 3 )(u 2 u 3 ) =, contradicting Corollary V.4.6, because σ( ) = if and only if σ is an odd permutation, by Proposition 4.5(ii). Exercise V.4.10(d) Determine the Galois group of: (b) x 3 10 over Q; over Q( 2). 2

char Q and char Q( (2)) 2, 3, and f splits in C with 3 distinct roots: 3 10 e 2πi 3. 3 10, 3 10 e 2πi 3, and f is irreducible in both Q and Q( 2) because none of the roots of f are contained in these fields. Therefore the discriminant, by Proposition 4.8, is 4p 3 27q 2, where f = x 3 + px + q. Taking p = 0 and q = 10 gives D = 2700, which is not the square of any element in either field. Thus, the Galois group of f over both fields must be S 3, by Corollary V.4.7. Exercise V.5.1 If K is a finite field of characteristic p, describe the structure of the additive group of K. K is a finite field of characteristic p, so F is a simple extension of its prime subfield, Z p (Corollary V.5.4), hence K is a vector space over Z p. By Theorem II.2.1(iii), K is isomorphic to a direct sum of, say m, copies of Z p. This m must coincide with the integer n of Corollary V.5.2 because p m = m = K = p n implies m = n. p=1 Z p Exercise V.5.3 If K = p n, then every element of K has a unique pth root in K. For any nonzero u K, u pn 1 = u, which implies u p = u. Hence, ( u pn 1 ) p = u p = u, so that u pn 1 is a pth root of u. This root is unique by the injectivity of the map u u p of Lemma V.5.5. Exercise V.5.4 If the roots of a monic polynomial f K[x] (in some splitting field of f over K) are distinct and form a field, then char K = p and f = x pn x for some n 1. The roots of f are finite, so if they form a field, R, it is finite and therefore has characteristic p. The order of this field is p n 1, so every nonzero u R satisfies u pn 1 = 1 R. Hence every nonzero root u of f is a root of x pn 1 1 R and therefore all roots of f are roots of x pn x. f and x pn 1 x are monic and have the same roots, thus they are equal. 1 R and R has characteristic p, which means p is the least such integer that 1 + + 1 p times is equal 0. 1 is a basis of K, so this implies p does the same to every element of K. Thus K has characteristic p. 3

Math 201C Assignment 7 Adam (Esteban) Navas May 27, 2010 Exercise V.6.2 If u F is purely inseparable over K, then u is purely inseparable over any intermediate field E. Hence if F is purely inseparable over K, then F is purely inseparable over E. If u E then it is purely inseparable, so consider u F \E. Since u F is purely inseparable over K, (x u) m F [x] is its irreducible polynomial in K[x]. K[x] E[x], hence this polynomial is in E[x]. But u / E, so that (x u) m is also irreducible in E[x]. Thus u F is purely inseparable over E. Arbitrariness of u implies F is purely inseparable over E. Exercise V.6.3 If F is purely inseparable over an intermediate field E and E is purely inseparable over K, then F is purely inseparable over K. Consider the case when char K = p 0, because if char K = 0 this question is trivial. Theorem 6.4(iii) implies there are integers n, m 0 such that if u F and v E, then u pn E and v pm K. Hence, for all u F, (u pn ) pm = u pn+m K, so that F is purely inseparable over K. Exercise V.6.5 If char K = p 0 and a K but a / K pn, then x pn a K[x] is irreducible for every n > 1. Let f(x) = x pn a and let u i be the roots of f in the splitting field F = K(u 1,..., u p n) of K. This splitting field is purely inseparable over K, so any u i F has minimal polynomial f = x pr b K[x] by Theorem 6.4(ii). Since f is the minimal polynomial of u i and u i is a root of f, we know that r n. If r = n then f(u) = f(u) = 0, which implies f(u) f(u) = a b = 0, i.e., that f(x) = f(x) = x pn a is irreducible. Suppose that r < n. Then u pr i = b K and u pn i = a K. Hence b pn r = (u pr i ) pn r = u pn i = a. But b K b pn r 1 K a = (b pn r 1 ) p K, so that a is an element of K p, contradicting a / K p. Exercise V.6.11 If f K[x] is irreducible of degree m > 0, and char K does not divide m, then f is separable. 1

By Theorem III.6.10, an irreducible polynomial in K[x] is separable if and only if f 0. Since f has degree m > 0, the highest degree term of f looks like ma m x m, where a m K is the coefficient of x m. Since char K m, m cannot be written as a product of char K and an integer n. Thus, f is never 0, so that f is separable. Exercise 1 (Additional) Let K be an infinite field. Show that if the extension F K has finitely many intermediate fields then F is a simple extension of K. We proceed by picking a u F such that [K(u) : K] is maximal and assuming F K(u) to derive a contradiction. If F K(u), then there exists v F \K(u). Since K is an infinite field with only finitely many intermediate fields, there exists a, b K, a b, such that K(u + av) = K(u + bv). If there were no such a and b, then the number of intermediate fields of F K would be infinite. Now, (a b)v = u u + av bv = (u + av) (u + bv) K(u + av). K(u + av) is a field and 0 (a b) K(u + av), so that v = (a b)(a b) 1 v K(u + av). Hence u = (u + av) av K(u + av) = K K(u) K(u + av). But then we have [K(u + av) : K] > [K(u) : K], contradicting maximality of [K(u) : K]. Exercise 2 (Additional) Find a primitive element of the extension Q( 2, i) Q. Find the minimal polynomial of this primitive element Q[x]. There are only finitely many intermediate fields of Q( 2, i) Q, hence Q( 2, i) is a simple extension of Q by Proposition V.6.15(ii)( ). There exists u C such that Q(u) = Q( 2, i). Let u = 1 2 + i 1 2. (1) and (4) of the following relations show that Q( 2) and Q(i) are subsets of Q(u): ( 1 u 2 = 2 + i 1 ) ( 1 2 + i 1 ) = 1 2 2 2 + i 1 2 = i (1) ( 1 u 3 = i 2 + i 1 ) = 1 + i 1 (2) 2 2 2 u 4 = u 2 u 2 = i 2 = 1 (3) u + u 7 = 1 + i 1 + u 3 u 4 = 1 + i 1 ( 1 + i 1 ) = 2 (4) 2 2 2 2 2 2 u Q( 2, i), so that Q(u) Q( 2, i). Q(u) Q( 2, i) because, by relations (1) and (4), Q( 2, i) is generated by elements of Q(u). Thus, Q(u) = Q( 2, i), so that u is primitive. The minimal polynomial of u is f(x) = x 4 + 1 Q[x] because f is irreducible and u 4 is the first power of u in Q. 2

Math 201C Assignment 8 Adam (Esteban) Navas May 31, 2010 Exercise V.7.2 Let F be a finite dimensional extension of a finite field K. The norm N F K F and the trace T K (considered as maps F K) are surjective. Since F K is a finite dimensional extension of a finite field K, by proposition V.5.10, F is finite and Galois over K, with Aut K F cyclic. Therefore by Theorem V.7.2, we have T F K (u) = σ 1 (u) + + σ n (u), where Aut K (F ) = r. Consider n = 1 + + 1, n times, an element of K. Then kn 1 + kn 1 = k(n 1 + + n 1 ) = kn 1 (1 + + 1) = kn 1 n = k. Thus, since σ i fix all elements of K, T K F (kn 1 ) = σ 1 (kn 1 ) + + σ n (kn 1 ) = kn 1 n = k, so that the trace is surjective. K is a finite field, so K = p n for some n 1. Let q = p n. The generator of Aut K (F ), σ(x) = x q, is a Z p -automorphism, so that N K F (x) = σ 1 (x) σ n (x) r 1 = σ i (x) = i=0 r 1 x qi i=0 = x r 1 0 q i = x qr 1 q 1 is a homomorphism. By Additional Exercise 3, K and F are cyclic. Let u 0 be a generator of F, with order q r 1. If [ N F K (u 0 ) ] k = 1, then k q r 1 is divisible by q 1 qr 1. But this only happens if q 1 k, which implies N F K (u 0 ) has order q 1 = p n 1, i.e., that N F K (u 0 ) is a generator of K. Thus, N F K (x) is surjective. Exercise V.7.7 If n is an odd integer such that K contains a primitive nth root of unity and char K 2, then K also contains a primitive 2nth root of unity. Let ζ be a primitive nth root of unity, so that {ζ, ζ 2,..., ζ n = 1} are roots of unity in K. Consider roots of x 2n 1 = (x n 1)(x n +1) K[x], R = { ζ, ζ 2,..., ζ n = 1} {ζ, ζ 2,..., ζ 1 = 1

1}, generated by ζ K. char K 2 so that 1 and 1 are distinct elements in K, and for all i, j < n, 1 = (ζ i ) n ( ζ j ) n = (ζ j ) n = 1, so that each element of R is distinct. Thus, ζ is a 2nth root of unity in K. Exercise V.7.8 If F is a finite dimensional extension of Q, then F contains only a finite number of roots of unity. If F is a finite dimensional extension of Q then, since Q is perfect, F Q is a separable extension. By Proposition V.6.15(i), this extension must therefore be simple, i.e., F = Q(u) for a primitive element u. Since this extension is simple, if F contains a primitive nth root of unity, it only contains other dth roots of unity such that d n, for some maximal n. If F contains such a primitive nth root of unity and n is odd, then by Lemma V.7.10(ii) and Exercise V.7.7, there are 2n roots of unity. If n is even then there are only n roots of unity, {1, e 2πi/n, (e 2πi/n ) 2,..., (e 2πi/n ) n 1 }. Lastly, if F has no primitive root of unity, 1 and 1 are the only roots of unity. Exercise 1 (Additional) Let F be a field and σ 1, σ 2,..., σ n be distinct automorphisms of F. Let a 1, a 2,..., a n F. Prove that if for all u F, then a 1 = a 2 = = a n = 0. a 1 σ 1 (u) + a 2 σ 2 (u) + + a n σ n (u) = 0 Suppose that this set of distinct automorphisms is dependent, i.e., that there exists distinct σ i and 0 a i F such that n > 1 is minimal. Hence for σ 1 (v) different from σ 2 (v) for some v F, we would have: a 1 σ 1 (u)σ 1 (v) + + a n σ n (u)σ n (v) = 0, and a 1 σ 1 (u)σ 1 (v) + + a n σ n (u)σ 1 (v) = 0 so that a 2 [σ 2 (v) σ1(v)]σ 2 (u) + + a n [σ n (v) σ 1 (v)]σ n (u) = 0, for all u F and not all zero coefficients, contradicting minimality of n. Exercise 2 (Additional) Let F K be a cyclic extension of degree n with σ being a generator of Aut K (F ). Prove that N K F = 1 if and only if there exists v F such that u = vσ(v) 1. (= ) Suppose u = vσ(v) 1, so that σ n (v 1 ) = v 1, σ(v 1 ) = σ(v) 1, and σ i (vσ(v) 1 ) = σ i (v)σ i+1 (v) 1. Therefore, N K F (u) = (vσ(v) 1 )(σvσ 2 (v) 1 )(σ 2 vσ 3 (v) 1 ) (σ n 1 vσ n (v) 1 ) = (v)(σ(v) 1 σv)(σ 2 (v) 1 σ 2 v)(σ 3 (v) 1 σ 3 v) (σ n 1 (v) 1 σ n 1 v)v 1 = vv 1 = 1 ( =) Suppose N(u) = 1. Then u 0. By linear independence of distinct automorphisms of a field, there must be a y F such that: 0 v = uy + (uσu)σy + (uσuσ 2 u)σ 2 y + + (uσu σ n 2 u)σ n 2 y) + (uσu σ n 1 u)σ n 1 y 2

Since (uσu σ n 1 u)σ n 1 y = N K F (u)σ n 1 y = σ n 1 y, u 1 v = σv. By a long calculation, this implies u = vσ(v) 1, where σ(v) 0 by injectivity of σ. Exercise 3 (Additional) Let F be any field and F be the multiplicative group of non-zero elements of F. Show that every finite subgroup of F is cyclic. Let G < F be a finite subgroup of F. Since F is a field, this subgroup is abelian, so that it is isomorphic to H = Z m1 Zmk, where m 1 m k. m k H = 0, so that every element u in G must be a root of x m k 1, which has at most mk distinct roots. Thus, k = 1 and G is cyclic. 3

Math 201C Assignment 8 Adam (Esteban) Navas June 5, 2010 Problem 1 Show that the extension Q is infnite dimensional over Q. Deduce that the Galois group Aut Q (Q) is infnite. Suppose by contradiction that Q Q is finite dimensional. Then [Q : Q] = k, for some k > 1. Consider the tower of field extensions Q Q( 2i 2) Q( 2 i 1 2) Q( 4 2) Q( 2) Q The irreducible polynomial of each extension Q( 2i 2) Q( 2 i 1 2) has degree 2, because x 2 2 i 1 is irreducible in Q( 2i 1 2) ( 2 i 2 is not an element of Q( 2 i 1 2)). Hence the dimension of Q( 2 i 2) Q is 2 i for all i. There must exist an n such that 2 n 1 k < 2 n. Since 2n+1 2 Q, we must have that Q is an extension of Q( 2n 2). So, consider the tower of extension fields, Q Q( 2 n 2) Q. This tower shows that [Q : Q( 2n 2)][Q( 2 n 2) : Q] = 2 n [Q : Q( 2n 2)] > k. But we now have a contradiction to our assumption that [Q : Q] = k. Thus, Q must be an infinite dimensional extension field of Q. Problem 2 Let E be the splitting field of (x 3 2)(x 2 3) over Q. Determine the Galois group G = Aut Q (E). Describe 3 nontrivial subgroups of G and the corresponding intermediate fields of the extension E Q. The Galois group of (x 3 2)(x 2 3) over Q is Z 2 S 3, because every Q-automorphism is determined by its action on the roots of the irreducible factors of (x 3 2)(x 2 3). The Galois group of x 2 3, say H 1, is Z 2 (char Q 2), and the Galois group of x 3 2 over Q, call it H 2, is S 3, by Corollary V.4.7, which yields the Galois group of the polyonomial. Corollary V.4.7 concludes such a Galois group for x 3 2 because the discriminant of x 3 2 is -108, which is not the square of any element of Q. We also have a product rather than a direct sum, because the intersection of H 1 and H 2 is not trivial (if σ H 1 sends 3 to 3, then it also sends 3 2e 2πi/3 to 3 2e 2πi/3, so that it is an element of both H 1 and H 2 ). Consider the intermediate fields Q( 3), Q( 3 2), and Q( 3 2, 3). The automorphisms in Aut Q (Q( 3)) are precisely those that fix 3, and act any way on the roots of the irreducible factor x 3 2. Hence Q( 3) corresponds to the subgroup {e} S 3. 1

Denote the automorphisms fixing 3 2 and permuting the (complex) roots of x 2 + 3 2x + ( 3 2) 2 by the permutation (23). Then Q( 3 2) corresponds to the subgroup Z 2 (23), because it is determined by its action on the roots of (x 2 3)(x 2 + 3 2x+( 3 2) 2 ). Lastly, Q( 3 2, 3) corresponds to the subgroup {e} (23) because each automorphism is determined by its action on the roots of x 2 + 3 2x + ( 3 2) 2. Problem 3 Let K be a finite field with K = q. Let K be an algebraic closure of K. Frobenius automorphism F r is the K-automorphism of K which is defined by the formula: F r(u) = u q. Let E be an intermediate field K E K such that [E : K] is finite. Show that F r generates the Galois group Aut K (E). K is a finite field, hence it has order q = p n for some prime p and integer n 1. Since E is a finite dimensional extension of K, say [E : K] = m, it must also be a finite field of characteristic p, of order E = p m, m n. Since E is a finite dimensional extension of a finite field, by Proposition V.5.10, E is Galois over K and Aut K E is cyclic of order [E : K] = m, by the Fundamental Theorem of Galois Theory. F r is a Z p -automorphism of E, by Lemma V.5.5, and F r fixes elements of K because u pn = u for all u K. Hence F r is also a K-automorphism of E. F r m = 1 because E = p m, and E is a splitting field of x pm x over Z p, so that it has p m distinct roots. If the order of F r is some integer k < m, then x pk x would have p m distinct roots, contradicting Theorem III.6.7. Thus, F r has order m, so that it generates Aut K E. Problem 4 Show that every field K of characteristic zero is perfect. It must be shown that all algebraic extensions of K are separable, which will be true if all irreducible polynomials in K[x] are separable. If f K[x] is irreducible, suppose it is not separable. This would imply that f has multiple roots. If f has a multiple root, c, then there is an integer m > 1 such that in a splitting field of f, f(x) = (x c) m g(x) = f (x) = m(x c) m 1 g(x) + (x c) m g (x) = (x c)[m(x c) m 1 g(x) + (x c) m 1 g (x)]. We therefore have f is not relative prime to f, by the contrapositive of Theorem III.6.10(ii), so that there is an h K[x] such that h f and h f. f is irreducible, so that h is an associate of f, with degree bigger than f. The only way for h to divide f is if f is the 0 polynomial, i.e., 0 f = a 1 + 2a 2 x + + na n x n 1 a 1 = 2a 2 = = na n = 0 Since char K = 0, a i = 0 for all 1 i n, therefore f is the constant polynomial a 0, contradicting f having a multiple root. Thus, f is separable, so that K is perfect. 2

Problem 5 Let K be a field of positive characteristic p. Let K pn be the set of all elements of the form u pn where u K. Show that K pn is a subfield of K. Prove that K is a purely inseparable extension of K pn. K pn is clearly a subset of K. The map φ : K K defined by φ(u) = u pn is a Z p - monomorphism of fields. φ is isomorphic onto image, K pn, hence it preserves the additive group structure of (K, +) and the multiplicative group structure of (K, ), so that K pn is a subfield of K. For every element u K, u pn Theorem V.6.4(iii). K pn, hence K is a purely inseparable extension of K pn, by Problem 6 Let K be an algebraic closure of the field K and σ Aut K (K). Consider the field F = {u K : σ(u) = u} Prove that every finite dimensional extension E of F is cyclic. Let Ẽ be the normal closure of E. If it can be shown that Ẽ is Galois over F and Aut F Ẽ is cyclic (hence abelian), then every subgroup of Aut F Ẽ will be normal. In particular, by Theorem V.2.5(ii), Aut E Ẽ Aut F Ẽ implies Aut F Ẽ/Aut E Ẽ = Aut F E. Quotients of cyclic groups are cyclic, so that Aut F E will be cyclic. It must be shown that (Aut F Ẽ) = F for Ẽ to be Galois over F. (Aut F Ẽ) F is trivial. For the reverse containment, if u Aut F Ẽ), then for the σ Aut F K, σ Ẽ : Ẽ K is an F -monomorphism of fields. Ẽ is a normal extension, so by Theorem V.3.14(iii), σ is actually an Ẽ-automorphism of Ẽ. Hence, σ(ẽ) = Ẽ σ(u) = u, so that Ẽ is Galois over F. Consider the subgroup of Aut F Ẽ generated by σ, σ < Aut F Ẽ, of order σ = n. The fixed field of σ is σ = {u Ẽ : σi (u) = u i = 1,..., n}. σ = F because if u F, then σ i (u) = u for all i; and if u σ, then σ(u) = u u F. Ẽ is a finite dimensional extension of F because E is, and Ẽ is Galois, hence by Lemma V.2.10(iii) all intermediate fields and all subgroups are closed. [Aut F Ẽ : σ ] = Aut F Ẽ <, σ so that by Lemma V.2.10(ii), [Aut F Ẽ : σ ] = [ σ : F ] = [F : F ] = 1. But, Lagrange s Theorem gives Aut F Ẽ = [Aut F Ẽ : σ ] σ = σ = n. Thus, Aut F Ẽ is cyclic, and by the first paragraph, Aut F E is cyclic for any finite dimensional extension E over F. Problem 7 We proved in class that if K is a Galois radical extension of K then Aut K ( K) is solvable. Using this result, prove that if F is a radical extension of K and E is an intermediate field of the extension F K then Aut K (E) is solvable. Let K 0 = (Aut K E). Then K K 0, E is Galois over K 0, Aut K E = Aut K0 E, and F is a radical extension of K 0 (K 0 is an intermediate field, so this follows by Exercise V.9.1). Hence we may assume WLOG that K = K 0, which means that E is Galois over K and hence by Lemma 3

V.2.13 that E is a stable intermediate field. Let N be the normal closure of F over K, so that by Lemma V.9.3 we have N is a radical extension of K. Since E is stable, σ E : Aut K N Aut K E is a homomorphism; and by Theorem V.3.8, N is a splitting field over K implies σ extends to a K-automorphism of N. Hence, σ E is surjective. The homomorphic image of a solvable group is solvable, so we must show that Aut K N is solvable. If K 1 = Aut K N, then N is a radical Galois extension of K 1 and Aut K1 N = Aut K N. Thus, by our assumption that any such extension (in particular, Aut K1 N) has a solvable Galois group, Aut K N is solvable. Problem 8 Let K be a field, F = K(x 1,..., x n ) be the field of rational functions in n indeterminates, and E F be the subfield of all symmetric rational functions in K(x 1,..., x n ) (See the Appendix of V.2). Show that {x 1,..., x n } is a transcendence basis of F over K. Show that the elementary symmetric functions f 1, f 2,..., f n form a transcendence basis of E. Do the functions f 1, f 2,..., f n also form a transcendence basis of F? {x 1,..., x n } is a transcendence basis of K(x 1,..., x n ) because it is algebraically independent: if f(x 1,..., x n ) = 0, then clearly f K[y 1,..., y n ] is 0. Suppose {x 1,..., x n, f/g} is a bigger algebraically independent set, where f, g 0 are in K[x 1,..., x n ]. Then h(y 1,..., y n+1 ) = g(y 1,..., y n )y n+1 f(y 1,..., y n ) K[y 1,..., y n, y n+1 ] is a nonzero polynomial with: h(x 1,..., x n, f/g) = g(x 1,..., x n ) f(x 1,..., x n ) g(x 1,..., x n ) f(x 1,..., x n ) = f(x 1,..., x n ) f(x 1,..., x n ) = 0, so that {x 1,..., x n, f/g} is algebraically dependent. By Theorem V.2.18 of the appendix, E = K(f 1,..., f n ). If h(f 1,..., f n ) = 0, then clearly h K[y 1,..., y n ] is the 0 polynomial, just as before. Hence {f 1,..., f n } is algebraically independent over K. Using the same trick as in the previous paragraph, suppose {f 1,..., f n, φ/ψ} is a bigger algebraically independent set, where φ, ψ K[y 1,..., y n ] and ψ 0. Then h(y 1,..., y n+1 ) = ψ(y 1,..., y n )y n+1 φ(y 1,..., y n ) K[y 1,..., y n, y n+1 ] is a nonzero polynomial with: h(f 1,..., f n, φ/ψ) = ψ(f 1,..., f n ) φ(f 1,..., f n ) ψ(f 1,..., f n ) φ(f 1,..., f n ) = φ(x 1,..., x n ) φ(x 1,..., x n ) = 0. Thus, {f 1,..., f n } is a transcendence basis of E over K. K(x 1,..., x n ) is algebraic over E because [K(x 1,..., x n ) : E] n!, and n = tr.d.(f/k) = tr.d.(f/e) + tr.d.(e/k) = 0 + tr.d.(e/k) tr.d.(e/k) = n. It was already shown that T = {f 1,..., f n } is algebraically independent over K. If T T, where T is a transcendence basis of F over K, then T = {x 1,..., x n } = n, so that T = T. Thus, {f 1,..., f n } is a transcendence basis of F over K. 4