Last Revised: December 7, 2001 Lecture Notes on Fields (Fall 1997) By George F. Seelinger NOTE: All references here are either made to Hungerford or to Beachy/Blair (2nd Edition). The references to Hungerford start with roman numerals while the references to Beachy/Blair are of the form x:y:z for positive integers x, y, and z. All internal references will be of the form x:y for positive integers x and y. 1 Field Extensions Definition 1.1 Let F be a field and let K F be a subring. Then we say K is a subfield of F if K is a field. In this case we also call⊆F an extension field of K and abbreviate this by saying F=K is a field extension. Recall the definition of a vector space over an arbitrary field. Definition 1.2 A vector space over a field K is an abelian group V whose group operation is denoted additively such that there exists a function : K V V with the following properties: ∗ × ! 1. (ab) x = a (b x) for all a; b K and x V . ∗ ∗ ∗ 2 2 2. (a + b) x = a x + b x for all a; b K and x V . ∗ ∗ ∗ 2 2 3. a (x + y) = a x + a y for all a K and x; y V . ∗ ∗ ∗ 2 2 4. 1 x = x for all x V ∗ 2 Then the following lemma tells us we can use some of the tools of linear algebra to help us study field extensions. Lemma 1.3 If F=K is a field extension, then F is a K vector space. Proof: By definition, F is an abelian group under addition, so we can define our vector addition to be the addition in F . Also, we can define our scalar multiplication : K F F to be given by k x = kx where the second multiplication is just multiplication∗ of× elements! in F . Then it is∗ easy to check that F satisfies the definition of a vector space with scalars K with these operations. Q.E.D. 1 So if F=K is a field extension, we define [F : K] = dimK (F ), the dimension of F as a K vector space. Therefore, [F : K] is the cardinality of any basis of F as a K-vector space. Review: Let V be a K-vector space for some field K. Let B be a non-empty subset of V . Then the K-span of B is the set of all finite sums of the form a1v1 + + anvn such that a1; : : : ; an K and v1; : : : ; vn B ··· 2 2 We say that B is linearly independent over K if for any n N and for any v1; : : : ; vn B 2 2 the only solution a1; : : : ; an K to the equation 2 a1v1 + ::: + anvn = 0 is the trivial one, a1 = a2 = = an = 0. Then a K-basis of V is a subset B V such that the K-span of B is V and···B is linearly independent over K. ⊆ Now let F=K be a field extension and choose u F . Then we can define K(u) to be the smallest subfield of F containing K and u. Hence2 for any subfield E of F such that K E and u E we have K(u) E. We call the extension K(u)=K a simple extension. ⊆ 2 ⊆ We can build more complicated extensions by taking a finite list of elements u1; : : : ; un F 2 and defining K(u1; : : : ; un) to be the smallest subfield of F containing the set u1; : : : ; un and K. In general, for any subset X F , we define K(X) to be the smallest subfieldf of Fg containing K and X. ⊆ First we try to understand simple extensions. Let F=K be a field extension and choose u F . Then there exists an evaluation homomorphism φu : K[x] F given by 2 n i n i ! φu( aix ) = aiu for every n N and every a0; : : : ; an K. (It is not hard to Pi=0 Pi=0 2 2 see that φu is a homomorphism if you note φu(p) = p(u) for every p K[x].) Furthermore, 2 it is straightforward to see that Imφu K(u) since K(u) is closed under addition and multiplication. ⊆ Definition 1.4 Let F=K be a field extension. Then an element u F is said to be alge- braic over K if there exists a non-zero p K[x] such that p(u) = 02. (This is equivalent to 2 saying ker(φu) = 0.) We say that u is transcendental over K if it is not algebraic over 6 K. (Hence u is transcendental over K if ker(φu) = 0.) We say that F is algebraic over K if every element of F is algebraic over K. If F is not algebraic over K, we say F is transcendental over K. Theorem 1.6 Let F=K be a field extension and let u F . 2 1. If u is transcendental over K then K(u) ∼= K(x). 2. If u is algebraic over K then K(u) = K[x]= p for some irreducible p K[x]. Also ∼2 nh 1i 2 n = [K(u): K] = deg(p) and 1; u; u ; : : : ; u − is a basis for K(u). f g 2 Proof: (1) Assume u is transcendental over K. Then for every nonzero g K[x] we have 2 g(u) = φu(g) = 0 must be an invertible element of K(u), hence for any f; g K[x] with g = 0 6 ~2 6 we have f(u)=g(u) K(u). Therefore we can define a homomorphism φu : K(x) K(u) ~ 2 ~ ! given by φu(f=g) = f(u)=g(u) for all f=g K(x). As φu is non-trivial, it must be a ~ 2 monomorphism. Therefore Imφu is a subfield of K(u) containing both u and K, hence must be equal to K(u). So our result follows from the first isomorphism theorem. (2) Assume u is algebraic over K. Then ker(φu) = p for some nonzero p K[x]. We claim that p must be irreducible, hence p is a maximalh ideali of K[x]. Indeed, if2 fg = p for some f; g K[x] of strictly lower degree,h theni f(u)g(u) = p(u) = 0. As F is a field we have 2 f(u) = 0 or g(u) = 0, which gives us f ker(φu) or g ker(φu). Therefore f = pq or g = pq for some q K[x]. This is a contradiction,2 hence p must2 be irreducible. 2 Now since p is irreducible, Im(φu) = K[x]= p is a field that contains K and u, hence ∼ h i we have Im(φu) K(u) Im(φu), so the first isomorphism theorem gives us that K(u) ∼= K[x]= p as required.⊆ ⊆ h i 2 n 1 Next we show that U = 1; u; u ; : : : ; u − is a basis of K(u) over K, where n = deg(p). f g For any w K(u) there exists an f K[x] such that φu(f) = w. By the Division Algorithm, 2 2 n 1 f = pq+r for some q; r K[x] such that r = a0 +a1x+ +an 1x − for some a0; : : : ; an 1 2 ··· − n 1 − 2 K. As pq ker(φu), we have w = φu(f) = φu(r) = a01 + a1u + + an 1u − . Therefore 2 n 1 ··· − U spans K(u). Also, if a01 + a1u + + an 1u − = 0 for some a0; : : : ; an 1 K, then n 1 i ··· − −n 12 i − aix ker(φu). As this has degree smaller than p, we must have − aix = 0. Pi=0 2 Pi=0 Therefore we have that ai = 0 for all 0 i n 1, so U is also linearly independent, hence a basis for K(u).≤ ≤ − Q.E.D. If F=K is an extension and u F is algebraic over K then we define the minimal polynomial of u to be the unique monic2 irreducible polynomial p K[x] such that p(u) = 0. 2 Note that the minimal polynomial of u is also the unique monic generator of ker(φu) where φu is the evaluation homomorphism. We define the degree of u over K to be the degree of the minimal polynomial of u. One important observation to make from this theorem is that if u F is algebraic over a subfield K of degree n then every element c K(u) can be written uniquely2 in the form 2 n 1 c = a0 + a1u + + an 1u − ··· − for some a0; : : : ; an 1 K. − 2 Example: If we consider F = Q(p3 2) then since x3 2 is irreducible over Q, by Theorem 1.6 the set 1; p3 2; p3 4 is a Q-basis of F , hence every element− of v F can be written uniquely f 3 g 3 2 as v = a0 + a1p2 + a2p4 for some a0; a1; a2 Q. 2 We would like to start comparing different extensions of a field K. First, we can use the evaluation homomorphism to classify the simple algebraic extensions up to isomorphism. 3 Proposition 1.7 Let E=K and F=K be two field extensions and let u E and v F be algebraic over K. Then u and v are roots of the same irreducible polynomial2 p K[x]2if and only if there is an isomorphism of fields : K(u) K(v) such that (u) = v 2and (a) = a for all a K.