301 Abstract Algebra 2:00 - 2:50 MWF Hentzel 432 Carver hentzel@iastate.edu Web Page http://www.math.iastate.edu/hentzel/class.301 Text: A First Course in Abstract Algebra, sixth Edition by John B. Fraleigh Course: Chapter 1,2,3,4,5 Monday, October 22. This material can be found in Section 8.2 page 393 starting with the words VECTOR SPACES through page 399. Main idea: A basis has enough to accomplish the task and not so many that they can be wasted. Key words: vector space, spanning set, linear independence, basis, dimension, field, Goal: The major goal is to show that if F is a subfield of G, then the dimension of F over Q divides the dimension of G over Q. The local goal is to learn what dimension is. Previous Assignment. (1) Find the inverse of 3+5 Sqrt[2] in Q[Sqrt[2]] 1 1 3 - 5 Sqrt[2] -------------- = -------------- * ---------------- 3 + 5 Sqrt[2] 3 + 5 Sqrt[2] 3 - 5 Sqrt[2] 3 - 5 Sqrt[2] = --------------- 9 - 50 = -3/41 + (5/41) Sqrt[2]. Check: (3 + 5 Sqrt[2])(-3/41 + (5/41) Sqrt[2]) = -9/41 +50/41 + (15/41 -15/41) Sqrt[2] = 1. (2) Find the minimal polynomial of Sqrt[2]+Sqrt[3] in (Q[ Sqrt[2] ])[ Sqrt[3] ] = Q[ Sqrt[2], Sqrt[3] ]. x = Sqrt[2] + Sqrt[3] x^2 = 2 + 2 Sqrt[6] + 3 x^2-5 = 2 Sqrt[6] (x^2-5)^2 = 24 x^4 -10 x^2 + 25 = 24 x^4 -10 x^2 + 1 = 0. (3) Find the inverse of Sqrt[2]+Sqrt[3] in Q[ Sqrt[2], Sqrt[3] ] Method (1): Since x^4 - 10 x^2 + 1 = 0 when x = Sqrt[2]+Sqrt[3] then 1 = x(-x^3 + 10x). The inverse of x is -x^3 + 10x. -(Sqrt[2]+Sqrt[3])^3 + 10(Sqrt[2]+Sqrt[3]) = -Sqrt[2]+Sqrt[3]. So the inverse of Sqrt[2]+Sqrt[3] is -Sqrt[2]+Sqrt[3]. Method (2). We consider the element in this form (a + b Sqrt[2]) + (c + d Sqrt[2]) Sqrt[3] which is obtained by adding Sqrt[3] to the field Q[ Sqrt[2] ]. Sqrt[2] + Sqrt[3] = (0 + 1 Sqrt[2]) + (1 + 0 Sqrt[2]) Sqrt[3] 1 1 Sqrt[2]-Sqrt[3] Sqrt[2]-Sqrt[3] ---------------- = --------------- * --------------- = --------------- Sqrt[2]+Sqrt[3] Sqrt[2]+Sqrt[3] Sqrt[2]-Sqrt[3] 2-3 = -Sqrt[2]+Sqrt[3]. Method (3). We work the problem with unknown coefficients. (Sqrt[2]+Sqrt[3])(a + b Sqrt[2] + c Sqrt[3] + d Sqrt[6]) = (+ 2 b + 3 c ) + (a + 3d)Sqrt[2]) + (a+2d)Sqrt[3] + (b+c)Sqrt[6] 2b+3c = 1 a+3d = 0 a+2d = 0 b+ c = 0 So a = d = 0 and c = 1 , b = -1. Definition 8.2.1 page 393. A vector space V over a field F consists of an abelian group V under addition together with a multiplication av for a in F and v in V satisfying. 1. av is in V 2. (ab)v = a(bv) 3. (a+b)v = av+bv 4. a(v+w) = av+aw 5. 1v = v The ubiquitous example of such a thing are n-tuples of real numbers for the vectors and the field being the real numbers themselves. a (v1,v2,v3) = (a v1,a v2,a v3). Properties one through five follow from the well known properties of the real numbers. One could restrict the above example to the integers. Let the scalars be Z and the components of the vector be from Z as well. Then properties one through five follow from the properties of the integers. BUT... This is not a vector space because Z is NOT a field. The behavior of vector spaces is pretty much ho-hum when F is a field. When F is not a field the situation becomes an endless bag of worms. Fortunately, for us, we need only to consider the case when F is a field. Notice that we say nothing in the definition about multiplying vectors together. The familiar cross product and dot product are not part of the fundamental structure of vector spaces. In our applications, there will be a product defined on the vectors, but it is not be needed now nor will it enter the discussion today. The current task is to define: "the dimension of the vector space over F." Properties one through 5 are enough to do that. The intended application for us will be to Fields of the form Q[ 2^(1/2) = { a + b 2^(1/2) | a,b are rationals }. This is a two dimensional vector space over the rationals. The number two comes because it takes two rationals to specify a number. The field: Q[ 2^(1/3) ] = {a + b 2^(1/3) + c 2^(2/3) | a,b,c are rationals} is a vector space of dimension three over the rationals. The three comes because it takes three rationals to specify a number. Definition. Let V be a vector space over a field F. The span of v1, v2, v3, ..., vn consists of all vectors which can be written in the form a1 v1 + a2 v2 + ... + an vn where a1, a2, ..., an are in F. The notation means the span of v1, v2, ..., vn. We introduce the terminology "finite dimensional" before we have introduced the term "dimension." Surprising, we can state whether a vector space will turn out to be finite dimensional without an exact knowledge of dimension. We will limit our discussion to "Let V =/= 0 be a finite dimensional vector space over a field F." We do this to avoid annoying details involving the 0 vector space and a basis of the empty set. Definition: Let V be a vector space over F. We say that V is "finite dimensional" over F if V has a finite spanning set. Definition. Let V be a vector space over a field F. Then vectors v1, v2, v3, ..., vn in V are said to be "linearly dependent" if there exists scalars a1, a2, a3, ..., an in F which are not all zero, such that ====================== a1 v1 + a2 v2 + ... + an vn = 0. A set of vectors which is not linearly dependent is said to be "linearly independent." Lemma: Let V be a vector space over a field F. Suppose v1, v2, v3, ..., vn are linearly dependent and v1 =/= 0, then some vector is a linear combination of the previous vectors and that vector can be deleted without changing the span. Proof: Since the v's were linearly dependent, there exist coefficients ai, not all of which are zero so that a1 v1 + a2 v2 + ... + an vn = 0 Let ak be the right most coefficient which is not zero. Then a1 v1 + a2 v2 + ... + ak vk = 0 We can solve this for vector vk yielding vk = -1/ak (a1 v1 + a2 v2 + ... + ak-1 vk-1). Thus vk is a vector which is a linear combination of the previous ones. The restriction that v1 =/= 0 is necessary because it has no previous vectors to be linear combinations of, and it certainly makes the set dependent since 1 v1 + 0 v2 + ... + 0 vn = 0. To prove the second part, if w = c1 v1 + c2 v3 + ... + cn vn, we can replace vk by vk = -1/ak (a1 v1 + a2 v2 + ... + ak-1 vk-1) omit and end up with w being a linear combination of v1 v2 ... (vk) ... vn. ck vk = -ck/ak (a1 v1 + a2 v2 + ... + ak-1 vk-1) and so omit |-----| w = d1 v1 + v2 v2 + ... + dk vk + ... + dn vn where: _ | ci - ck ai/ak for i = 1 ... k-1 di = | | ci for i = k+1 ... n. |_ Thus w is in the span of the vectors which remain after vk is deleted. Definition: A basis of a vector space is a linearly independent spanning set. Theorem 8.2.17 page 397. Let V =/= 0 be a finite dimensional vector space over a field K. Every spanning set of V can be reduced to a basis of V. Remark: The traditional literature declares that every vector space has a basis. When asked for the basis of the zero vector space, V = {0}, the traditional answer is that the basis of the zero space is the empty set of vectors { }. To me this is too devious an explanation. I avoid the difficulty by looking at bases for V =/= 0. Proof: Suppose V = < v1, v2, ..., vn>. Not all of the vi's can be zero since V =/= 0. We can certainly delete all of the vi's which are zero with out changing the span. Now the problem reduces the case where all of the vi's are not zero. If v1, v2, ..., Vn are linearly dependent, then one vector is a linear combination of the previous and may be removed without changing the span. Continue removing vectors until the remaining vi's are linearly independent. We cannot remove all the vectors since V =/= 0. Thus at some place the remaining vectors have to be linearly independent and are a basis of V. Theorem 8.2.18 Let V =/= 0 be a finite dimensional vector space over a field F. Then any linearly independent set can be extended to a basis. Or said more precisely. If V = < v1, v2, ..., vn> and W = {w1,w2, ..., wk} is a linearly independent set of vectors of V, then W can be extended to a basis of V. Proof: The set {w1,w2,w3,...,wk,v1,v2, ..., vn} is a dependent set which spans V. Since w1 =/= 0, we can reduce this to a basis of V by removing vectors with are linear combinations of the previous ones. The resulting basis is of the form {w1,w2, ..., wk, and perhaps some v's}. Corollary 8.2.20. Let V =/= 0 be a finite dimensional vector space over a field F. Then any two bases of V have the same number of elements. Proof: Let v1, v2, ..., vr and w1, w2, ..., ws be two bases of V. Using the previous argument in a slightly more sophisticated way, w1, v1, v2, ..., vr is certainly dependent and a spanning set. It can be reduced to a basis by removing one or more of the v's. the resulting basis is of the form w1, {some v's } We continue by adding the next w which forces the set to be dependent. We reduce this to a basis and again, since the w's are independent, the removed elements are again v's. The new basis is of the form w1,w2, {some still fewer v's } We can continue this process until all the w's are exhausted. In the process we have removed at least s of the v's. Thus r <= s. By symmetry, we could have put the v's first and removed the w's and this time we would have proved that s <= r. The only conclusion is that r = s and any two bases have the same number of elements. Definition: The dimension of a vector space V is the number of elements in a basis of V. Assignment: Page 399 Exercises 8.2 Problems 1,2,3,4,6,8,10