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Math 123—Algebra II Math 123|Algebra II Lectures by Joe Harris Notes by Max Wang Harvard University, Spring 2012 Lecture 1: 1/23/12 . 1 Lecture 19: 3/7/12 . 26 Lecture 2: 1/25/12 . 2 Lecture 21: 3/19/12 . 27 Lecture 3: 1/27/12 . 3 Lecture 22: 3/21/12 . 31 Lecture 4: 1/30/12 . 4 Lecture 23: 3/23/12 . 33 Lecture 5: 2/1/12 . 6 Lecture 24: 3/26/12 . 34 Lecture 6: 2/3/12 . 8 Lecture 25: 3/28/12 . 36 Lecture 7: 2/6/12 . 9 Lecture 26: 3/30/12 . 37 Lecture 8: 2/8/12 . 10 Lecture 27: 4/2/12 . 38 Lecture 9: 2/10/12 . 12 Lecture 28: 4/4/12 . 40 Lecture 10: 2/13/12 . 13 Lecture 29: 4/6/12 . 41 Lecture 11: 2/15/12 . 15 Lecture 30: 4/9/12 . 42 Lecture 12: 2/17/16 . 17 Lecture 31: 4/11/12 . 44 Lecture 13: 2/22/12 . 19 Lecture 14: 2/24/12 . 21 Lecture 32: 4/13/12 . 45 Lecture 15: 2/27/12 . 22 Lecture 33: 4/16/12 . 46 Lecture 17: 3/2/12 . 23 Lecture 34: 4/18/12 . 47 Lecture 18: 3/5/12 . 25 Lecture 35: 4/20/12 . 48 Introduction Math 123 is the second in a two-course undergraduate series on abstract algebra offered at Harvard University. This instance of the course dealt with fields and Galois theory, representation theory of finite groups, and rings and modules. These notes were live-TEXed, then edited for correctness and clarity. I am responsible for all errata in this document, mathematical or otherwise; any merits of the material here should be credited to the lecturer, not to me. Feel free to email me at [email protected] with any comments. Acknowledgments In addition to the course staff, acknowledgment goes to Zev Chonoles, whose online lecture notes (http://math. uchicago.edu/~chonoles/expository-notes/) inspired me to post my own. I have also borrowed his format for this introduction page. The page layout for these notes is based on the layout I used back when I took notes by hand. The LATEX styles can be found here: https://github.com/mxw/latex-custom. Copyright Copyright © 2012 Max Wang. This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This means you are free to edit, adapt, transform, or redistribute this work as long as you • include an attribution of Joe Harris as the instructor of the course these notes are based on, and an attribution of Max Wang as the note-taker; • do so in a way that does not suggest that either of us endorses you or your use of this work; • use this work for noncommercial purposes only; and • if you adapt or build upon this work, apply this same license to your contributions. See http://creativecommons.org/licenses/by-nc-sa/4.0/ for the license details. Math 123|Algebra II Max Wang Lecture 1 | 1/23/12 Claim 1.4. Let F,−! K be a field extension, α 2 K. Consider the evaluation ring homomorphism given by Example (Fields). ' : F [X] −! K 1. Q, the rational numbers X 7−! α n 2. Fp = Z=pZ ,! Fq, where q = p Then α is transcendental over F iff ' is injective. 3. C(t), C(t1; : : : ; tn), the complex function fields of one or more variables (where addition and multipli- cation are defined pointwise as usual) Proof. ' is injective iff ker ' = f0g; this means that α does not satisfy any polynomials with coefficients in F , While in group theory, we often studied a group by ex- and hence is transcendental. amining its various subgroups, our primary tool in field theory will be to take a ground field F and to study its Recall from ring theory that an ideal is any subgroup extensions. of a ring that is closed under multiplication. We denote Definition 1.1. Let F be a field. A field extension of F an ideal generated by some element x by (x). is given by a field K with Definition 1.5. Let F,−! K be a field extension, α 2 K F / K algebraic over F . Since F [X] is a PID1, we have where F is included in K and is closed under K's field ker ' = (f) operations and inverse. We denote the field extension K=F . for some f 2 F [X] (we assume that f is monic). We call Definition 1.2. Two field extensions K=F and K0=F of f the irreducible polynomial satisfied by α=F . We also a ground field F are isomorphic define deg α = deg f K=F =∼ K0=F F if 9' : K ! K0 such that we have the diagram The degree of an algebraic element is also dependent on the ground field; for example, ' K / K0 p O O πi=4 Example. Let K = C and α = i = e . In Q, α satisfies the polynomial X4 + 1, and we have ? ? F id / F deg α = 4 Q Definition 1.3. Let F,−! K be a field extension. We say that an element α 2 K is algebraic over F if α satisfies In Q(i) (the rationals with i adjoined), α satisfies the a polynomial equation in K polynomial X2 − i, and we have n n−1 anα + an−1α + ··· + a1a + a0 = 0 deg α = 2 Q(i) where ai 2 F . Dividing by an, we can assume the poly- nomial to be monic. If α does not satisfy any such poly- Definition 1.6. Let F,−! K be a field extension, nomial, it is transcendental over F . α 2 K. Recall our evaluation map ' : F [X] ! K. We denote Note that whether or not an element α is algebraic n depends not only on the extension field K but also on F [α] = im ' = fβ 2 K : β = anα + ··· + a0; ai 2 F g the ground field F . For example, which is the smallest subring in K that contains both F Example. Let K = C. Then πi is transcendental over and α. Similarly, let Q but is algebraic over R. F (α) Recall that denote the smallest subfield of K containing F and α. F [X] = fa Xn + ··· + a X + a : a 2 F g n 1 0 i We call these ring adjunction and field adjunction respec- is the polynomial ring with coefficients in F . tively. 1Principal ideal domain; that is, a ring in which all ideals are generated by a single element. Recall that all fields and all rings F [X] are PIDs. 1 Math 123|Algebra II Max Wang Observation 1.7. If α is transcendental over F , then we The characteristic of F is defined by have ( ∼ 0 ker ' = f0g F [α] = F [X] char(F ) = p ker ' = (p) and F (α) =∼ F (X) Note that the map ' is unique for each field F and hence commutes with inclusion; so, all field extensions of a given where F (X) denotes the field of rational functions with ground field share the ground field’s characteristic. coefficients in F . If α is algebraic over F , ker ' = (f) is nonzero. Then Definition 2.2. If F,−! K is a finite field extension, we have then the degree of the extension is the dimension of K as an F -vector space, denoted F [α] = F (α) =∼ F [X]=(f) deg(K=F ) = [K : F ] This makes F (α) a finite-dimensional vector space over F with a basis of Note. If we have 1; α; α2; : : : ; αn−1 F,−! K,−! K0 where n = deg α, since these powers of α are those ze- then F 0 : : roed by the irreducible polynomial. [K F ] ≥ [K F ] and also Definition 1.8. We say that a field extension F,−! K [K0 : F ] ≥ [K0 : K] is algebraic over F if every α 2 K is algebraic over F . We say F,−! K is finite over F if K is a finite-dimensional Claim 2.3. [K : F ] = 1 () K = F . F -vector space. Note that a field extension that is finite Proof. Immediate. over its ground field is also algebraic over that ground field; however, the converse is not true in general. Proposition 2.4. Let F,−! K be a field extension with [K : F ] = 2. If char(F ) 6= 2, then Let F,−! K be a field extension, α; β 2 K. We will 2 ask three questions about the subfields of K generated by K = F (δ); δ 2 F α and β. Proof. Choose any α 2 K − F . Then since [K : F ] = 2, α and 1 are linearly independent. Then 1; α; α2 are de- 1. When is F (α) = F (β)? This is true, for instance, pendent, and hence 9a ; a ; a such that if β = α + 1. We can also consider less trivial ex- 0 1 2 2 amples; for instance, taking F = Q, K = C, α the a2α + a1α + a0 root of the polynomial X3 − X + 1, and β = α2, we We know a 6= 0, so dividing by a , also have Q(α) = Q(β) (although this is far from 2 2 immediately clear). 2 α + b1α + b0 = 0 2. When is F (α) =∼ F (β) as extensions of F ? Since char(F ) 6= 2, we have an element 2−1, so we can set ∼ b 3. When is F (α) = F (β) as extensions of F via an δ = α + 1 isomorphism α $ β? If α and β are both transcen- 2 dental, the the isomorphism is obvious.
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