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 © 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 ∈ K. Consider the evaluation ring homomorphism given by Example (Fields). ϕ : 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 ϕ is injective. 3. C(t), C(t1, . . . , tn), the complex function fields of one or more variables (where addition and multiplication are defined pointwise as usual) Proof. ϕ is injective iff ker ϕ = {0}; 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 ∈ 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 ϕ = (f) operations and inverse. We denote the field extension K/F . for some f ∈ 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 ∃ϕ : K → K0 such that we have the diagram The degree of an algebraic element is also dependent on the ground field; for example, ϕ K / K0 √ 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 ∈ 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 ∈ F . Dividing by an, we can assume the polynomial 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 ∈ K. Recall our evaluation map ϕ : 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 ϕ = {β ∈ K : β = anα + ··· + a0, ai ∈ F } 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] = {a Xn + ··· + a X + a : a ∈ F } 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.

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Observation 1.7. If α is transcendental over F , then we The characteristic of F is defined by have ( ∼ 0 ker ϕ = {0} F [α] = F [X] char(F ) = p ker ϕ = (p) and F (α) =∼ F (X) Note that the map ϕ 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 ϕ = (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 ∈ 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 ∈ K. We will 2 ask three questions about the subfields of K generated by K = F (δ), δ ∈ F α and β. Proof. Choose any α ∈ 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 ∃a , 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. If they are Then we have both algebraic, then the isomorphism holds iff the b2 δ2 + (b − 1 ) = 0 monic polynomials satisfied by α and β over F are 0 4 2 equal, yielding b1 2 where b0 − 4 ∈ F . Hence, K = F (δ) with δ ∈ F , as desired. F (α) =∼ F [X]/(f) =∼ F (β) Note that we could abolish the restriction on charac- Lecture 2 — 1/25/12 teristic by demanding only that some quadratic polynomial of δ to be in F . Definition 2.1. Let F be a field. There exists a canoni- Proposition 2.5. Let F,−→ K,−→ L where L/F is cal ring homomorphism finite. Then : : : ϕ : Z −→ F [L F ] = [L K][K F ]

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Proof. Choose a basis α1, . . . , αn for L/K and another Lecture 3 — 1/27/12 basis β1, . . . , βm for K/F . We claim that the pairwise products αiβj form a basis for L/F . Example. Let α be the real cube root of 2, and let 2πi/3 First, we will show these αiβj span. Let γ ∈ L. We ω = e . Deﬁne can write 2 γ = a1α1 + ··· + anαn α1 = α α2 = αω α3 = αω where ai ∈ K. But then we can write These are the roots of the polynomial x3 − 2. Note that this polynomial is irreducible over Q, so we have ai = bi1β1 + ··· + bimβm

[ (αi) : ] = 3 where bij ∈ F . Then we have Q Q X From the point of view of algebra, these three roots are γ = bij(αiβj) indistinguishable (i.e., the ﬁeld extensions made by ad- Next, we show independence. Suppose we have joining αi are all isomorphic). We want to distinguish between our three extensions X bij(αiβj) = 0 Q(αi) by considering them as subﬁelds of

Defining ai as above, we get Q(α1, α2, α3) = Q(α, ω) X aiαi = 0 One way to make such a distinction would be to note that Q(α1) ⊂ R whereas Q(α2), Q(α3) 6⊂ R. However, we But by the independence of the αi in L/K, we have would like to obtain this knowledge without relying on X the existence of R or C. bijβj = 0 ω satisfies the irreducible polynomial j 3 x − 1 2 for every i, which yields bij = 0, ∀i, j by the independence = x + x + 1 = 0 x − 1 of the βj in K/F . This completes the proof. Corollary 2.6. Let F,−→ K be a finite field extension. which yields Then ∀α ∈ K, [Q(ω) : Q] = 2 deg α|[K : F ] F A field extension of degree 3 cannot contain an element because we can write of degree 2, so ∀i, ω∈ / Q(αi). Thus, we learn that

F,−→ F (α) ,−→ K [Q(α, ω) : Q] = 6

Theorem 2.7. Let F,−→ K be a field extension, and let and also that L ⊆ K be the subset of elements that are algebraic over [ (α, ω) : (α )] = 2 F . Then L is a subfield of K. Q Q i and Proof. Take α, β ∈ L. We have the sequence of exten- [ (α, ω) : (ω)] = 3 sions Q Q F,−→ F (α) ,−→ F (α, β) Finally, we can also conclude that the Q(αi) must be dis- It is obvious that if β/F is algebraic, then so is β/F (α). tinct fields because ω is their ratio, and ω is not in any Then F (α, β) is a finite extension over F , and hence every Q(αi). We end up with the following picture: γ ∈ F (α, β) is algebraic over F . Example. Let F = Q, K = C. Set L = Q, the algebraic Q(α, ω) closure of . We can factor polynomials completely in Q 3 2 this field. 2 2

We would like to know if the algebraic closure of a field Q(ω) Q(α1) Q(α2) Q(α3) always exists. That is, given a field F , does there exist an extension F,−→ K such that any polynomial with co- 3 3 2 3 efficients in K factors completely in K (or, equivalently, has a root in K). Q

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√ √ √ Note that, although we set out to obtain our results 2, 3, and 6). Alternatively, since 1, α, α2 are inde- without relying on the complex numbers, we still required pendent, we know that the desired irreducible polynomial knowledge of C in order to construct our field extensions. is not quadratic (and we know it cannot be cubic because √ our extensions have even degrees). Example.√ Let us consider Q( 2, i)/Q. We know that We could, instead, denote f(x) as the irreducible poly- ( 2) and (i) have degree 2 as extensions of (their Q Q Q nomial over Q satisfied by α, and consider the other roots generators satisfy quadratic polynomials). Thus, we know √ √ √ of f beyond 2 +√ 3. Note√ that algebra can’t tell the that Q( 2, i)/Q has degree either 2 or 4. Again, this difference between 2 and − 2; they are both just num-√ question reduces to whether the two single-adjunction bers satisfying x2 − 1. A similar statement holds for 3. fields are equal. √ √ We√ might√ guess, then,√ that√ the other roots are 2 − 3, We know any number α ∈ Q(i) can be written√ α = − 2 + 3, and − 2 − 3. So we simply try a + bi with a, b ∈ Q. Thus, we would have Q( 2) = Q(i) √ √ √ √ as field extensions over Q if we had (x − 2 − 3)(x − 2 + 3) √ √ √ √ 2 2 2 ? (x + 2 − 3)(x + 2 + 3) α = a − b + 2abi = 2 √ √ = ((x − 2)2 − 3)((x + 2)2 − 3) However, this equation cannot be satisfied; hence, the = (x2 − 1)2 − 8x2 fields are distinct, and we have 4 2 √ = x − 10x + 1 [Q( 2, i) : Q] = 4 Note that this doesn’t tell us that our polynomial is irre- √ Note that there is a third extension, Q(i 2). This is ducible (but in fact it is). √ √ again a quadratic extension (has degree 2) and this is dis- Consider our picture for ( 2, 3): tinct for the same reason the other two are distinct. We Q √ √ assert without proof that these are all the intermediate Q( 2, 3) extensions. √ 2 2 2 Example. We can replace i with 3 in the above exam- √ √ √ ple to yield identical results. Let us now attempt to find Q( 2) Q( 3) Q( 6) the irreducible polynomial over Q satisfied by √ √ 2 2 2 α := 2 + 3 Q (We know that α is algebraic because√ it is an element of a finite√ extension.)√ Note first that√ 1, 2 are a basis for If we consider√ our√ field extensions as vector spaces, we Q( 2)/Q, 1, 3 are a basis for Q( 3)/Q, and hence find that Q( 2, 3) is a four-space, and the intermediate √ √ √ extensions are all two-planes;√ therefore,√ in a very strong 1, 2, 3, 6 sense, most elements of Q( 2, 3) must be single gener- √ √ ators of that field. form a basis for Q( 2, 3)/Q. We will solve this problem using two different ap- proaches. First, let us write out some of the powers of Lecture 4 — 1/30/12 α: Definition 4.1. Let us formalize the notion of α0 = 1 constructions with straightedge and compass, which al- √ √ lows us to bridge our algebra with planar geometry. In α1 = 2 + 3 √ making such constructions, we begin with a pair of points α2 = 5 + 2 6 p, q ∈ R2. We then define two basic constructions: √ 4 α = 49 + 20 6 1. Draw the line Lp,q = pq.

Note that 1, α2, α4 are linearly dependent, and in partic- 2. Draw the circle Cp(q) with center at p passing ular, we find that α satisfies through q. Given two lines L ,L , we can find their point of intersec- x4 − 10x2 + 1 1 2 tion; given a line L and a conic C or two conics C1,C2, we We could check irreducibility by checking that α is not can find their two points of intersection. The intersections in any of the three intermediate extensions (generated by of these lines and conics are called constructible.

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Construction 4.2. Begin with p, q ∈ R2. We construct Definition 4.7. Begin with the two points (0, 0), (1, 0) ∈ 2 the point r ∈ Lp,q equidistant to both p and q as follows: R . We call a point (a, b) constructible if we can construct it starting with our two points using our various construc- 1. Draw the circle C (q) through q about p. p tions. Similarly, we say a point a ∈ R is constructible if (a, 0) is constructible. Finally, an angle θ ∈ [0, 2π) 2. Draw the circle Cq(p) through p about q. is constructible if we can construct two lines L, L0 with 0 3. Draw the line L between the two intersection points ∠(L, L ) = θ. Note that θ is constructible if sin θ and of these circles. cos θ are constructible. Let p, q ∈ K2 ⊂ 2 where K is some subfield of . 4. r is the intersection L ∩ Lp,q R R Then Lp,q is defined by a linear equation with coefficients Construction 4.3. Begin with a line L and a point in K. Similarly, the circle Cp(q) is defined by a quadratic p∈ / L. We construct a perpendicular to L through p polynomial in K. as follows: If we have two lines L, L0 defined by linear equations with coefficients in K, their point of intersection L ∩ L0 1. Draw any circle about p intersecting L at two points is in K2. If instead we have a line L and a circle C de- q and q0. fined by equations with coefficients in K, their intersection 2. Find the midpoint r between q and q0 on L. points in L ∩ C have coordinates that live in a quadratic extension of K. To see this, we can parametrize L 3. Draw the line Lp,r; this is our desired perpendicular. L = {(t, α + βt) : t ∈ R} α, β ∈ K

Construction 4.4. Begin with a line L and a point and then apply the equation for C to (t, α + βt) to get p ∈ L. We construct a perpendicular to L through p a quadratic polynomial in t alone, with coefficients in K. as follows: This results in the following conclusion: Proposition 4.8. If a is constructible, then there exists 1. Draw any circle C around p, intersecting L at q p a tower of fields and r. Q ⊂ K1 ⊂ · · · ⊂ Kn 3 a 2. Draw the circle Cq(r) and the circle Cr(q). such that 3. Connect the intersections of these two circles via a [Ki : Ki−1] = 2 line L0; this line passes through the midpoint p and and hence is perpendicular to L. n [Kn : Q] = 2 Construction 4.5. Begin with a line L and a point Moreover, we have, for some r ∈ N, p∈ / L. We construct a line parallel to L as follows: deg a = r 1. Draw the line L⊥ perpendicular to L through p. Q Example. We can use this proposition to conclude that 2. Draw the line L0 perpendicular to L⊥ through p; π it is not possible to trisect an arbitrary angle. Take 3 ; this is parallel, as desired. π we can ask whether θ = 9 is constructible. Define Construction 4.6. Begin with a pair of points p, q and α = 2 cos θ = eπi/9 + e−πi/9 a line L with a point r ∈ L. We construct a line segment in L with one endpoint r with length equal to d(p, q) as Then we have follows: α3 = eπi/3 + 3eπi/9 + 3e−πi/9 +e−πi/3 | {z } 1. Draw the circle Cp(q). 3α and hence 2. Draw the line Lp,r. α3 − 3α − 1 = 0 0 3. Draw the line L through p parallel to L; let s be Thus, α satisfies a cubic polynomial with no linear factors 0 the point L ∩ Cp(q). and which therefore is irreducible. So 00 4. Draw the line L parallel to Lp,r through s. deg α = 3 Q 0 00 0 5. Let r be the intersection L ∩ L; the segment r, r which, by our proposition, means that α and θ are not 0 has d(r, r ) = d(p, q). constructible.

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Proposition 4.9. Let L ⊂ R be the set of constructible Lecture 5 — 2/1/12 numbers. Then L is a subfield of R. Definition 5.1. Let F be a field The polynomial ring Proof. Let a, b ∈ L. We know that over F is given by 1. a + b ∈ L since we can simply extend a segment of n F length a by one of length b. F [X] = {anx + ··· + a1x + a0 : ai ∈ F } ⊂ F 2. −a ∈ L because if we can construct (a, 0), we can just as easily construct (−a, 0). Note that the polynomial ring does not map injec- tively to functions on F ; this is pointedly the case where 3. ab ∈ L. To show this, we first construct a triangle F is a finite field. of side length 1 along the x-axis and side length a vertically. We then construct the similar triangle n Definition 5.2. Let f = anx + ··· + a0. The derivative with side length b along the x-axis; this will have of f is given by side length ab vertically. 0 n−1 1 f = n · anx + ··· + a1 4. a ∈ L. We use similar triangles again, this time beginning with a triangle of side length a and 1 and scale it down so that the a side has length 1. where n is the image of n under the canonical map Z → F . and so L is a field. Example. Let √ Proposition 4.10. If a is constructible, then a is con- 2 structible. f = x + x = x(x − 1) ∈ F2[X]

Proof. To construct it, we Then f 0 = 1 because (x2)0 = 0. Note that any field 1. Draw a circle with diameter a + 1 and divide the of positive characteristic admits nonconstant polynomi- diameter L into two segments of lengths a and 1 at als whose derivatives are zero. a point p. Theorem 5.3. Let F,−→ K, f, g ∈ F [X] ,−→ K[X]. 2. Draw a line perpendicular to L√through p. The Then any identity in F [X] holds in F [X] iff it holds in height of this line in the circle is a. K[X]. This includes: Thus, not only is L a field, but it is closed under square root. 1. ∃q, r ∈ F [X] : g = fq + r with deg r < deg f when carried out in F [X] iff g = fq + r in K[X]. Note Theorem 4.11. a is constructible iff there exists a tower that q, r are unique. of fields ⊂ K ⊂ · · · ⊂ K 3 a Q 1 n 2. f |g ∈ F [X] ⇐⇒ f |g ∈ K[X]. such that [Ki : Ki−1] = 2 3. gcdF [X](f, g) = gcdK[X](f, g). Theorem 4.12. Let F be a field, f ∈ F [X] irreducible. Then ∃K/F such that f has a root in K. 4. Let h ∈ F [X]. Then h | f and h | g ∈ F [X] ⇐⇒ h|f and h|g ∈ K[X]. Proof. Simply take F [X]/(f). f is maximal since it is irreducible, so K is a field. Thenx ¯, the equivalence class Claim 5.4. Let F,−→ K be a field extension, f, g ∈ of x mod (f), satisfies polynomial f(¯x) = 0. F [X]. If f is irreducible over F and f, g have a common factor in K[X], then f |g ∈ F [X]. Example. Let F = F2 = Z/(2). The polynomial 2 f(x) = x + x + 1 Proof. Since f, g ∈ K[X] have a common factor h, then is the unique irreducible quadratic polynomial in this h is a common factor of f, g ∈ F [X]. But f is irreducible; field. Then we can form a field thus, we must have f = h, which means f | g, as desired. 2 ∼ F2[X]/(x + x + 1) = F4 This is the unique field of four elements. (Note that it is Lemma 5.5. Let α ∈ F , f ∈ F [X], f(α) = 0. Then α not Z/4, which is a ring but not a field.) is a multiple root (i.e., (x − α)2 |f) iff f 0(α) = 0.

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Proof. Suppose ﬁrst that (x − α)2 |f. Then ∃g ∈ F [X] : We have f = (x − α2)g. Taking derivatives, we have 0 1 α α + 1 f 0 = (x − α)2g0 + 2(x − α)g 0 0 0 0 0 1 0 1 α α + 1 Then clearly, f 0(α) = 0. α 0 α α + 1 1 Now suppose f 0(α) = 0. Then ∃h ∈ F [X] : f = α + 1 0 α + 1 1 α (x − α)h, and so f 0 = (x − α)h0 + h. Then we have

0 0 Deﬁnition 5.9. Let R be a ring. We deﬁne the group of 0 = f (α) = (α − α)h (α) + h(α) = h(α) units of R to be This means α is a root of h, which yields (x − α)|h. But R× = {x : ∃x−1, xx−1 = x−1x = 1} since f = (x − α)h, we have

× (x − α)|f Note that for a ﬁeld F , F = F − {0}.

r as desired. Observation 5.10. Let F be a field of order q = p . Then F × is a finite abelian group of order q − 1. So Corollary 5.6. If f ∈ F [X] is irreducible, f 0 6= 0, then ∀α ∈ F , α 6= 0, we have αq−1 = 1, and hence every such f has no repeated roots in any extension F,−→ K. α is a root of the polynomial Proof. Suppose that f had a repeated root in some ex- xq−1 − 1 tension K/F . Then in K[X], f, f 0 are not relatively prime (they have a common factor). But then they also have a Multiplying by x, we get that every element of F is a root common factor in F [X]. Then f | f 0, but deg f 0 < deg f. 0 of the polynomial This implies f = 0, a contradiction. xq − x Proposition 5.7. Let F be a finite field. Then Since |F | = q and deg(xq − x) = q, there are no repeated |F | = #F = pr roots; thus, Y xq − x = (x − α) for some p ∈ P, r ∈ N. α∈F Proof. Let ϕ : → F be the canonical homomor- Z × phism. We have ker ϕ = (p) for some prime p and Lemma 5.11. Let K be any field, G ⊂ K a finite subgroup of K×. Then G is cyclic. im ϕ = Z/(p) = Fp. So F is a finite-dimensional vec- r tor space over Fp, which implies #F = p . Proof. First, consider any a, b ∈ G of orders α and β re- Theorem 5.8. There exists a unique field of order pr, spectively. If gcd(α, β) = 1, then ord(ab) = αβ. Suppose which we denote that G is not cyclic. Then by the above (and since G is Fpr = Fq a finite abelian group),

Example. Let F = F2. There are four polynomials of n := lcm{ord(α) : α ∈ G} < |G| 2 degree 2 over F2. Three of them factor: x , x(x − 1), and 2 (x − 1) . The remaining one, (Actually, the lcm properly divides |G|, but the strict inequality is all we need.) But this means that the poly- x2 + x + 1 nomial xn − 1 has |G| > n roots. However, a polynomial cannot have more roots than its degree, and hence, G is the unique irreducible polynomial of degree 2 in F2. Then must be cyclic. 2 F2[X]/(x + x + 1) = F4 Example. Note that the last deduction in the above is the unique field of four elements because every field of proof requires that K be a field. For instance, we have four elements must be a quadratic extension of F2 and there is only one irreducible quadratic polynomial in F2. × (Z/8) = {1, 3, 5, 7} =∼ Z/2 × Z/2 Let us write out the multiplication table of F4, given as But x2 − 1 has four roots in /8. F4 = {0, 1, α, α + 1} Z

7 Math 123—Algebra II Max Wang

pk pr Proof (of Theorem). First, we will show that ∃K/Fp a Then x −x factors completely in Fpr since x −x does. r field extension of order p for any r ∈ N. Let L/Fp be Take q pk any extension in which the polynomial f = x − x factors {x ∈ Fpr : x − x = 0} ⊂ Fpr 0 q−1 completely. Note that f = qx − 1 which is −1 for This is a subfield of order pk. x = 0 and q − 1 for x 6= 0. Since f 0(α) 6= 0 for every q α ∈ L, it has q distinct roots. We claim that Proposition 6.2. The irreducible factors of x − x/Fp are exactly the irreducible polynomials over p whose de- q F K := {α ∈ L : f(α) = α − α = 0} ,−→ L gree k divides r. is a subfield (where by the above, we have #K = q). Proof. Let f ∈ Fp[X] be irreducible of degree k |r. f fac- q q q If a, b ∈ K, then a = a and b = b, so (ab) = ab, and tors completely in Fpr ⊃ Fpk = Fp[X]/(f), which means hence ab ∈ K. Moreover, we have (a + b)q = aq + bq = it has a common root with xq − x. But since f is irre- a + b, so a + b ∈ K. So K is indeed a field, as desired. ducible over Fp, this means that Now we want to show that K is unique. Suppose we f |xq − x ∈ have two extensions K,K0 with #K = #K0 = pr = q. Fp ∼ 0 q We claim that K = K . Now let f ∈ p[X] be irreducible and f |x − x ∈ p. × F F We know that K is cyclic; let α ∈ K be a generator We know that f factors completely in pr . Choose α a × F of K . So root of f in Fpr . We have K = {0, 1, α, α2, . . . , αq−2} Fp ,−→ Fp(α) ,−→ Fpr In particular, we have K = Fp(α). Now let f be the irreducible polynomial satisfied by α/Fp. By definition, we and so deg f = k |r, as desired. have q Example. Let us consider the finite fields of characteris- f |x − x ∈ K 2 tic 2. We have F4 = F2(α), where α is a root of x +x+1, q q 0 But then f | x − x ∈ Fp, and hence also f | x − x ∈ K . the unique irreducible quadratic polynomial over F2. We Since xq −x factors completely in K0, f factors completely can factor 0 0 0 0 0 in K , so f has a root α ∈ K . Then K = p(α ), and F 4 2 hence x − x = x(x + 1)(x + x + 1) ∼ ∼ 0 K = Fp[X]/(f) = K We also have F2 ,−→ F8. What are the cubic polyno- as desired. mials over F2? We have four which factor 1, 1, 1 x3 x2(x + 1) x(x + 1)2 (x + 1)3 Lecture 6 — 2/3/12 two which factor 1, 2

Before we proceed, we recall some general facts about 2 2 ﬁnite ﬁelds. x(x + x + 1) (x + 1)(x + x + 1)

r and two which do not factor • F finite =⇒ ∃p ∈ P, r ∈ N : #F = p = q. x3 + x + 1 x3 + x2 + 1 • #F = q =⇒ ∀x ∈ F, xq − x = 0. We can write • F finite =⇒ F × cyclic. x8 − x = x(x + 1)(x3 + x + 1)(x3 + x2 + 1) • ∃!F : #F = pr. Consider also F2 ,−→ F16. Of the quartic polynomi- Proposition 6.1. Fpr has a subfield isomorphic to Fpk iff k |r. als in F2, 5 factor into linear factors, 1 factors into two quadratics, 4 factor into one linear and one cubic factor, 3 Proof. First, let Fpk be a subfield. Then Fpr is a m- factor into a quadratic factor and two linear factors, and dimensional vector space over Fpk . Hence, three are irreducible. We can factor 16 2 4 4 3 r k m km x − x = x(x + 1)(x + x + 1)(x + x + 1)(x + x + 1) p = #Fpr = (p ) = p (x4 + x3 + x2 + x + 1) and thus k |r. Now suppose instead that k |r. Then ∃m : pr = (pk)m. Definition 6.3. A polynomial f ∈ F [X] is said to split So we have completely in an extension K/F if f factors into linear k r (xp − x)|(xp − x) factors in K[X].

8 Math 123—Algebra II Max Wang

Theorem 6.4 (Primitive Element Theorem). Let F be a Definition 7.2. α1, . . . , αn ∈ K are a transcendence field, char(F ) = 0, and F,−→ K a finite extension. Then base for K/F if they are algebraically independent and ∃α ∈ K : K = F (α). α is called a primitive element for K/F (it generates the entire extension). F (α1, . . . , αn) ,−→ K Note that this theorem also holds for all finite fields F of arbitrary characteristic. is an algebraic extension. That is, α1, . . . , αn are a maximal collection of algebraically independent elements. Proof. We know K = F (α1, . . . , αk) for some elements α1, . . . , αk ∈ K. By induction on k, we can assume that We will for the duration of this lecture assume that F,−→ F (α1, . . . , αk−1) has a primitive element β. But any transcendence basis is finite. then our problem reduces to the base case, of showing that Theorem 7.3. Let α1, . . . , αm ∈ K be a maximal tran- F,−→ F (β, αk) = F (α1, . . . , αk) scendence basis and β1, . . . , βn ∈ K be algebraically inde- has a primitive element. So, we will try to show that if pendent. Then n ≤ m. K = F (α, β), then K = F (γ) for some γ ∈ K. In particular, we claim that for all but finitely many c ∈ K, β +cα The proof of this theorem involves repeatedly replac- is primitive. ing the αi with the βj in the transcendence basis; how- Let f, g be the irreducible polynomials satisfied by α ever, it is omitted. and β, respectively, over F . Let K0/K be an extension Corollary 7.4. Any two transcendence bases have the in which f and g split completely. Let α := α1, . . . , αm same cardinality. be the roots of f; β = β1, . . . , βn, the roots of g. Let

γ = β + cα Definition 7.5. The transcendence degree of K/F is defined as the cardinality of any transcendence basis for Now let L = F (γ). We claim that α ∈ L; it will im- K/F . mediately follow that β = γ − cα ∈ L, and hence L = K. Define a polynomial h ∈ L[X] by Definition 7.6. Let F,−→ K be a field extension. h(x) = g(γ − cx) We say that K/F is purely transcendental if K = F (α1, . . . , αn) for algebraically independent α1, . . . , αn. We know that the roots of g are βi; thus, the roots of h are given by Note that all transcendental extensions can be decom- γ − β β − β posed as x = i = i + α c c p.t. alg. F ,−→ F (α1, . . . , αn) ,−→ K We want to choose c so that gcd(f, h) = x − α; this will imply that α ∈ L, as desired. But we can compute Theorem 7.7 (Luroth). Any transcendental subfield of 0 0 this gcd in K [X] instead But f splits completely over K ; C(t) is purely transcendental. this means it suffices to show that none of the αj (except for α) is a root of h. This is the case whenever Theorem 7.8 (Catelnuovo-Enriquez). If K ⊂ C(t, s) has β − β transcendental degree 2, then K is purely transcendental. c 6= i α − α i Theorem 7.9 (Clemens-Griffiths). The above does not which completes the proof. hold for 3; that is, ∃K ⊂ C(t, s, u) with transcendental degree 3 that is unpure. Lecture 7 — 2/6/12 Example. The field Definition 7.1. Let F,−→ K be a field extension. The 2 2 elements α1, . . . , αn ∈ K are algebraically independent if K = C(x)[y]/(y − x − 1) they do not satisfy any f ∈ F [X1,...,Xn] (i.e., if there is no f such that f(α1, . . . , αn) = 0). Alternatively, we is purely transcendental over C; however, have a tower of extensions 2 3 L = C(x)[y]/(y − x − 1) F,−→ F (α1) ,−→ F (α1, α2) ,−→ · · · ,−→ F (α1, . . . , αn) where each extension is transcendental. is not pure.

9 Math 123—Algebra II Max Wang

One important context of this discussion is that of in- Note that we do not mean to say that the values of tegrating functions. If f(x) ∈ C(x) (the ﬁeld of rational the polynomials are equal; for example, for R = Fp, functions over C), then p p x1 − x1 6= x2 − x2 Z f(x) dx Rather, we want to say that the polynomials themselves are equivalent. can be calculated using partial fractions. Example. Start with any monomial. Then the sum of Meanwhile, the elements of its orbit under the action of Sn (which we refer to as the orbit sum) is symmetric. For instance, Z dx Z dx k √ = starting with x1 , we have 2 x + 1 C y k k x1 + ··· + xn 2 2 where C is the curve y = x + 1. We can parametrize This particular sum is called the power sum. If we start C: “every” line through C meets C exactly once, so we with, say, x1x2, we instead get parametrize by the slope t through the point (0, 1). Solv- X ing for the parametrization gives xixj i

2 x, of where x 7→ 2t and y 7→ 1+t . 1−t2 1−t2 n How, then, do we solve Y p(x) = (x − ui) Z dx i=1 √ n n−1 n−2 x3 + 1 = x − s1x + s2x − · · · ± sn Theorem 8.3. Let R be a ring. Any symmetric poly- This integral spurred huge mathematical progress. A key nomial g ∈ R[u , . . . , u ] is expressible as a polynomial fact in this problem is that L is not purely transcendental. 1 n of the elementary symmetric polynomials s1, . . . , sn in Consider surfaces in C × C. u1, . . . , un. Z = {(x, y) : y2 = x2 + 1} Example. We can write the second power sum 2 2 W = {(x, y) : y2 = x3 + 1} x1 + ··· + xn Z is a sphere with punctures; integration along a path is in terms of elementary symmetric polynomials as 2 path-independent. W is a torus with punctures; integra-  n  tion is path-dependent. 2 2 X X x1 + ··· + xn =  xi − 2 xixj i=1 i

= s1s2 − 3s3 f(x1, . . . , xn) = f(xσ(1), . . . , xσ(n)), ∀σ ∈ Sn

10 Math 123—Algebra II Max Wang

Proof. Say g(u1, . . . , un) is symmetric. Let us deﬁne with si = si(u1, . . . , un). We want to deﬁne a function

g0(u1, . . . , un−1) = g(u1, . . . , un−1, 0) Y 2 D(u1, . . . , un) = (ui − uj) i

sn |g(u1, . . . , un) − Q(s1, . . . , sn−1) where s1 = α + β and s2 = αβ. We have Then h is the quotient of this division, and it must be ∆ = (α − β)2 symmetric since everything else in the equation is symmetric. Then by inducting on the degree of g with n fixed, = α2 − 2αβ we are done. 2 = s1 − 4s2 Remark. Let R = Z. Note that the power sums which is the well-known discriminant for monic quadratic k k polynomials. {x1 + ··· + xn} Now consider generate the ring of Sn-invariant polynomials in the vari- 3 2 ables x1, . . . , xn over Q, but not over Z as the elementary p(x) = x − s1x + s2x − s3 polynomials do. which has a much more complicated discriminant (which Observation 8.4. Let F,−→ K be a field extension. Let we will give without computation), f ∈ F [X], and say that f splits completely in K, with roots α1, . . . , αn. Since we can write 3 2 2 3 2 ∆ = −4s1s3 + s1s2 + 18s1s2s3 − 4s2 − 27s3 n Y f(x) = (x − αi) (In this formula, we have actually taken si = |si|.) Note i=1 that if s1 = 0, this reduces to n n−1 n−2 = x − s1x + s2x − · · · ± sn 3 2 ∆ = −4s2 − 27s3 we have si = si(α1, . . . , αn) ∈ F s1 We can arrive at this version by substituting x 7→ x − 3 for general cubics. Then if p(u1, . . . , un) is any symmetric polynomial, we have p(α1, . . . , αn) ∈ F . Definition 8.6. The Vandermonde matrix is given by Definition 8.5. Define p ∈ F [X] by   n n−1 n−2 1 1 ··· 1 p(x) = x − s1x + s2x − · · · ± sn  x1 x2 ··· xn  n V =   Y  x2 x2 ··· x2  = (x − u )  1 2 n  i xn−1 xn−1 ··· xn−1 i=1 1 2 n

11 Math 123—Algebra II Max Wang

Observation 8.7. Note that the determinant vanishes Proposition 9.3. whenever x = x for some i 6= j. It is given by i j 1. Let K/F be a splitting ﬁeld for f ∈ F [X]. Then Y [K : F ] < ∞. det V = ± (xi − xj) i

aj n det(xi ) Y f(x) = (x − α ) det(xj) i i i=1 will be symmetric. The collection of such polynomials Take g ∈ F [X] irreducible such that ∃β ∈ K : g(β) = 0. form an additive basis of the symmetric polynomials. Since K = F (α1, . . . , αn) Lecture 9 — 2/10/12 we can write β = p(α1, . . . , αn) Definition 9.1. Let F,−→ K be a field extension, f ∈ F [X]. We say that K is a splitting field for f/F for some p ∈ F [X1,...,Xn]. if Let {p1, . . . , pk} be the orbit of p under the action of Sn on F [X1,...,Xn], with p1 = p. Set 1. f splits completely in K as βi = pi(α1, . . . , αn) n Y f(x) = (x − αi), αi ∈ K and define a polynomial i=1 k Y h(x) = (x − β ) ∈ K[X] 2. K is the minimal extension in which f splits com- i i=1 pletely; that is, We claim that h ∈ F [X]. We can write

K = F (α1, . . . , αn) n n−1 n−2 h(x) = x − s1x + s2x − · · · ± sn Proposition 9.2. Every f ∈ F [X] has a splitting ﬁeld. where the

Proof. We know that f splits completely in some ex- si = si(β1, . . . , βk) tension L/F . Then take the roots α1, . . . , αn of f; the = s (p (α , . . . , α ), . . . , p (α , . . . , α )) extension i 1 1 n k 1 n F,−→ F (α1, . . . , αn) Since the si are symmetric on the β1, . . . , βk and the pi are symmetric on the α , . . . , α , s is symmetric with re- is a splitting field. 1 n i spect to S on α , . . . , α . Hence, we can express h as a We can also determine the splitting field algorithmi- n 1 n polynomial in the elementary symmetric polynomials over cally. Choose an irreducible factor f0 of f. Let α1, . . . , αn, and thus in the coefficients of f. But these coefficients are in F , and we have h ∈ F [X] as desired. K1 = F [X]/(f0) If g irreducible in F but has root in common with h Then ∃α ∈ K1 : f(α) = 0. We replace f by f/(x − α) in K, g | h ∈ K[X]. But then g | h ∈ F [X]. Since h splits and repeat this process as necessary. completely in K, so does g.

12 Math 123—Algebra II Max Wang

Example. Any quadratic extension is a splitting field Lemma 10.2. Let F,−→ K and F,−→ K0 be field ex- K = F (α) where α satisfies a quadratic polynomial tensions. Then we have the following 2 f = x + ax + b, and its inclusion in K enables f to 1. Let f ∈ F [X], ϕ : K → K0 an F -isomorphism. If split completely. α ∈ K is a root of f, then so is ϕ(α) ∈ K0. √ 3 2πi/3 i Example. Let α = 2 and ω = e , set αi = α ω, and 0 2. Suppose that K = F (α1, . . . , αn). Let ϕ, ϕ : K → recall our observations about field extensions concerning 0 0 K be F -isomorphisms. If ∀i, ϕ(αi) = ϕ (αi), then these values. ϕ = ϕ0. 3 ∼ K = Q[X]/(x − 2) = Q(α1) ⊂ R ⊂ C 3. Let f ∈ F [X] be irreducible, with a root α ∈ K and α ∈ K0 in each extension. Then ∃!ϕ : F (α) → 3 This is not a splitting field over Q, since x − 2 only has F (α0) an F -isomorphism sending α 7→ α0. one root in K, as we have previously shown. If, however, we include other roots in the extension, then we can make Proof. Omitted. K into a splitting field. Definition 10.3. Let F,−→ K be a field extension. The Note that we do need to check that K does not con- Galois group of the extension K/F is defined as tain other roots; for instance, if ω ∈ K, since ω satisfies x3−1 2 Gal(K/F ) = Aut(K/F ) x−1 = x +x+1, then that polynomial would split completely in K. Fortunately, we have also shown previously the group of F -automorphisms of K. that ω∈ / K. Definition 10.4. A field extension F,−→ K is called a Theorem 9.5. Let F be a field of characteristic zero. Galois extension if Then any two splitting fields of f ∈ F [X] are isomorphic. | Gal(K/F )| = [K : F ] Proof. Say K ,K any two splitting fields for f/F . By 1 2 Definition 10.5. Let K be a field, H a group of au- the primitive element theorem, we have K = F (γ) for 1 tomorphisms of K. The fixed field of H is the set of some γ ∈ K . Let g ∈ F [X] be irreducible polynomial 1 elements of K that are fixed by every element of H, satisfied by γ/F Extend K2 by a field L such that g has a root γ0, and let K0 = F (γ0) ⊂ L. Then KH = {α ∈ K : σ(α) = α, ∀σ ∈ H}

∼ ∼ 0 Note that KH ⊆ K is a subfield. K1 = F [X]/(g) = K Theorem 10.6. Let K be a field, H a finite group of 0 via an F -isomorphism sending γ 7→ γ . Since K1 is a automorphisms on K. Let F = KH . Let β ∈ K, 0 0 1 splitting field of f, so then is K . But then K ,K2 ⊂ L {β , . . . , β } its H-orbit. Then the irreducible polynomial 0 ∼ 1 r are both splitting fields for f/F , and hence K = K2 of β1/F is because L can contain at most one splitting field. g(x) = (x − β1) ··· (x − βr)

Lecture 10 — 2/13/12 It follows that β1 is algebraic over F with

We shall assume from this point onward that all fields in degF β1 = r question have characteristic zero. and that degF β1 ||H|. Definition 10.1. Recall that we call two field exten- Proof. We want to show that g as defined is irreducible. sions K/F and K0/F isomorphic, or more specifically We can write F -isomorphic g(x) = (x − β ) ··· (x − β ) K/F =∼ K0/F 1 r = xr − b xr−1 + · · · ± b if ∃ϕ : K → K0 such that we have the diagram 1 r Each bi is a symmetric polynomial in the βj. Since H ϕ H / 0 permutes the βj, it fixes each bi, so bi ∈ K . KO KO Now let h ∈ F [X] with β1 as a root. For each i = 1, . . . , r, we can find σi ∈ H with σi(β1) = βi. Since ? ? F id / F the σi are F -automorphisms, then σi(β1) = βi must also be roots of h. Then g |h ∈ K[X], which means An F -isomorphism from K/F to itself is called an g |h ∈ F [X] F -automorphism. These are the symmetries of the field extension K. It follows that g must be irreducible as desired.

13 Math 123—Algebra II Max Wang

Observation 10.7. Let F,−→ K be algebraic but infi- Proposition 10.11. Let K be a field, H a finite group nite. Then we can construct an infinite tower of fields of automorphisms on K. Then KH ,−→ K is Galois and H = Gal(K/KH ). F < F1 < F2 < ··· < K Proof. By definition, H ⊆ Gal(K/KH ). By the above H H To do so, start by taking α1 ∈ K − F and F1 = F (α1). lemma, we know |G(K/K )| | [K : K ], and the fixed H H Since α1 is algebraic over F ,[F1 : F ] < ∞, so F1 < K. field theorem gives |H| = [K : K ]. Then G(K/K ) ⊆ We can then take α2 ∈ K − F1 and F2 = F1(α2). Simi- H, and we have equality as desired. larly, [F2 : F1] < ∞, so F2 < K. Continuing in this way, we get our desired tower. Observation 10.12. Let F,−→ K be a finite extension with primitive element γ1 ∈ K. Let f ∈ F [X] Theorem 10.8 (Fixed Field Theorem). Let K be a field, be the irreducible polynomial satisfied by γ1 with roots H H a finite group of automorphisms on K, F = K its γ1, . . . , γr ∈ K. fixed field. Then We know that there is a unique F -isomorphism σi : : [K F ] = |H| F (γ1) → F (γi) with γ1 7→ γi. Since K = F (γ1), it follows Proof. Let n = |H|. We know that K/F is algebraic and that K = F (γi) for each i, so σi is an F -automorphism of K. Since every F -automorphism must take γ1 7→ γi degF β |n for every β ∈ K. This tells us that (since γi are all the roots of f in K), we have [K : F ] < ∞ Gal(K/F ) = {σi} By the primitive element theorem, ∃γ ∈ K : K = F (γ). Since γ generates K, if σ ∈ H fixes γ, we would have with | Gal(K/F )| = r. σ = id. So the stabilizer of γ is {1} ⊂ H, and hence its Theorem 10.13. Let F,−→ K be a finite field extension orbit Hγ has order with G = Gal(K/F ). The following are equivalent |Hγ| = n 1. K/F is a Galois extension (i.e., |G| = [K : F ]) Then degF γ = n, which means 2. F = KG [K : F ] = n 3. K is a splitting field over F as desired. We use the second condition to show that an element Definition 10.9. Let F,−→ K be a field extension. An α ∈ K is actually also in F ; we use the third condition to intermediate field L is a field determine that an extension is Galois. F ⊆ L ⊆ K Proof. Showing that 1 ⇐⇒ 2 is quick. The fixed field theorem gives us |G| = [K : KG]. Since F,−→ KG ,−→ We say an intermediate field is proper if L 6= F and K, L 6= K. |G| = [K : F ] ⇐⇒ F = KG Note that every L-automorphism of K is also an F - Now we will show that 1 ⇐⇒ 3. By the primi- automorphism of K, and hence tive element theorem, we can choose γ1 ∈ K such that K = F (γ ). Let f ∈ F [X] be the irreducible polynomial Gal(K/L) ⊆ Gal(K/F ) 1 satisfied by γ1. We know that Lemma 10.10. Let F,−→ K be a finite field extension, [K : F ] = deg γ = deg f G = Gal(K/F ). Then F 1

Let γ1, . . . , γr ∈ K be the roots of f in K. Then we have |G||[K : F ] |G| = r by our previous observation. Proof. By the fixed field theorem, we know that |G| = If r = |G| = [K : F ], then since deg f = [K : F ], [K : KG]. Since F -automorphisms are the identity on F , f splits completely in K and hence K = F (γ1) = F ⊆ KG, and hence we have the tower of extensions F (γ1, . . . , γn) is a splitting field. Conversely, if K is a splitting field, the same reasoning applied in reverse yields G F,−→ K ,−→ K |G| = r = [K : F ]. Then Corollary 10.14. |G| = [K : KG]|[K : F ] 1. Every finite extension K/F is contained in a Galois as desired. extension.

14 Math 123—Algebra II Max Wang

2. If K/F is Galois, L an intermediate field, then K/L Observation 10.17. Note that if L and L0 are interme- is also Galois, and diate fields and H and H0 are their corresponding subgroups, L ⊂ L0 iff H ⊃ H0. In particular, F corresponds Gal(K/L) ⊆ Gal(K/F ) to Gal(K/F ) and K corresponds to {1}. If we have L corresponding to H, the since K/L is Observation 10.15. Let F,−→ K be a Galois extension Galois and H = Gal(K/L), we have with G = Gal(K/F ). Let g ∈ F [X] split completely in K [K : L] = |H| with roots β1, . . . , βr. Then We also know that |G| = [K : F ] = [K : L][L : F ] and 1. G acts on the roots {βi} by permuting them. |G| = |H|[G : H], so we also have

2. If K is a splitting field of g/F , we claim that [L : F ] = [G : H] the action of G is faithful.2 We know that K = Corollary 10.18. A finite field extension F,−→ K has F (β1, . . . , βr), and we get faithfulness from the fact that σ ∈ G is determined entirely by its mapping of finitely many intermediate fields. the generators β1, . . . , βr. It follows that G,−→ Sr. Lecture 11 — 2/15/12 3. If g is irreducible over F , the action of G is transi- 3 √ √ tive. Since g is irreducible, we know that g must Example. Let F = Q. Take α = 3, β = 5 and let be the irreducible polynomial satisfied by β1. Since K = F (α, β). K is the splitting field of F = KG, then {β } = Gβ , which is the statement i 1 2 2 of transitivity. (x − 3)(x − 5)

It follows that if K is a splitting ﬁeld of g/F and g is and hence is a Galois extension. We have irreducible in F , then G,−→ Sr embeds transitively. | Gal(K/F )| = [K : F ] = 4

We will now state and prove the fundamental theorem and hence Gal(K/F ) is either C4 the cyclic group or V the of Galois theory, which provides for a bijective correspon- Klein four group. We know that F (α), F (β), and F (αβ) dence between intermediate fields and subgroups of the are three distinct intermediate fields of K/F , and hence Galois group. Having build up significant machinery con- correspond to proper subgroups of Gal(K/F ). Since cerning Galois extensions, this proof will be trivial. [K : L] = 2 for each of these intermediate subgroups L, they correspond to subgroups of order 2. However, C4 has Theorem 10.16 (Fundamental Theorem of Galois The- only one element of order 2, and hence Gal(K/F ) = V , ory). Let F,−→ K be a Galois extension, G = which has three elements (and hence three subgroups) of Gal(K/F ). Then there is a bijective correspondence be- order 2. tween These are the only proper subgroups of Gal(K/F ), which means that F (α), F (β), and F (αβ) are the only {H : H ≤ G} ←→ {L : F,−→ L,−→ K} proper intermediate fields. In one direction, the bijection maps Note that given

H 7−→ KH F,−→ L,−→ K where K/F is Galois, we are only guaranteed that K/L is and in the inverse direction, it takes Galois, but not that L/F is. To determine whether L/F is Galois, we have the following: L 7−→ Gal(K/L) Theorem 11.1. Let F,−→ K be a Galois extension with Proof. Let H ≤ G and L = KH . By the fixed field G = Gal(K/F ). Let L be the fixed field KH for a sub- 4 theorem, H = Gal(K/L). Now suppose that L is an group H ≤ G. Then L/F is Galois iff H E G. If so, intermediate field, H = Gal(K/L). Then since K/F is then H ∼ Galois, so is K/L, and equivalently, L = K . Gal(L/F ) = G/H 2A group action on X is faithful if ∀g ∈ G, ∃x ∈ X : gx 6= x; in other words, if g fixes X, g = e. 3A group action on X is transitive if Gx = X for any x ∈ X; in other words, if X has a single orbit under G. 4A subgroup N ≤ G is normal if ∀n ∈ N, ∀g ∈ G, gng−1 ∈ N. In other words, a normal subgroup is a subgroup that is invariant under conjugation.

15 Math 123—Algebra II Max Wang

Proof. Let 1 ∈ L be a primitive element for L/F , Note that a1 = α1 + α2 + α3 ∈ F . Hence, α1 and α2 g ∈ F [X] the irreducible polynomial for 1. Since K generate α3, and we have the tower of extensions is a splitting field, 1 ∈ K, g splits completely with roots F,−→ F (α1) ,−→ F (α1, α2) = F (α1, α2, α3) = K 1, . . . , r. Note that L/F is Galois iff L is a splitting field, which is the case iff Let L = F (α1). Since f/F is irreducible,

∀i ∈ {1, . . . , r}, i ∈ L [L : F ] = 3

We will show that this holds iﬀ H E G. and we can factor Since G is transitive on { , . . . , }, we know that 1 r f(x) = (x − α )q(x) ∀i ∈ {1, . . . , r}, 1

where q is the quadratic polynomial with roots α2, α3. ∃σi ∈ G : σi(1) = i Either q is irreducible over L or it is not; if it is, then

Fix i and consider σi. We have F (i) = L iff i ∈ L (since [K : L] = 2 [K : F ] = 6 deg 1 = deg i). This is the case iff the stabilizer of F F Otherwise, i is H. Meanwhile, the stabilizer of σ(1) is the conjugate group σHσ−1. The condition that H = σHσ−1 is L = K [K : F ] = 3 exactly the condition that H E G, which is our desired Example. Let F = Q, f(x) = x3 + 3x + 1, which is irre- conclusion. ducible over Q. The derivative f 0(x) = 3x2 + 3 is strictly Suppose that L/F is Galois. Then i ∈ L for each i. greater than zero, so f is strictly increasing and hence has An F -automorphism σ ∈ G takes 1 7→ i for some i, and only one real zero α1. α1 cannot generate the complex hence maps roots of f, so [K : F ] = 6 where K is the splitting field of f. σ : F (1) = L −→ L = F (i) Example. Let F = Q, f(x) = x3 − 3x + 1, also irre- 2 The restriction σ|L is hence an F -automorphism of L. ducible over Q. If α1 is a root, then α1 − 2 is another This restriction operation induces a group homomor- root. We can generate the third root as given above, and phism hence in this case, [K : F ] = 3. ϕ : G −→ Gal(L/F ) By its action on the roots of the cubic f, the Ga- We have lois group G = Gal(K/F ) is a transitive subgroup of S3. There are two such groups: S3 itself, and the alternating ker ϕ = {σ ∈ G : σ|L = id} = H (and cyclic) group A3. If [K : F ] = 3, then G = A3; if [K : F ] = 6, then G = S3. The key distinction is whether We also have that or not q is irreducible over L. To decide this, we will use the value |G/H| = [G : H] = | Gal(L/F )| δ = (α1 − α2)(α1 − α3)(α2 − α3) ∈ K which means im ϕ = Gal(L/F ), and by the First Isomor- which is the square root of the discriminant D of f.5 Note phism Theorem, that δ 6= 0 since the roots are distinct (we assume F has characteristic zero). G/H =∼ Gal(L/F ) Theorem 11.2. Let F be a field, f ∈ F [X] an irreducible as desired. cubic polynomial, K the splitting field of f over F , and G = Gal(K/F ). Let D be the discriminant of f; then Let us now apply the machinery of Galois theory to √ the study of cubic polynomials over a field F . Consider 1. If δ = D ∈ F , then [K : F ] = 3 and G = A3. √ 3 2 2. If δ = D/∈ F , then [K : F ] = 6 and G = S3. f(x) = x − a1x + a2x − a3

= (x − α1)(x − α2)(x − α3) Proof. Permuting the roots of f multiplies δ by the sign of the permutation. If δ ∈ F , it is fixed by G. Then every where αi ∈ K are the roots of f in the splitting field K σ ∈ G must be even, meaning G = A3 and [K : F ] = 3. of f over F . Otherwise, G = S3 and [K : F ] = 6. 5This is slight abuse of terminology, where we previously defined the discriminant as a polynomial in the elementary symmetric polynomials.

16 Math 123—Algebra II Max Wang

The alternating group A3 has no proper subgroups (it • C4: rotations of the square. |C4| = 4. S4 has three is the cyclic group of order 3). Hence, if G = A3, there conjugate subgroups isomorphic to C4. are no intermediate fields, and our lattice is simply • D2: reflections of the square. |D2| = 4. D2 C S4 is K = F (α , α , α ) normal. 1 2 3 √ √ Example.√ Q( a, b) is a quartic extension as long as 3 √ b∈ / Q( a). Its Galois group must be D2; it is given by the automorphisms F (√ √ a 7→ a This must be the case since [K : F ] = 3 is prime. If in- √ √ b 7→ b stead we have G = S , we have four proper subgroups to 3 (√ √ consider, namely hyi, hxyi, hx2yi, all of order 2, and hxi a 7→ − a √ √ of order 3. Our lattice of fields is then b 7→ b (√ √ K = F (α1, α2, α3) a 7→ a √ √ b 7→ − b 2 2 2 3 (√ √ a 7→ − a F (α1) F (α2) F (α3) √ √ b 7→ − b 3 3 3 F (δ) √ √ We can also view K = Q( a, b) as the splitting field of 2 F a quartic polynomial. Choose any√ element√ of K not in one of the subfields, such as α = a + b. Take its orbit The corresponding lattice of Galois groups is √ √ √ √ √ √ √ √ { a + b, a − b, − a + b, − a − b} {1} The irreducible polynomial of α over Q is the product of the monomials with these elements as roots. 2 2 2 3 Consider now the dihedral group D4. It has three hyi hxyi hx2yi normal subgroups of order 4:

hxi = A3 • D4 acts on the set of two diagonals of a square. Let HD = Z2 × Z2 be the subgroup preserving the G = S3 diagonals.

• D4 acts on the set of two edge symmetries of a Lecture 12 — 2/17/16 square. Let HA = Z2 × Z2 be the subgroup preserving opposite edges. We now turn to the study of quartic polynomials. Let • Let HO = 4 be the subgroup preserving orienta- F be any ﬁeld, f ∈ F [X] an irreducible quartic polyno- Z tion. mial. Let K/F be the splitting extension of f over F , G = Gal(K/F ). In K, we can write Meanwhile, it has only one subgroup of order 2: • The group {1, −1}, representing rotation by 180◦. f(x) = (x − α1)(x − α2)(x − α3)(x − α4) Example. Consider the following tower of ﬁeld exten- Then G acts faithfully on the roots {αi}, and we have sions: G,−→ S4 transitively. p √ We can enumerate the transitive subgroups of S4: Q( 4 + 5)

• S4: symmetries of the tetrahedron. |S4| = 24. √ ( 5) • A4: rotations of the tetrahedron. |A4| = 12. A4CS4 Q is normal.

• D4: symmetries of the square. |D4| = 8. S4 has Q three conjugate subgroups isomorphic to D4.

17 Math 123—Algebra II Max Wang

p √ Note that ( 4 + 5)/ is not Galois; we can permute αα0 is invariant under this group but not under all of D ; √Q Q√ 4 p p hence, it must generate a nontrivial subﬁeld Q( 4 + 5) to Q( 4 − 5). Let √ q √ q √ H α = 4 + 5 α0 = 4 − 5 Q( 11) = K A

Take K = (α, α0). We have a quartic irreducible poly- Q Finally, consider the ﬁxed ﬁeld of the subgroup HO pre- nomial given by serving orientation, for which we have f(x) = (x − α)(x + α)(x − α0)(x + α0) √ H Q( 55) = K O = x4 − 8x2 + 11

Now consider the following square, with the vertices la- We also have the ﬁxed ﬁeld of the only normal subgroup beled by the roots ±α, ±α0: of order 2,

α α0 T = HD ∩ HA

which has corresponding intermediate ﬁeld −α0 −α √ √ T Q( 5, 11) = K The subgroup HD comprises the following automorphisms: We get the lattice ( α 7→ α 0 α0 7→ α0 K = Q(α, α ) ( 2 α 7→ −α 2 2 α0 7→ α0 √ √ Q(α) Q(α0) ( 5, 11) ( Q α 7→ α 2 2 2 2 α0 7→ −α0 2 √ √ √ ( α 7→ −α Q( 5) Q( 11) Q( 55) 0 0 2 α 7→ −α 2 2

2 α is invariant under this group but not under all of D4; Q hence, it must generate a nontrivial subﬁeld √ H Q( 5) = K D which can also be written Note that this speciﬁc correspondence is dependent on K the labeling of our square. The subgroup HA comprises 2 the following automorphisms: 2 2 ( α 7→ α Q(α) Q(α0) KT α0 7→ α0 ( 2 2 2 2 α 7→ α0 2 α0 7→ α KHD KHA KHO ( 2 α 7→ −α0 2 2 α0 7→ −α Q ( α 7→ −α 0 0 α 7→ −α and with corresponding Galois group lattice

18 Math 123—Algebra II Max Wang

{1} The fixed field KD2∩G is splitting field of a cubic, called the resolvent cubic. To construct this field, we want to 2 2 find an element of K that is invariant under D but no 2 2 larger subgroup of G. Take ? ? T β = α α + α α β = α α + α α 2 2 1 1 2 3 4 2 1 3 2 4 2

HD HA HO β3 = α1α4 + α2α3

D2∩G These are in K Every permutation of the αi permutes the βj (S4/D2 = S3), and hence G g(x) = (x − β1)(x − β2)(x − β3) ∈ F [X] Note that Q(α) and Q(α0) do not correspond to normal subgroups of D4; they are not Galois over Q. We also have

D2∩G Example. We can ﬁnd a ﬁeld extension with Galois K = K(βj) group C4 by a similar construction to the above. Let q √ q √ α = 2 + 2 α0 = 2 − 2 Lecture 13 — 2/22/12 √ 0 0 Then αα = 2 ∈ Q(α). Hence, Q(α) = Q(α ), and our Let Galois group will be C4. 2πi/n ζn = e Recall our initial setup for general quartic polynomi- be an nth root of unity. We will assume that n is some als. G must be one of the transitive subgroups of S4; we can ask whether G is contained in each of them. Let prime p. We know that the irreducible polynomial for ζp Y over Q is D = (α − α )2 i j xp−1 + ··· + x + 1 i

19 Math 123—Algebra II Max Wang

× Example. Consider ζ = ζ17. Since h3i = F17, we know Let α1, α2 denote the orbit sums. Then {α1, α2} is a that G-orbit, and our study of ﬁxed ﬁelds tells us that the G = hσ3i irreducible polynomial for α1 and α2 is where G = G(K/F ). Let σ = σ3. The subgroups of G (x − α1)(x − α2) are hσi ⊃ hσ2i ⊃ hσ4i ⊃ hσ8i ⊃ hidi To compute this polynomial, we want to determine which corresponds to the tower of extensions s1(α) = α1 + α2 s2(α) = α1α2

i i 2 2 2 4 2 8 2 Since s is the sum of all ζ with ζ 6= 1, the irreducible F = Khσi ,−→ Khσ i ,−→ Khσ i ,−→ Khσ i ,−→ K 1 polynomial of ζ gives 8 Now let θ = 2π/17. We claim that F (cos θ) = Khσ i. s1(α) = −1 Since ζ + ζ−1 = 2 cos θ, we know that cos θ ∈ K. More- over, ζ satisfies the polynomial We can compute s2(α) by our previous lemma. Since s2(α) ∈ Q and since expanding α1α2 results in 64 sum- −1 2 i (x − ζ)(x − ζ ) = x − 2(cos θ)x + 1 ∈ F (cos θ) mands of the form ζ , we know that each ζi appears four times, which yields Thus, s2(α) = −4 [K : F (cos θ)] ≤ 2 and [F (cos θ) : F ] ≥ 8 Then our polynomial is 8 So F (cos θ) is either Khσ i or K. But F (cos θ) ⊆ , R (x − α )(x − α ) = x2 + x − 4 whereas K 6⊆ ; thus, 1 2 R √ hσ2i 8 Its discriminant is D = 17, and hence K = F ( 17). F (cos θ) = Khσ i In the same way, we can determine the quadratic ex- as desired. tension over a field F that is contained in F (ζp) for any odd prime p. Lemma 13.4. Let ζ = ζp. Let Proposition 13.5. Let p 6= 2 be prime. Let L be the α = c ζ + c ζ2 + ··· + c ζp−1 unique quadratic extension over contained in (ζ ). If 1 2 p−1 √ Q Q p p ≡ 1 (mod 4), then L = Q( p); if p ≡ 3 (mod 4), then be a linear combination with c ∈ . If α ∈ , then √ i Q Q L = Q( −p). c1 = c2 = ··· = cp−1 and α = −c1. Note that we know the extension is unique by the Ga- × Proof. Since ζ satisfies lois correspondence, since the cyclic group Fp for p prime has exactly one subgroup of index 2. xp−1 + ··· + x + 1 Proof. Analogous to the example. we can solve for ζp−1 and rewrite Theorem 13.6 (Kronecker-Weber Theorem). Every Ga- p−2 α = (−cp−1)1 + (c1 − cp−1)ζ + ··· + (cp−2 − cp−1)ζ lois extension of Q with an abelian Galois group is contained in a cyclotomic extension Q(ζn). Since {1, ζ, . . . , ζp−2} are a basis for K/F , we must have Observation 13.7. Let all coefficients except −cp−1 equal to zero. This gives our desired result. a1 a2 ak n = p1 p2 ··· pk Example. Again, take ζ = ζ17. Taking iterative powers be the prime decomposition of some n ∈ . Then the 3 N of ζ , we get the powers fields (ζ ai ) intersect only in and Q pi Q 1, 3, −8, −7, −4, 5, −2, −6, −1, −3, 8, 7, 4, −5, 2, 6 Y (ζ ai ) = (ζn) Q pi Q Recall that σ = σ3 is a generator for the Galois group Theorem 13.8. Let F be a field with ζp ∈ F for some p G = Gal(K/F ). The orbit Gζ (under the action of au- prime, and let b ∈ F with b 6= 0. Then the polynomial tomorphism) is the set {ζi : ζi 6= 1}. Let H = hσ2i. H splits Gζ into two H-orbits, g(x) = xp − b

{ζ, ζ−8, ζ−4,...}{ζ3, ζ−7, ζ5,...} is either irreducible or splits completely in F .

20 Math 123—Algebra II Max Wang

Proof. Let K/F be the splitting ﬁeld of g and suppose KF 0 there is a root β of g not in F . Then [K : F ] > 1, which G means ∃σ ∈ Gal(K/F ) : σ 6= id. Since β generates K, 0 r K F σ(β) = ζp β for 0 < r < p. Meanwhile, σ(ζp) = ζp. Then G

k kr K ∩ F 0 σ (β) = ζp β

This attains every root ζiβ, and hence the action of G is transitive on the roots of g. Thus, g is irreducible over F , F which completes our proof. Proof. Since K/F is Galois, K is the splitting field of a 0 0 Theorem 13.9. Let F ⊆ C be a subfield containing ζp polynomial f ∈ F [X]. Then KF /F is the splitting field of f ∈ F 0[X], and hence KF 0/F 0 is Galois. Consider the for p prime, and let F,−→ K be a√ Galois extension with degree [K : F ] = p. Then K = F ( p α) for some α ∈ F . restriction map ϕ : Gal(KF 0/F 0) −→ Gal(K/F ) Proof. See Artin. σ 7−→ σ|K Lecture 14 — 2/24/12 This is a homomorphism with kernel 0 0 ker ϕ = {σ ∈ Gal(KF /F ) : σ|K = id} Definition 14.1. Let K1,K2 ⊆ K be subfields. The 0 composite field of K1 and K2, denoted K1K2, is the small- Note that σ ∈ ker ϕ is the identity on both F and K by est subfield of K containing both K1 and K2. We can also construction. Thus, ker ϕ = {id}, and hence ϕ is injec- define the composite as the intersection of all subfields tive. 0 0 H K ⊆ K containing both K1,K2 ⊆ K as subfields. Let H = im ϕ, K the corresponding fixed field in K containing F . Since H also fixes F 0, we know that Proposition 14.2. Let F,−→ K be a field extension, H 0 and let K1/F and K2/F be finite field extensions of F K ⊇ K ∩ F contained in K. Then Meanwhile, the group Gal(KF 0/F 0) fixes KH F 0. By the Galois correspondence, this tells us that [K1K2 : F ] ≤ [K1 : F ][K2 : F ] KH F 0 = F 0 Proof. Let {α1, . . . , αn} be a basis for K1/F , and H 0 H 0 {β1, . . . , βm} a basis for K2/F . Then we have which means K ⊆ F and hence K ⊆ K ∩ F . Then KH = K ∩ F 0, and the Galois correspondence yields K K = K (β , . . . , β ) = F (α , . . . , α , β , . . . , β ) 1 2 1 1 m 1 n 1 m H = Gal(K/K ∩ F 0)

The βj span K1K2 over K1, so which completes our proof. 0 [K1K2 : K1] ≤ m = [K2 : F ] Corollary 14.5. Let K/F be a Galois extension, F /F any finite extension. Then with equality iff the βj are independent over K1. [K : F ][F 0 : F ] [KF 0 : F ] = [K ∩ F 0 : F ] Corollary 14.3. If [K1 : F ] = n, [K2 : F ] = m, with gcd(n, m) = 1, then Proposition 14.6. Let F be a field, K1/F and K2/F Galois extensions. Then K1 ∩ K2/F is Galois. [K K : F ] = [K : F ][K : F ] 1 2 1 2 Proof. Let f ∈ F [X] be an irreducible polynomial with root α ∈ K ∩ K . Since α ∈ K and K is a splitting Proposition 14.4. Let F,−→ K be a Galois extension, 1 2 i i field over F , f splits in each. Then f splits in K ∩ K , F,−→ F 0 any extension. Then KF 0/F 0 is a Galois ex- 1 2 which is therefore Galois as desired. tension with Galois group Proposition 14.7. Let F be a field, K1/F and K2/F 0 0 ∼ 0 Gal(KF /F ) = Gal(K/K ∩ F ) Galois extensions. Then K1K2/F is Galois, and

We have the diagram Gal(K1K2/F ) = {(σ, τ) : σ|K1∩K2 = τ|K1∩K2 }

21 Math 123—Algebra II Max Wang

K1K2 Observation 15.2. Let QS be the ﬁeld of solvable numbers over Q. The ﬁeld of constructible numbers QC is given by those elements α ∈ ¯ obtained by a tower of K1 K2 Q extensions with [Ki : Ki−1] = 2; it is clear, then, that

K1 ∩ K2 QC ⊆ QS ¯ Now choose some β ∈ QS ⊆ Q. Then we claim that α ∈ ¯ is solvable iff it is solvable over F = (β). If F Q Q α is solvable over F , we can simply append the tower ...... F ,−→ F (α) to the tower ,−→ F , which yields solvabil- Proof. We know that K1 is the splitting field of some Q ity over . If instead α is solvable over , it is trivially polynomial f1 ∈ F [X], and similarly K2 for f2 ∈ F [X]. Q Q solvable over F . Although some extensions in our tower Then K1K2 is the splitting field of f1f2 (eliminating mul- might collapse, none will decompose because they all have tiple roots). Hence, K1K2/F is Galois. Consider the homomorphism prime degree. This yields the following diagram:

ϕ : Gal(K1K2/F ) −→ Gal(K1/F ) × Gal(K2/F ) F¯ σ 7−→ (σ| , σ| ) ¯ K1 K2 Q

Its kernel is trivial on K1 and on K2 and hence on the FS composite, so ϕ is injective. Since QS (σ| )| = σ| = (σ| )| K1 K1∩K2 K1∩K2 K2 K1∩K2 FC we know that im ϕ ⊆ H. QC Note that for each σ ∈ Gal(K1/F ), there are precisely F | Gal(K2/K1 ∩ K2)| elements τ ∈ Gal(K2/F ) satisfying

σ|K1∩K2 = τ|K1∩K2 . Then Q |H| = | Gal(K /F )|| Gal(K /K ∩ K )| 1 2 1 2 where the horizontal edges represent equality. | Gal(K2/F )| = | Gal(K1/F )| ¯ ¯ | Gal(K1 ∩ K2/F )| Theorem 15.3. QS ( Q. Specifically, suppose α ∈ Q has irreducible polynomial f ∈ Q[X], and let K be the Then we have |H| = | Gal(K1K2/F )| = [K1K2 : F ], splitting field of f/Q. Then if Gal(K/Q) is A5 or S5, which yields then α is not solvable. im ϕ = H Proof. Let G = Gal(K/ ). WLOG, we can assume which completes the proof. Q √ G = A5. For if G = S5, then Gal(√K/Q( D)) = A5, Corollary 14.8. Let K1/F and K2/F be Galois exten- and we can simply replace Q with Q( D). sions. If K1 ∩ K2 = F , then If α is solvable, then we have a tower ∼ Gal(K1K2/F ) = Gal(K1/F ) × Gal(K2/F ) = KFn Conversely, if K/F is a Galois extension with Fn Gal(K/F ) = G1 × G2 a product of two subgroups, then KFn−1 there are two Galois extensions K1/F and K2/F such that K = K1K2 and K1 ∩ K2 = F . Fn−1 . Lecture 15 — 2/27/12 . . . Definition 15.1. Let F be a field, α algebraic over F . KF We say that α is solvable if α ∈ K for some K that can 1 be obtained by a tower of field extensions F1

K0 = F,−→ K1 ,−→ K2 ,−→ · · · ,−→ Kn = K K Q where Ki/Ki−1 is Galois of order pi prime.

22 Math 123—Algebra II Max Wang

Our theorem then reduces to the following lemma: Proof of Claim. Consider the splitting ﬁeld K/Q for such a polynomial. We know that Gal(K/Q) would be tran- Lemma 15.4. Let K/F be Galois with Gal(K/F ) = A5, sitive in S5 (i.e., would contain a 5-cycle). We want to L/F also Galois with Gal(L/F ) = Zp. Then we have the construct f such that it also contains a transposition. In following diagram other words, we want f to have exactly 3 real roots and 2 complex conjugate roots. KL Take A5 2 2 L Zp f0(x) = x(x − 4)(x + 4) 5 Zp K = x − 16x

F A5 This polynomial is negative in (−∞, −2) and changes sign at −2, 0, and 2. Its local maximum in (−2, 0) is 15, and its local minimum in (0, 2) is −15. Thus, we can add a Proof of Lemma. We have only two possibilities for constant term −15 < c < 15 and preserve the form of our Gal(KL/K). First, suppose Gal(KL/K) = {e}. Then roots. By the Eisenstein criterion for p = 2, KL = K, so Gal(KL/F ) = A5. But this yields a surjec- 5 tion A5 Zp, which is impossible. f(x) = f0(x) + 2 = x − 16x + 2 It must be the case, then, that Gal(KL/K) = Zp. Then [KL : K] = p, and it follows that [KL : L] = 60. is irreducible. Moreover we know we have an injection This completes the proof of our theorem. Gal(KL/L) ,−→ Gal(K/F ) Lecture 17 — 3/2/12 and so we must have Gal(KL/L) = Gal(K/F ) = A5, as desired. Definition 17.1. A representation of a finite group G is a finite-dimensional complex vector space V with an action of G on V ; that is, a map It remains still to be shown that there exists such a polynomial f as we assumed in our hypothesis. We will ρ0 : G × V −→ V show that ∃f ∈ Q[X] irreducible and of degree 5 with Galois group S5. To do so, we make use of the following satisfying lemma: ∀g, h ∈ G, ∀v ∈ V, ρ0(g, ρ(h, v)) = ρ0(gh, v)

Lemma 15.5. If G ≤ Sp is transitive (or equivalently, if G contains a p-cycle σ) and G contains a transposition Equivalently, we have a homomorphism τ = (i, j), then G = Sp. ρ : G −→ GL(V ) = Aut(V )

Proof of Lemma. We begin by relabeling the set of letters Deﬁnition 17.2. A morphism of representations V , W {1, . . . , p} so that σ takes 1 7→ 2 7→ · · · 7→ p 7→ 1. Now of G is a linear map ϕ : V → W that commutes with the replace σ by σj−i. Then σ carries i 7→ j. We can relabel action of G; that is, the following diagram our letters again so that ϕ V / W σ = 1 7→ 2 7→ · · · 7→ p 7→ 1 g g τ = 1 7→ 2 7→ 1 ϕ V / W which makes σ = (1, . . . , p) and τ = (1, 2). Then commutes ∀g ∈ G. Note that on the left, we have written στσ−1 = (2, 3), and repeated conjugation yields g for ρV (g), and similarly for ρW (g) on the right, so more ∀i ∈ {1, . . . , p},G 3 (i, i + 1) verbosely, we have

ϕρV (g) = ρW (g)ϕ These transpositions generate Sp, as desired. We can also think of the map as respecting conjugacy: Claim 15.6. ∃f ∈ Q[X] irreducible and of degree 5 with ϕ = gϕg−1. A morphism is, in particular, a G-module Galois group S5. homomorphism.

23 Math 123—Algebra II Max Wang

Deﬁnition 17.3. A subrepresentation of a representa- Given a ﬁnite group G, our goal in our study of repre- tion V of a group G is a vector subspace W ⊆ V that is sentations is to classify, to describe, and to construct all invariant under G; that is, representations of G.

∀g ∈ G, g(W ) = W Definition 17.4. We say that a representation V of G is irreducible if it has no nontrivial subrepresentations; that The direct sum of two representations V , W of G is also is, a representation, given by @W ( V,W 6= 0 : ∀g ∈ G, g(W ) = W g · (v, w) = (gv, gw) Theorem 17.5. Every representation of a group G is a direct sum of irreducible representations. Recall that the dual space V ∗ = Hom(V, ). To define C Proof. The theorem follows immediately from the fol- the dual representation, we want the pair v, v∗ to be asso- lowing lemma: ciated with ρ(g)v, ρ∗(g)v∗; that is, we want ρ∗ to preserve g the dual relationship. We know that the map V −→ W Lemma 17.6. Let V be any representation, W ⊂ V a tg proper subrepresentation. Then ∃W 0 ⊂ V a subrepresen- induces the dual map W ∗ −→ V ∗, which is defined by tation such that V = W ⊕ W 0. g V / W We supply two proofs for this lemma.

λ t First Proof of Lemma. We want an inner product h that g(λ) = λg is preserved by the action of G; that is, ∀g ∈ G, C h(v, u) = h(gv, gu) However, if we have Recall that an inner product is a positive definite Hermi- g h V −→ V −→ V tian form; that is, it satisfies thent (h ◦ g) = tg ◦ th. Then we cannot define the action 1. Conjugate symmetry. h(v, u) = h(u, v). ρ∗ by the transpose; instead, we take 2. Linearity in the first argument. ∗ t −1 ∗ ∗ ρ (g) = ρ(g ) : V −→ V h(v + w, u) = h(v, u) + h(w, u) which is a valid representation and respects duality. h(λv, u) = λh(v, u)

Example. Let G be any finite group. 3. Positive-definiteness. h(v, v) ≥ 0 with equality iff v = 0. 1. Let V = C. The trivial representation is given by taking g ≡ id; that is, gv = v. Take h0 to be any inner product on V . By averaging over G, 2. Let V = C; then Aut(V ) = C∗. We get another X one-dimensional representation via the character h(v, u) = h0(gv, gu) homomorphism g∈G we get such an inner product. Then the subspace W ⊥ ∗ χ : G −→ C taken with respect to h is our desired complementary subrepresentation. 3. Let V be a vector space with basis {eg : g ∈ G}. 0 The regular representation is given by Second Proof of Lemma. Let W ⊂ V be any complementary linear subspace, and take : g eh 7→ egh 0 p0 : V = W ⊕ W −→ W 4. Suppose G acts on a set S. The associated to be any projection map. We want a projection p re- permutation representation is a vector space V specting conjugacy. Once again, we simply average over given by a basis {es : s ∈ S}. The action of G all of G, and define on V is given by X −1 p = g ◦ p0 ◦ g g : es 7→ eg(s) g∈G

n For instance, Sn acts on C by permuting the coor- Then ker p gives our desired complementary subrepresen- dinates. tation.

24 Math 123—Algebra II Max Wang

This proves the theorem. This is irreducible because the action of S3 on U is faithful, and hence no subspace is invariant. Example. This decomposition into irreducibles does not Now let W be any representation of G. We will ap- hold in general for inﬁnite groups; consider, for instance, proach the problem of describing W by considering the ( ) eigenvalues and eigenvectors of the action of S3 ⊃ A3 1 n V . Let σ = (1 2 3) ∈ A , and let ω = e2πi/3 be a primi- =∼ : n ∈ ⊂ GL 3 Z 0 1 Z 2 tive cube root of unity. σ has eigenvectors (1, ω, ω2), with eigenvalue ω, and (1, ω2, ω) with eigenvalue ω2. It is easy to see that span(v, w) = V . Lecture 18 — 3/5/12 Now let τ ∈ S3 be any transposition. We have

Before continuing with our study of representation theory, τστ −1 = σ2 we take note of the following two preliminaries: Suppose that v ∈ W is a σ-eigenvector with eigenvalue Theorem 18.1. Let V be a finite-dimensional complex ω. We claim that w = τ(v) is also an eigenvector, with vector space, g : V → V a linear map with eigenvalue ω2. For we have n g = id σ(w) = σ(τ(v)) 2 for some n ∈ N. Then g is diagonalizable; that is, V has = τ(σ (v)) an eigenbasis of g-eigenvectors. = τ(ω2(v)) 2 Note that if G is finite, gn = e for some n for every = ω τ(v) g ∈ G, so we can always diagonalize g’s representation as = ω2w a linear map. So τ acts by exchanging the ωi-eigenspaces of σ. If W is Theorem 18.2. Define irreducible, then W = span(v, w) = V .

X k Suppose instead that v has eigenvalue 1. Then w = τk(x) = xi τ(v) is also an eigenvector with eigenvalue 1, since τ only transposes the ω- and ω2-eigenspaces. We have the fol- be the power sums over the variables x1, . . . , xn. The lowing cases: set τ1, . . . , τn generates the ring of symmetric polynomials over Q. 1. w = v. If W is irreducible, then W = span(v) = U. We will devote most of this lecture to studying the rep- 2. w = −v. If W is irreducible, then W = span(v) = 0 resentations of the simplest nonabelian group, S3. From U . the outset, we have two obvious irreducible representations: 3. w, v linearly independent. Then ∼ ∼ 1. U, the trivial representation, where U = C and S3 span(v + w) = U acts as the identity. span(v − w) =∼ U 0

2. U 0, the alternating representation, where U 0 =∼ C This demonstrates that U, U 0, and V are the only irre- and S3 acts as the sign character. ducible representations of S3. We also have the permutation representation, where S 3 In general, this example illustrates that to understand acts by permuting the coordinates of C3. However, this representation contains a copy of U, namely a representation V of a ﬁnite group G, we want to know the eigenvalues of each element g ∈ G. We can use symmetric polynomials (since the eigenvalues are expressed in U =∼ {(x, x, x) : x ∈ C} the characteristic polynomial) to convey this information. Taking the complementary representation, we get a third Since the power sums generate the ring of symmet- irreducible representation ric polynomials, it will be enough to know for each g the P k sums λi , for this knowledge can be used to retrieve the k 3. V , the so-called standard representation, given by λi themselves. But if g has eigenvalues λi, then g has k λi as its eigenvalues. This then motivates the following V = {(x, y, z) : x + y + z = 0} deﬁnition:

25 Math 123—Algebra II Max Wang

Deﬁnition 18.3. If V is a representation of G, we deﬁne Proof. If w = ϕ(v), then for any h ∈ G, the character map 1 X hw = hg(v) |G| χ : G −→ C g∈G g 7−→ tr(g) 1 X = g(v) |G| that is, we associate g with the sum of its eigenvalues. g∈G Observe that χ(g) depends only on the conjugacy class = ϕ(v) of g ∈ G (since conjugation changes only the eigenvector, = w not the eigenvalue). Hence, we can think of the character as a map and if v ∈ V G, then χ : C −→ C 1 X ϕ(v) = v = v |G| where C is the set of conjugacy classes of G. g∈G 2 The character table for S3 is given by and hence ϕ = ϕ, as desired.

1 3 2 We thus have S3 e (1 2) (1 2 3) dim V G U 1 1 1 z }| { U 0 1 −1 1   1 0 V 2 0 −1  .   ..  2 ϕ =   Note that χV (τ) is 0 because it is a transposition ω ↔ ω  0  and hence has zeros along the diagonal. On the other    ..  hand, we know that σ has eigenvalues ω and ω2, and 0 . hence χ (σ) = ω + ω2 = −1. V and hence G Lecture 19 — 3/7/12 dim V = tr(ϕ : V → V ) 1 X = tr(g : V → V ) Let V be a representation of G. Recall that the character |G| g∈G of V is given by 1 X = χV (g) : |G| χV G −→ G g∈G g 7−→ tr(g : V → V ) This dimension is the number of copies of the trivial rep- Definition 19.1. The invariant subspace of G of a representation in the decomposition of V . In particular, if resentation V is given by V is irreducible (and not the trivial representation), the sum G X V = {v ∈ V : gv = v, ∀g ∈ G} χV (g) = 0 g∈G It is easy to see that V G ⊆ V is a subrepresentation. Definition 19.4. The space of representation morphisms Definition 19.2. The space of endomorphisms of a vec- between representations V and W of G is denoted tor space V is −1 HomG(V,W ) = {α : V → W | α = ρW (g) ◦ α ◦ ρV (g)} End(V ) = Hom(V,V ) This is a subspace of Hom(V,W ). Note also that We want to know how we can determine V G, or at Hom(V,W ⊕ W 0) = Hom(V,W ) ⊕ Hom(V,W 0) least determine its dimension. To do so, we once again The space U = Hom(V,W ) forms a natural represen- rely on averaging over G to obtain G-invariance. tation of G, given by Theorem 19.3. The map ϕ : V −→ V given by ρV (g) : Hom(V,W ) −→ Hom(V,W ) 1 X −1 ϕ(v) = g(v) α 7−→ g αg |G| g∈G The key observation to make here is that

G G is a projection map V − V . Hom(V,W ) = HomG(V,W )

26 Math 123—Algebra II Max Wang

Lemma 19.5 (Schur’s Lemma). Let V , W be irreducible where β is linear. We have a natural bijection of representations of G. Then {bilinear maps V ⊕ W −→ U} ( 1 V =∼ W l dim(HomG(V,W )) = 0 otherwise {linear maps V ⊗ W −→ U} Note that if U = , we have Proof. Let ϕ ∈ HomG(V,W ). Both ker ϕ ⊆ V and C ∗ im ϕ ⊆ W are subrepresentations. Thus, either ϕ = 0 (V ⊗ W ) = {bilinear maps V ⊕ W −→ C} or ϕ is an isomorphism (due to the irreducibility of V and W ). It remains to be shown that the space of iso- The tensor product of two representations V and W morphisms is 1-dimensional in the second case. Suppose yield a natural representation given by V = W . Then the map g(v ⊗ w) = gv ⊗ gw

ϕ : V −→ V ∈ HomG(V,V ) has an eigenvector with eigenvalue λ. Then ϕ − λI has a kernel, so ϕ − λI = 0. Lecture 21 — 3/19/12 Definition 19.6. Let V and W be vector spaces. The Let V be a representation of a finite group G with basis tensor product is a vector space V ⊗ W with a bilinear e1, . . . , en. We will notate map V ⊗k = V ⊗ · · · ⊗ V | {z } ϕ : V ⊕ W −→ V ⊗ W k (v, w) 7−→ v ⊗ w Definition 21.1. The k-th symmetric power of V , denoted We provide three equivalent definitions for the tensor Symk V ⊂ V ⊗k product (although we will not prove their equivalence) is the subspace invariant under Sk. For v1, . . . , vk ∈ V , 1. Let {v1, . . . , vn} be a basis for V , {w1, . . . , wm} a we write basis for W . Then we define V ⊗ W as the vector 1 X v ··· v = v ⊗ · · · ⊗ v ∈ V ⊗k space with basis {vi ⊗ wj} and extend the map ϕ 1 k k! σ(1) σ(k) by bilinearity. This definition readily yields σ∈Sk Then v ··· v ∈ Symk V , and dim(V ⊗ W ) = dim V · dim W 1 k

{ei1 ··· eik : 1 ≤ i1 ≤ · · · ≤ ik ≤ n} 2. Let U be the vector space with basis is a basis for Symk V . {v ⊗ w : v ∈ V, w ∈ W } Deﬁnition 21.2. The k-th alternating power of V , denoted We let U 0 be the subspace spanned by ∧k V ⊂ V ⊗k 0 0 (v + v ) ⊗ w − (v ⊗ w + v ⊗ w) is the subspace skew-invariant under Sk; that is, where (λv) ⊗ w − λ(v ⊗ w) σ(v) = εσv for σ ∈ Sk, with 0 0 v ⊗ (w + w ) − (v ⊗ w + v ⊗ w ) εσ = sgn(σ) v ⊗ (λw) − λ(v ⊗ w) For v1, . . . , vk ∈ V , we write We deﬁne the tensor product as the quotient 1 X v ∧ · · · ∧ v = ε v ⊗ · · · ⊗ v ∈ V ⊗k 1 k k! σ σ(1) σ(k) V ⊗ W = U/U 0 σ∈Sk Then v ∧ · · · ∧ v ∈ ∧k V , and and we take (v, w) 7→ v ⊗ w. 1 k {ei ∧ · · · ∧ ei : 1 ≤ i1 < ··· < ik ≤ n} 3. We take V ⊗ W to be the universal object for bilin- 1 k ear maps V ⊕ W → U. That is, every bilinear map is a basis for ∧k V . Note that we have < rather than α : V ⊕ W → U factors uniquely through V ⊗ W . ≤ because a wedge product where any of the terms are equal will be zero, since transposing them must alternate ϕ β V ⊕ W −→ V ⊗ W −→ U the sign.

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Proposition 21.3. Let V and W be representations of It follows from this, and the first projection formula, a finite group G. Then the following formulas hold that, if V and W are irreducible representations of a finite group G, then 1. χV ⊕W = χV + χW 1 X dim Hom (V,W ) = χ ∗ (g) G |G| V ⊗W 2. χV ⊗W = χV · χW g∈G 1 X 3. χV ∗ = χV = χ (g) · χ (g) |G| V W g∈G 2 2 χV (g) + χV (g ) ( 4. χSym2 V (g) = 1 V =∼ W 2 = 0 otherwise 2 2 χV (g) − χV (g ) 5. χ 2 (g) = ∧ V 2 To describe this equation, we can define a Hermitian inner product on CC by Proof. Suppose g has eigenvalues {λi} as an endomor- 1 X phism of V and {µj} as an endomorphism of W . Then (χ, ψ) = χ(g) · ψ(g) |G| {λi}∪{µj} and {λi·µj} are the eigenvalues for g on V ⊕W g∈G and V ⊗ W respectively, which proves (1) and (2). We Then we have proven that know that {λi} are n-th roots of unity, for n the order of −1 ∗ C g. Hence, {λi = λi} are the eigenvalues for g on V , Theorem 21.5. In the inner product space of C given which proves (3). For g on Sym2 V , the eigenvalues are by (χ, ψ), the characters of the irreducible representations {λiλj}i≤j, so of a finite group G are orthonormal. X This has a number of consequences. χSym2 V (g) = λiλj i≤j Corollary 21.6. There are only finitely many irreducible 2 representations, at most the number | | of conjugacy P λ + P λ2 C = i i classes of G. 2 χ (g)2 + χ (g2) We will show shortly that, in fact, equality holds be- = V V 2 tween the number of irreducible representations and the number of conjugacy classes of G. 2 Similarly, g on ∧ V has eigenvalues {λiλj}i

28 Math 123—Algebra II Max Wang

Corollary 21.9. Let V be an irreducible representation, So every irreducible representation Vi appears in R ex- U any one-dimensional representation. Then U ⊗ V is actly dim Vi times. Therefore, irreducible. X |G| = dim R = dim(V )2 Proof. We have i i 1 X (χ , χ ) = χ (g) · χ (g) Note also that, for g 6= e, we have U⊗V U⊗V |G| U⊗V U⊗V g∈G 1 X 0 = χR(g) = χ (g) · χ (g) · χ (g) · χ (g) |G| U U V V X g∈G = aiχVi (g) i 1 X −1 = χU (g) · χU (g) · χV (g) · χV (g) X |G| = dim ViχVi (g) g∈G i = (χV , χV ) X = χVi (e)χVi (g) = 1 i which yield irreducibility, as desired. These formulas are useful in filling out the character table for a given group. Theorem 21.10 (Fixed Point Theorem). If V is the permutation representation associated with the action of G Example. Let G = S3 and suppose we want to deter- on a finite set X, then mine the character of V ⊗ V (where V is the standard representation). Then we simply take the square of χ , χ (g) = #{x ∈ X : gx = x} V V yielding that is, the character of g is the number of elements in X fixed by g. 1 3 2 S3 e (1 2) (1 2 3) Proof. Recall that V has basis {ex : x ∈ X} and an el- U 1 1 1 ement g ∈ G permutes the basis vectors according to its U 0 1 −1 1 permutation of X. Then g has 1’s on the diagonal exactly V 2 0 −1 when g(ex) = ex and 0’s elsewhere; hence, its character V ⊗ V 4 0 1 is exactly the number of fixed points. Let us now apply these theorems by writing out the Observation 21.11. Let R be the regular representation 0 character tables for S4 and A4. Let U and U denote the of G. Recall that it has basis {eg : g ∈ G}, with standard and alternating representations, whose characters we can deduce immediately. g : eh 7→ egh

By the above theorem, the character of R is given by 1 6 8 6 3 S4 e (1 2) (1 2 3) (1 2 3 4) (1 2)(3 4) ( U 1 1 1 1 1 |G| g = e 0 χR(g) = U 1 −1 1 −1 1 0 otherwise Now let V be the standard representation given by We ﬁrst note that R is not irreducible if G 6= {e}. Now if V are the irreducible representations and a their mul- i i V = {(x, y, z, w) ∈ C : x + y + z + w = 0} tiplicity in the decomposition of R, we have with C4 = U ⊕ V . We know from the ﬁxed point the- ai = (χR, χVi ) 4 orem that C has character χ 4 = (4 2 1 0 0), and by 1 X C = χ (g) · χ (g) subtracting the character of U, we get |G| R Vi g∈G 1 1 6 8 6 3 = (|G| · χ (e)) S e (1 2) (1 2 3) (1 2 3 4) (1 2)(3 4) |G| Vi 4 U 1 1 1 1 1 0 = χVi (e) U 1 −1 1 −1 1 = dim Vi V 3 1 0 −1 −1

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∼ We can check that V is irreducible by taking the inner the quotient S4/V4 = S3. Then we see that W is the product of the character with itself: standard representation of S3 pulled back along this quotient: 1 ρW (χ , χ ) = 1(32) + 6(12) + 8(02) S / GL V V 24 4 O 2 2 2 ρV + 6((−1) ) + 3((−1) ) π 1 " = (9 + 6 + 6 + 3) S4/V4 24 = 1 Note that, in general, if N C G is a normal subgroup, a representation ρ : G −→ GL(V ) is trivial on N iﬀ it Note that the sum of the squares of the dimensions (which factors through the quotient are listed in the ﬁrst column, for e) is only G / GL(V ) O 12 + 12 + 32 = 11 < 24 and hence we are not done enumerating our irreducible " G/N representations. There must be additional representations whose dimensions, when squared, sum to 24 − 11 = Now consider A4. Note that (1 2) and (1 2 3 4) are 13. But this can only be partitioned as simply not found in A4, and (1 2 3) splits into two conjugacy classes, (1 2 3) and (1 3 2). We can easily check 22 + 32 = 13 that U and V remain irreducible, while U 0 collapses into 0 We get another irreducible representation by taking the U, and similarly U ⊗ V collapses into U ⊗ V = V . tensor product U 0 ⊗ V . 1 4 4 3 A4 e (1 2 3) (1 3 2) (1 2)(3 4) 1 6 8 6 3 U =∼ U 0 1 1 1 1 S e (1 2) (1 2 3) (1 2 3 4) (1 2)(3 4) 4 V 3 0 0 −1 U 1 1 1 1 1 U 0 1 −1 1 −1 1 This only accounts for V 3 1 0 −1 −1 2 2 U 0 ⊗ V 3 −1 0 1 −1 1 + 3 = 10

We know that U 0 ⊗ V is irreducible, and it is distinct squared degrees, so we are short two one-dimensional rep- since its trace is distinct. Our ﬁnal irreducible, which we resentation. Note that we have will call W , can be determined purely from orthogonality S4 / / S3 relations, in particular, the formula O O

X ? ? χVi (e)χVi (g) = 0 A4 / / A3 i ∼ with A3 = /3. From this and the details below on since we know that χW (e) = dim W = 2. Z abelian groups, we get

1 6 8 6 3 S e (1 2) (1 2 3) (1 2 3 4) (1 2)(3 4) 1 4 4 3 4 A e (1 2 3) (1 3 2) (1 2)(3 4) U 1 1 1 1 1 4 U =∼ U 0 1 1 1 1 U 0 1 −1 1 −1 1 V 3 0 0 −1 V 3 1 0 −1 −1 W 1 ω ω2 1 U 0 ⊗ V 3 −1 0 1 −1 1 W 1 ω2 ω 1 W 2 0 −1 0 2 2 Observation 21.12. Let G be an abelian group. Note The character allows us to determine the form of W . that for an arbitrary g ∈ G, the map g : V −→ V deter- 6 Since (1 2)(3 4) is an involution with trace 2, and since its mined by a representation ρ is not, in general, a morphism eigenvalues must be roots of unity, we decompose 2 = 1+1 of representations, since and hence (1 2)(3 4) acts as the identity. Note that (1 2)(3 4) generates the Klein four group, and we have g(h(v)) 6= h(g(v)) 6An involution is a function that is its own inverse.

30 Math 123—Algebra II Max Wang in general. However, if g ∈ Z(G),7 then the above equal- Now take the forward direction. Assume that ∀h ∈ G, ity does hold, and hence g is a morphism. But G = Z(G) v ∈ V , we have in abelian groups. If V is irreducible, then by Schur’s ϕ(hv) = hϕ(v) lemma, every g ∈ G acts on V by a scalar multiple of 1 X 1 X the identity. Then every subspace of V is invariant, so α(g) · ghv = α(g) · hgv |G| |G| dim V = 1, and we will have g∈G g∈G 1 X 1 X χ (g) = ζ α(g) · h−1ghv = α(g) · gv V ord(g) |G| |G| g∈G g∈G The irreducible representations ρ of G are all therefore Suppose that V is the regular representation and v = ee. elements of the dual group, which is the group of homo- 1 X −1 1 X morphisms α(g) · h ghee = α(g) · gee ∗ |G| |G| ρ : G −→ C g∈G g∈G 1 X 1 X α(g) · e −1 = α(g) · e |G| h gh |G| g Lecture 22 — 3/21/12 g∈G g∈G Note that all summands on each side of the equation Theorem 22.1. Let V1,...,Vk be the irreducible repre- will be linearly independent. If we choose h such that −1 sentations of a ﬁnite group G. Then {χV1 , . . . , χVk } form α(h gh) 6= α(g), then we will have α(g) · eh−1gh on the C −1 a basis for the space of class functions C . LHS and α(h gh)·eh−1gh on the RHS. Then ϕ is not a G- module homomorphism, and this completes our proof. Proof. Recall that, when we consider a representation V of G, we can average the elements of G (considered as ele- When we average over g, we use ϕα,V with the con- ments of End(V )) to obtain a G-module homomorphism. stant function α : g 7→ 1; this projects V onto V1, taking For instance, we know that V1 to be the trivial representation (which is irreducible

for any group G). Using α = χVi , ϕα,V similarly becomes 1 X a projection V − V . ϕ = g : V → V i |G| g∈G C Claim 22.3. If α ∈ C and (α, χVi ) = 0, ∀i, then α = 0.

G In other words, we claim that the χ span C , and is a projection V − V . For all representations V and Vi C since they are independent by orthogonality, they form a functions α ∈ CG, let us deﬁne basis. 1 X ϕ = α(g) · g : V → V Proof. Let V be irreducible, n = dim V . By Schur’s α,V |G| g∈G lemma, the only G-module homomorphisms are scalar multiples of the identity, so for some λ ∈ C, we have To generalize our method of averaging over G, we have the following claim: ϕα,V = λI We wish to determine λ. We have Claim 22.2. ϕα,V is a G-module homomorphism iﬀ 1 C λ = tr(ϕ ) α ∈ C . n α,V 1 X = α(g) · χ (g) Proof. Consider the reverse direction; we want to show n V that ∀h ∈ G, v ∈ V , ϕ(hv) = hϕ(v). We have g∈G |G| = (α, χ ∗ ) 1 X n V ϕ(hv) = α(g) · ghv |G| = 0 g∈G P 1 X Then ϕα,V = α(g) · g = 0 for all representations = α(hgh−1) · hgh−1hv g∈G |G| V . Consider V = R the regular representation. In R, g∈G {ρR(g)}g∈G are linearly independent elements of End(V ). 1 X since α ∈ C = α(g) · hgv So it must be the case that α(g) = 0 for all g ∈ G, and C |G| g∈G hence, α = 0 as desired. = hϕ(v) This completes the proof. 7We denote Z(G) as the center of G, the set of elements of G which commute with all elements; i.e., Z(G) = {z ∈ G : ∀g ∈ G, zg = gz}.

31 Math 123—Algebra II Max Wang

Claim 22.4. Let V be any representation of G. Then which gives our desired result. V ⊗ V = Sym2 V ⊕ ∧2 V Let us now take our theory of characters and apply Proof. Let {ei} be any basis for V . Let g ∈ S2 be it by computing the character table of S5. Let U be the the nontrivial element, which acts as a transposition trivial representation, U 0 the alternating representation, ei ⊗ ej 7→ ej ⊗ ei, and by bilinearity, v ⊗ w 7→ w ⊗ v and V the standard representation. We immediately get for any v, w ∈ V . g has order 2 and hence its eigenval- the characters for these representations and for V ⊗ U 0. ues are −1 and 1. Its eigenspaces are therefore precisely 2 2 We then begin searching for the other irreducible repre- Sym V and ∧ V . But, setting n = dim V , we have sentations by taking tensor products, symmetric powers, n(n + 1) n(n − 1) and alternating powers of these irreducibles. Consider dim Sym2 V + dim ∧2 V = + 2 2 V ⊗ V . We have = n2 = dim V ⊗ V

1 10 20 30 24 15 20 S5 e (1 2) (1 2 3) (1 2 3 4) (1 2 3 4 5) (1 2)(3 4) (1 2)(3 4 5) U 1 1 1 1 1 1 1 U 0 1 −1 1 −1 1 1 −1 V 4 2 1 0 −1 0 −1 V ⊗ U 0 4 −2 1 0 −1 0 1 V ⊗ V 16 4 1 0 1 0 1

Note that ever, that V ⊗ V decomposes as 1 (χV ⊗V , χV ⊗V ) = (256 + 160 + 20 + 24 + 20) 120 V ⊗ V = Sym2 V ⊕ ∧2 V = 4 and hence, χV ⊗V is not irreducible. We do know, how- Computing these characters, we have

1 10 20 30 24 15 20 S5 e (1 2) (1 2 3) (1 2 3 4) (1 2 3 4 5) (1 2)(3 4) (1 2)(3 4 5) U 1 1 1 1 1 1 1 U 0 1 −1 1 −1 1 1 −1 V 4 2 1 0 −1 0 −1 V ⊗ U 0 4 −2 1 0 −1 0 1 ∧2 V 6 0 0 0 1 −2 0 Sym2 V 10 4 1 0 0 2 1

We can check that ∧2 V is irreducible by taking its posed into two squares either as norm; we have 50 = 12 + 72 1 or as (χ∧2 V , χ∧2 V ) = (36 + 24 + 60) 120 50 = 52 + 52 = 1 Suppose we have another one-dimensional irreducible rep-

2 resentation. Any one-dimensional representation is a ho- It follows from the decomposition of V ⊗ V that Sym V momorphism must be the direct sum of three irreducible representa- ρ : S −→ ∗ tions. We have thus accounted for ﬁve irreducibles, and 5 C the sum of squared degrees thus far is whose kernel will be a normal subgroup whose image is abelian. The only subgroup of S5 with abelian quotient 2 2 2 2 2 1 + 1 + 4 + 4 + 6 = 70 is A5, so every one-dimensional representation of S5 must ∼ factor through a representation of S5/A5 = C2. Then U We are still short 50 square degrees. This can be decom- and U 0 are the only such representations.

32 Math 123—Algebra II Max Wang

So we have two irreducible representations remaining, We can check that the remaining irreducible representa- both of degree 5. Consider again Sym2 V . Sym2 V decom- tion has degree 5; call it W . Taking poses as three irreducibles, and we can determine that one is U and another is V by computing χW = χSym2 V − (χU + χV )

(χSym2 V , χU ) = 1 (χSym2 V , χV ) = 1 we obtain our full character table as

1 10 20 30 24 15 20 S5 e (1 2) (1 2 3) (1 2 3 4) (1 2 3 4 5) (1 2)(3 4) (1 2)(3 4 5) U 1 1 1 1 1 1 1 U 0 1 −1 1 −1 1 1 −1 V 4 2 1 0 −1 0 −1 V ⊗ U 0 4 −2 1 0 −1 0 1 ∧2 V 6 0 0 0 1 −2 0 W 5 −1 −1 1 0 1 −1 W ⊗ U 0 5 1 −1 −1 0 1 1

Lecture 23 — 3/23/12 That is, if any v ∈ V can be written uniquely as a sum of elements in these copies of W . In this case, we write Suppose that H ≤ G is a subgroup of a ﬁnite group. We G would like to somehow relate the representations of G to V = IndH W those of H. One direction of this relationship is easy: Example. Deﬁnition 23.1. Let V be a representation of G. By restricting the action of G to H ≤ G, we can naturally 1. G acts on its left H-cosets G/H by left multipli- restrict V to a representation of H denoted cation. Let V be the permutation representation of G this action, with basis {eσ}σ∈G/H . Denote [e] as the W = ResH V identity coset [e] = H. The subspace W = he[e]i is G We see that ResH is an operator mapping invariant under H, and moreover, G ResH : {reps of G} −→ {reps of H} σ he[e]i = heσi , ∀σ ∈ G/H Note that by restricting a representation to a sub- L Then we have V = σ∈G/H σW , and hence V is group H, new invariant subspaces may be created. Con- induced from W , which is the trivial representation trarily, distinct representations on G may become isomor- on H. phic on H.

What we want now is the relationship in the other di- 2. Let RG be the regular representation of G, which rection; from a representation W of H ≤ G, we want to has basis {eg}g∈G. Take RH to be the subspace induce a representation of G. given by basis {eh}h∈H . A similar argument easily Observation 23.2. Suppose that V is a representation demonstrates that RG is induced from RH . of G, W ⊂ V an H-invariant subspace. Note that, for all Theorem 23.4. Let H ≤ G, W a representation of H. g ∈ G, the subspace Then there exists a unique (up to isomorphism) induced G gW = {g · w : w ∈ W } representation V = IndH W . depends only on the left coset gH of g, since We thus have a corresponding operator (gh)W = g(hW ) = gW G IndH : {reps of H} −→ {reps of G} Thus, for σ ∈ G/H a left coset of H, we will write Note that this is not an inverse to ResG ! In general, a σW = g W for g ∈ σ any representative. This moti- H σ σ representation is not induced by its restriction. vates the next deﬁnition. Deﬁnition 23.3. A representation V of G is induced Proof. First, we show uniqueness. Let us start with a from a representation W of H ≤ G if representation V of G, W ⊂ V invariant under H, and M M V = σW V = σW σ∈G/H σ∈G/H

33 Math 123—Algebra II Max Wang

We claim that the action of G on V is determined entirely group. We start with the characters and compose them by the action of H on W and on the group H ≤ G itself. with conjugation by (1 2)—this has the eﬀect of switching Choose a representative gσ ∈ σ for each coset (taking the characters in the last two columns. g[e] = e). Given some g ∈ G and σ ∈ G/H, say gσ = τ. Note that, since the characters form a basis, one of Then we can write g · gσ = gτ · h for some h ∈ H. the irreducible representations, say Y , must diﬀere in the We know that each v ∈ V can be written last two columns of the character table. Then by com- X posing χY with the outer automorphism of conjugation v = g w σ σ with (1 2), we get another character χZ , which is identi- σ∈G/H cal to χY but with the last two columns switched. This 2 for w ∈ W . But we have demonstrates that we must, indeed, have ∧ V = Y ⊕ Z σ and not Y ⊕ Y or Z ⊕ Z, because doubling the characters

g(gσwσ) = gτ hwσ in the last two columns of Y could not yield two 1’s (and the same is true for Z). P So g(v) = gτ hwσ. This indeed determines a unique Thus, our character table is the group action on V , and this construction also proves existence. 1 20 15 12 12 A5 e (1 2 3) (1 2)(3 4) (1 2 3 4 5) (2 1 3 4 5) Let us compute the character table of A5 ≤ S5. Note U 1 1 1 1 1 that the odd permutations are not present in A5; more- V 4 1 0 −1 −1 over, (1 2 3 4 5) breaks up into two conjugacy classes, W 5 −1 1 0 0 (1 2 3 4 5) and (2 1 3 4 5). Y 3 0 −1 α 1 − α We know that the trivial and alternating represen- Z 3 0 −1 1 − α α ∼ 0 tations on S5 collapse in A5, and we have U = U . ∧2 V 6 0 −2 1 1 Then two other pairs of irreducibles also collapse, namely V ∼ V ⊗ U 0 and W ∼ W ⊗ U 0. We can restrict the = = deduced from the fact that χY + χZ = χ∧2 V . We can use remaining four irreducibles of S5 to A5 to get orthogonality relations to solve for α; this yields √ 1 20 15 12 12 1 + 5 A e (1 2 3) (1 2)(3 4) (1 2 3 4 5) (2 1 3 4 5) α = ϕ = 5 2 U 1 1 1 1 1 V 4 1 0 −1 −1 and so we have W 5 −1 1 0 0 ∧2 V 6 0 −2 1 1 1 20 15 12 12 A5 e (1 2 3) (1 2)(3 4) (1 2 3 4 5) (2 1 3 4 5) Taking the norms of χU , χV , and χW conﬁrms their irre- U 1 1 1 1 1 ducibility. ∧2 V then cannot be irreducible since the sum V 4 1 0 −1 −1 of square degrees would exceed |A5|. Currently, we have W 5 −1 1 0 0 accounted for Y 3 0 −1 ϕ ϕˆ 12 + 42 + 52 = 42 Z 3 0 −1ϕ ˆ ϕ Hence, we have remaining Lecture 24 — 3/26/12 60 − 42 = 18 = 32 + 32 Deﬁnition 24.1. For every g ∈ G, we get an automor- We can check that phism of G given by conjugation with g

(χ 2 , χ 2 ) = 2 ∧ V ∧ V G −→ G We then claim that ∧2 V = Y ⊕ Z where Y and Z are h 7−→ ghg−1 our remaining degree 3 irreducible representations. To show this, consider the automorphism of A5 given This yields a group homomorphism by conjugation with (1 2). Observer that, while this is an inner automorphism for S5, it is an outer automorphism G −→ Aut(G) for A5 since (1 2) ∈/ A5. This automorphism ﬁxes the conjugacy classes e, (1 2 3), and (1 2)(3 4) and exchanges which partitions Aut(G) into a normal subgroup Inn(G) the classes (1 2 3 4 5) and (2 1 3 4 5). A group automor- of inner automorphisms given by conjugation and a com- phism additionally acts on the set of representations of a plementary collection of outer automorphisms.

34 Math 123—Algebra II Max Wang

Observation 24.2. Let τ : G → G be an automorphism, acter table is

V any representation of G. We can obtain another rep- ±1 ±2 ±m τ D2n e h h ··· h g resentation V by composing ρV with τ. U 1 1 1 ··· 1 1 0 τ ρV U 1 1 1 ··· 1 −1 G / G / GL(V ) −1 2 −2 m −m = V1 2 ζ + ζ ζ + ζ ··· ζ + ζ 0 2 −2 4 −4 −1 V2 2 ζ + ζ ζ + ζ ··· ζ + ζ 0 ρ τ . V . If τ is an inner automorphism, then We return now to the study of induced representations. We begin with some basic properties. V τ = V, ∀V G Proposition 24.3. The operator IndH satisﬁes the fol- If τ is outer, then it may permute the irreducible repre- lowing properties: sentations nontrivially, e.g., for G = A3. 1. Linearity. Let us explore the representations of the dihedral G G G groups, G = D2n, the group of isometries of the regu- IndH (W1 ⊕ W2) = IndH (W1) ⊕ IndH (W2) lar n-gon. We have the following picture for any dihedral group: 2. Transitivity. For H ≤ K ≤ G, ∼ Cn = Zn / D2n / / Z2 G G K IndH (W ) = IndK (IndH (W )) Choose ζ = ζn any primitive nth root of unity. Let h be a rotation through 2π/n, g any reﬂection. Then hn = e, 3. Push-Pull. Let U be a representation of G, W a 2 g = e, gh = hg, and D2n = hh, gi. representation of H. Let V be any irreducible representation. V has an k G G G eigenbasis with respect to h, with eigenvalues ζ . Sup- U ⊗ IndH (W ) = IndH (ResH (U) ⊗ W ) pose that v ∈ V is an h-eigenvector, Proof. We will prove only the push-pull property. U ⊗ k G L hv = ζ v IndH W =: V . U⊗W0 is H-invariant V = σ∈G/H σ(U⊗ W0) Then Proposition 24.4. Let U be a representation of G, W a h(gv) = g(h−1v) representation of H ≤ G. Then −k = g(ζ v) G G HomH (W, ResH U) = HomG(IndH W, U) = ζ−k(gv) Proof. We claim that any H-module homomorphism So gv ∈ V is an h-eigenvector with eigenvalue ζ−k. The ϕ : W → U can be uniquely extended to G-module ho- G subspace hv, gvi ⊂ V is thus G-invariant, and hence momorphismϕ ˆ : IndH W → U. Recall that

G M V = hv, gvi V := IndH W = σW σ∈G/H For now we will only consider n odd. Then dim V = 2 unless k = 0. So suppose that h acts as the identity. If Choose any v ∈ σW , which will have the form v = gσw g = 1, we get the trivial representation U; if g = −1, we for some representative gσ. We deﬁneϕ ˆ on σW by 0 get the alternating representation U . −1 ϕˆ(v) = gσϕg (v) Note that as with our picture of S = D , if h has σ 3 6 −1 k −k = g ϕg (g w) eigenvectors v1 and v2 with eigenvalues ζ and ζ , then σ σ σ g switches them. Thus, = gσϕ(w) ! = ϕ(gσw) ζk 0 0 1 h 7−→ g 7−→ = ϕ(v) 0 ζ−k 1 0 and clearly we have independence of our choice of repre- n−1 k sentative gσ. This completes the proof. Since we have m = 2 pairs of nontrivial ζ , our char-

35 Math 123—Algebra II Max Wang

Corollary 24.5 (Frobenius Reciprocity). If U is a rep- Note that G/{±1} = {1¯,¯i, ¯j, k¯}. Then the second quo- resentation of G, W a representation of H ≤ G, then tient map kills 1¯ and one of {¯i, ¯j, k¯}, sending the remaining two to the nonidentity element of Z/2, which has char- (χW , χ G )H = (χ G , χU )G acter −1. Thus, we get ResH U IndH W Q +1 −1 ±i ±j ±k This is equivalent to the claim U 1 1 1 1 1 G G Ui 1 1 1 −1 −1 dim(HomH (W, ResH U)) = dim(HomG(IndH W, U)) Uj 1 1 −1 1 −1 Uk 1 1 −1 −1 1 Lecture 25 — 3/28/12 V 2 −2 0 0 0 We get the final irreducible using orthogonality rela- Definition 25.1. Let G be any finite group. Define tions (also note that if all representations were one- dimensional, the group would be abelian). nX o R(G) = aiVi : Vi rep of G / hV ⊕ W − V − W i We can identify this representation V as H. We make H into a two-dimensional complex vector space via com- This is the free abelian group on the isomorphism classes plex multiplication on the right. Identifying C = h1, ii, of irreducible representations of G; note that we can take {1, j} as a basis, whereupon we have 1 = (1, 0) i = (i, 0) j = (0, 1) k = (0, −i) c R(G) ' Z The elements of Q act via left multiplication on H. Since where c is the number of conjugacy classes of G. Addition H is associative, this is indeed a representation, and we on R(G) is given by ⊕, and multiplication ⊗. can check it against our character table. Definition 25.5. A representation V of G is a real Theorem 25.2 (Artin). The representations of G in- representation if we can write duced from cyclic subgroups of G generate a subgroup of finite index in R(G). V = V0 ⊗R C

with an action of G on V0 that extends by linearity to the Theorem 25.3 (Brauer). We say a group H = A × B is action of G on V . In other words, for some choice of iso- elementary if A is cyclic and B is a p-group with p |A|. ∼ - morphism GL(V ) = GLn C, the representation ρ factors The representations of G induced from elementary sub- through GLn R: groups of G generate R(G). ρ ∼ G / GL(V ) = GLn C Deﬁnition 25.4. The vector space of quaternions is 7 given by " * GLn R H = {α1 + α2i + α3j + α4k : αi ∈ R} Let us look at the construction V = V0 ⊗R C more where multiplication is given by closely. Say V0 is a real vector space of dimR V0 = n. Then, taking C as a two-dimensional real vector space, i2 = j2 = k2 = ijk = −1 V = V0 ⊗R C is also a real vector space of dimR V = 2n. However, V also has the structure of a complex vector Note that space of dimC V = n, given by ∼ 4 H = R i(v ⊗ λ) = v ⊗ iλ and hence is a real vector space. That is, if we write

Let us compute the character table for the group of V0 = {a1v1 + ··· + anvn : ai ∈ R} quaternions we have Q = {±1, ±i, ±j, ±k} V = V0 ⊗R C = {a1v1 + ··· + anvn : ai ∈ C}

We know that {±1}CG is normal, and we have a sequence V is called the complexiﬁcation of V0. We can see from of quotient maps this that any g : V → V ∈ G can be represented either as a complex matrix or as a real matrix with twice the G / / G/{±1} =∼ Z/2 × Z/2 / / Z/2 dimension, if V is a real representation.

36 Math 123—Algebra II Max Wang

Example. Consider G = Z/3 acting on But note that B0 cannot be positive deﬁnite over C, since, for instance, 3 V = {(x, y, z) ∈ C : x + y + z = 0} B0(iv, iw) = −B0(v, w) This representation is real because it breaks into V ⊗ 0 R C We do not want B to be trivially zero anywhere, so we for demand nondegeneracy. As it turns out, this is true if 3 V0 = {(x, y, z) ∈ : x + y + z = 0} R and only if V is real (since B0 can be positive deﬁnite on the real vector space V0). We know that V0 ' R; in this interpretation, G acts via rotation through 120◦. Note that, setting v = (1, ω, ω2) and w = (1, ω2, ω), we have Lecture 26 — 3/30/12

V = C hvi ⊕ C hwi Let V be an irreducible representation of a finite group G. We know there exists a G-invariant positive definite However, neither of these component vector spaces is real, Hermitian form H : V ×V → C on V . That is, H satisfies so V is irreducible as a real representation. H(gv, gw) = H(v, w), ∀g ∈ G, ∀v, w ∈ V Observation 25.6. Note that real representations have We wish to determine whether there also exists a G- real character, but the converse is not true. For instance, invariant nondegenerate symmetric bilinear form B : the representation V of the quaternion group has real V × V → on V . Every bilinear form gives a map character, but is not real. C ∗ αB : V −→ V Definition 25.7. A symmetric bilinear form on a vector space V is a map v 7−→ B(v, ·)

That B is nondegenerate means that αB is an isomor- B : V × V → C phism; that B is G-invariant means such that, ∀u, v, w ∈ V , ∀λ ∈ C, B(gv, gw) = B(v, w), ∀g ∈ G, ∀v, w ∈ V 1. B(u + v, w) = B(u, w) + B(v, w) Theorem 26.1. An irreducible representation V of a ﬁ- nite group G admits a nondegenerate G-invariant sym- 2. B(λv, w) = λB(v, w) metric bilinear form iﬀ V is a real representation.

3. B(u, v) = B(v, u) Proof. We begin with the following claim: Lemma 26.2. B is G-invariant iff α is a G-module A bilinear form yields a map B homomorphism. ∗ αB : V −→ V Proof. To be a G-module homomorphism, by the defini- v 7−→ B(v, ·) tion of the dual representation, αB must satisfy t −1 which takes a vector v to B pre-parametrized with v as αB(gv) = g αB(v) an argument. We say that B is nondegenerate if this map t −1 αB(gv) and g αB(v) are respectively maps αB is an isomorphism. w 7−→ B(gv, w) w 7−→ B(v, g−1w) Recall that if V is a representation of G, then V admits a positive definite Hermitian inner product H which Clearly, these are equal iff B is G-invariant. is G-invariant; we took any H0 and then averaged, yield- Note that ing 2 1 X Hom(V,V ∗) = V ∗ ⊗ V ∗ = Sym V ∗ ⊕ ∧2 V ∗ H(v, w) = H (gv, gw) |G| 0 g∈G so every bilinear form is a sum of a symmetric bilinear form with a skew-symmetric bilinear form. Since V is ir- We would like to know whether we can also find a non- reducible, this means that B must either be symmetric or degenerate symmetric bilinear form B on V . Note that, skew-symmetric. We then have have one of the following if we try to choose B and average, we get 0 three cases: 1 X ∼ ∗ B(v, w) = B (gv, gw) 1. V =6 V , so there does not exist any nondegenerate |G| 0 g∈G G-invariant bilinear form on V

37 Math 123—Algebra II Max Wang

2. V =∼ V ∗ through a nondegenerate G-invariant sym- Then λ = λ¯ and hence λ is real (and positive). metric bilinear form Multiplying B by a scalar, we can assume λ = 1. Then ϕ : V → V is a real linear map, but ϕ2 = I. Hence, it ∼ ∗ 3. V = V through a nondegenerate G-invariant skew- has has eigenvalues 1 or −1 So we can decompose symmetric bilinear form + − We also know that V = V ⊕ V

∗ 1 X 2 where V + is the ϕ2-eigenspace of eigenvalue 1, and V − is dim(HomG(V,V )) = χV (g) |G| + g∈G that of eigenvalue −1. Note that v ∈ V means ϕ(v) = v, so If the character is real, then this dimension must be ϕ(iv) = −iϕ(v) 2 nonzero because each summand χ (g) = χ (g)2 will V V Then we have iV + = V −, and therefore we can write be positive. Hence, a real character implies that we are V = V + ⊗ , as desired. in either case 2 or case 3. Meanwhile, a complex character R C means we will be in case 1. We have actually proven a stronger theorem, which Our theorem then reduces to the following lemma: we state as: Lemma 26.3. V is real iff we have case 2. Theorem 26.4. An irreducible representation V of G is Proof. First suppose that V is real. This direction is easy. one and one of the following: We have V = V0 ⊗ C, so we can find a positive definite R 1. Complex. χV is not real, and V does not admit a symmetric bilinear form B0 on V0 and make it G-invariant G-invariant nondegenerate bilinear form. by averaging. We then simply extend this by linearity to a form B on V . 2. Real. V is a real representation, χV is real, and Now consider the reverse direction. Suppose we have a V admits a G-invariant nondegenerate symmetric nondegenerate G-invariant symmetric bilinear form B on bilinear form. V . Let H be any positive definite G-invariant Hermitian form. As with B, this yields a map 3. Quaternionic. V is a real representation, χV is real, and V admits a G-invariant nondegenerate skew- ∗ αH : V −→ V symmetric bilinear form. v 7−→ H(v, ·) Note that unlike over the complex numbers, when we Note, however, that unlike αB which is linear, the map work over the real numbers, we can’t easily determine the αH is conjugate linear. number of representations. About all we can state in gen- Take the composition eral is that the complex representations come in cojugate pairs. α−1 ϕ : V −−→αB V ∗ −−−→h V which is an automorphism of V . Now consider ϕ2 : V → Lecture 27 — 4/2/12 V , which is a complex linear G-module homomorphism. Definition 27.1. A module M over a ring8 R is an By Schur’s lemma, abelian group with a scalar multiplication map ϕ2 = λI Moreover, we can see by our construction that, ∀v, w ∈ V , R × M −→ R ϕ satisfies (r, x) 7−→ rx

H(ϕ(v), w) = B(v, w) satisfying the usual axioms ∀r, s ∈ R, ∀x, y ∈ M, = B(w, v) r(sx) = (rs)x = H(ϕ(w), v) • = H(v, ϕ(w)) • 1 · x = x So for ϕ2, we have • (r + s)x = rx + sx H(ϕ2(v), w) = H(v, ϕ2(w)) • r(x + y) = rx + ry H(λv, w) = H(v, λw) Example. 8Any ring R we consider will be commutative with identity unless otherwise speciﬁed.

38 Math 123—Algebra II Max Wang

1. R is a module over R, r · s = rs. Note that for free modules, we have

2. The product Rm ⊕ Rn = Rm+n n m n mn R = {(x1, . . . , xn) : xi ∈ R} R ⊗ R = R

is a module given by the rule Deﬁnition 27.9. Let M be an R-module, x1, . . . , xn ∈ M. We have a map ϕ : Rn → M given by r(x1, . . . , xn) = (rx1, . . . , rxn)

A module of this form is called a free module. (a1, . . . , an) 7−→ a1x1 + ··· + anxn

Definition 27.2. A submodule N ⊂ M over R is a sub- We say that x1, . . . , xn are generators of M if ϕ is sur- group closed under scalar multiplication. jective. We say M is finitely-generated if there exists a finite set of generators. R × M −→ M ∪ ∪ Example. Let us consider some modules over Z. R × N −→ N 1. Z, Zn, and the algebraic integers are all Z-modules. Definition 27.3. If N ⊂ M is a submodule, the quotient 2. Let M be any -module. Then module M/N is the quotient group together with scalar Z multiplication rule mx = x + ··· + x rx = rx | {z } m Observation 27.4. The submodules of the module M = R are the ideals in R. That is, the module structure is determined by the group structure. We thus get a bijection between : Definition 27.5. An R-module homomorphism ϕ abelian groups and Z-modules. M → N is a group homomorphism that commutes with scalar multiplication 3.2 Z ⊂ Z is a submodule of Z; moreover, ϕ(rx) = rϕ(x) 2Z =∼ Z Observation 27.6. The kernel ker(ϕ) is a submodule of as -modules, since we have M, and likewise the image im(ϕ) is a submodule of N. Z ∼ Definition 27.7. Let M, N be R-modules. The direct Z −→ 2Z sum of modules is given by m 7−→ 2m M ⊕ N = {(x, y) : x ∈ M, y ∈ N} We get Z/2 as a the quotient module Z/2Z. with scalar multiplication given by 4. We claim that Z/2 ⊗ Z/3 = (0). For we have r(x, y) = (rx, ry) 2(1 ⊗ 1) = 2 ⊗ 1 = 0 Definition 27.8. Let M, N be R-modules. The tensor 3(1 ⊗ 1) = 1 ⊗ 3 = 0 product of modules is a R-module M ⊗ N with an associated bilinear map Then ϕ : M ⊕ N −→ M ⊗ N 1 ⊗ 1 = 3(1 ⊗ 1) − 2(1 ⊗ 1) = 0 (x, y) 7−→ x ⊗ y Note that this situation could never occur in a vector space. such that every bilinear map ψ : M × N → P for an R-module P factors uniquely through ϕ: Definition 27.10. Let S be any set. Then the free module generated by S is given by ψ M × N / P ; S R = {a1s1 + ··· + ansn : ai ∈ R, si ∈ S} ϕ α % In general, given x , . . . , x ∈ M, we say they generate M ⊗ N 1 n M if {x } where α is an R-module homomorphism. R i − M

39 Math 123—Algebra II Max Wang

Lecture 28 — 4/4/12 Observation 28.5. Note that modules exhibit patholog- ical behavior as compared to vector spaces. Definition 28.1. Let R be a ring. Define 1. Not every module has a basis; for instance, Z/n as Mn(R) = {n × n matrices (aij) : aij ∈ R} a Z-module. = Hom(Rn,Rn) 2. Even if a module M has a basis, (e.g., a free mod- 2 ' Rn ule M =∼ Rn), it is not the case that every linearly independent set can be extended to a basis. For Let GL (R) ⊂ M (R) be the subset of invertible matri- n n instance, take M = Z as a Z-module; the vector ces; that is, v1 = 2 cannot be made a part of a basis. That is, we have a Z-module homomorphism GLn(R) = {A ∈ Mn(R) : ∃B ∈ Mn(R), AB = In} = Aut(Rn) 2x Z −→ Z Definition 28.2. Let A ∈ M (R) with entries (a ). The n ij that is injective without being surjective. determinant of the matrix A is given by Similarly, if M = Z × Z, the vectors v1 = (1, 1), X det A = sgn(σ) · a1,σ(1) ··· an,σ(n) v2 = (1, −1) are independent, but cannot be ex-

σ∈Sn tended to a basis. Observe that 3. It is also not the case that every generating set contains a basis. For instance, v1 = 2 and v2 = 3 det(AB) = det(A) · det(B) generate M = Z but do not contain a basis.

Theorem 28.3. A ∈ Mn(R) is an isomorphism (i.e., These pathologies make it impossible for modules to have invertible) iff det A is a unit in R. a well-defined notion of dimension. −1 Proof. First, suppose A is invertible. Then AA = In, In general, a module M is free iff it has a basis (not so we have necessarily a finite one). det(A) · det(A−1) = 1 Definition 28.6. If a module M has a basis v , . . . , v from which it follows that det A is a unit. 1 n (i.e., M = Rn), we say that n is the rank of M. Now let cof(A) be the matrix of cofactors of A. We know that Note that a module can have rank 0 without being A · cof(A) = det(A) · I the zero module (for instance, in Z/n). since det A is a unit, it has an inverse b ∈ R, and hence Lemma 28.7. If a module M has a basis, then any two b · cof(A) is our desired inverse to A. bases of M have the same cardinality. It follows also that n if n 6= m, then Rn =6∼ Rm. We will denote the basis of R by {e1, . . . , en}, where e = (0,..., 1,..., 0) Let M be a finitely-generated free module with basis i ∼ k | {z } v1, . . . , vk. Then we have a natural isomorphism M = R i given by Recall that ϕ : Rk −→ M Definition 28.4. We say that v1, . . . , vk ∈ M generate a e 7−→ v module M if every v ∈ M is a linear combination of the i i vi. Equivalently, the vi generate M if the map 0 0 If v1, . . . , vk is another basis, we can write k ϕ : R −→ M 0 X vi = aijvj ei 7−→ vi

for some aij ∈ R. Then we have is surjective. We say the vi are linearly independent if X k ϕ / aivi = 0 =⇒ ai = 0, ∀i R M or equivalently, if ϕ is injective. A linearly independent P ϕ0 generating set is a basis for M. Rk / M

40 Math 123—Algebra II Max Wang

where P = (aij) is called the change of basis matrix. We now pose the following question: given a homo- Now we can identify morphism ϕ, how can we ﬁnd bases that make this matrix as simple as possible? Equivalently, given a matrix, n m Hom(R ,R ) = Mm×n(R) can we ﬁnd Q ∈ GLm(R) and P ∈ GLn(R) such that Q−1AP has a particularly simple form? Were V and W For suppose M and N are free modules with ranks m and vector spaces, we would have such simple matrix forms; n respectively, ϕ : N → M an R-module homomorphism. however, for modules, in general, the answer here is no. Let us choose respective bases v1, . . . , vm and w1, . . . , wn. In the case of R = Z, however, we have a nicer picture: Then we get an m × n matrix Theorem 29.1. Let V and W be free modules over Z, n ∼ R / N ϕ : V → W a Z-module homomorphism. Then there exist bases {v }n for V and {w }m for W such that the ϕ j j=1 i i=1 matrix representation A of ϕ is the block matrix Rm ∼ / M D 0 which gives us Hom(N,M) = Mm×n(R). 0 0

Definition 28.8. The cokernel of an R-module homo- where D has the form morphism ϕ : M → N is defined as   coker ϕ = N/ im ϕ d1 0    d2  D =  .  We would like now to determine the kernel and coker-  ..  nel of a homomorphism ϕ : Rn → Rm. The issue is that,   0 dk while the kernel is again a free module (as we will show), the cokernel need not be. For instance, such that ∀i < k, di |di+1. 1 2 3 Proof. For every m×n matrix A, we want Q and P such 4 5 6 : 3 −→ 3   Z Z that Q−1AP has the desired form. We will construct Q 7 8 9 and P using a set of elementary matrices (analogous to those from linear algebra). These have the form Lecture 29 — 4/6/12   1 Let V and W be free modules with bases v1, . . . vn and  .   ..  w1, . . . , wm. We can express an R-module homomorphism    1 c  ϕ : V → W as a matrix A = (aij) with    ..     .    0     a1,1 ··· a1,n a1,j  1  .     .    ..    .    .   . .. .     .    Avj =  . . .  1 =  .  1         0     .   am,1 ··· am,n . am,j where c ∈ Z is in the i, j-th place; this adds the ith column of A scaled by c to the jth column (via multiplication That is, we have on the right); X   ϕ(vj) = aijwi 1  .   ..  Now suppose that we choose different bases {v0 } and {w0}   j i  0 1  for V and W . Then in terms of these bases, ϕ corresponds    .   ..  to the matrix   −1   ϕ ' Q AP  1 0   .  0  ..  where P is the change of basis matrix for {vj} → {vj}   0 and Q is likewise for {wi} and {wi}. 1

41 Math 123—Algebra II Max Wang which is the identity matrix with the ith and jth columns Recursing on the submatrix M completes our swapped; this interchanges columns i and j of A; and fi- proof. nally   We have shown that for -modules V and W and any 1 Z homomorphism ϕ : V → W , we can represent ϕ, for some  .   ..  choices of bases, by    1       −1      d1 0  1        d2    .     0   ..  A =   ..       .   1      0 dk    which negates the ith column (note that unlike in a field, 0 0 we cannot scale by arbitrary constants, since only ±1 are units; other constants would yield an uninvertible deter- Note that A takes ei 7→ d1e1 for i ≤ k and ei 7→ 0 for minant). Note that we have analogous row operations i > k. Thus, we have resulting from left-multiplication. ker(ϕ) =∼ Rn−k and im(ϕ) =∼ Rk First, want to arrive at a matrix which has the form 0   Specifically, the image of ϕ is spanned by vectors diwi for d 0 ··· 0 i ≤ k (where {w0} ∈ W is the changed basis) Combined,   i 0  these observations give us: .  . M    Corollary 29.2. If ϕ : V → W is a homomorphism of 0 free Z-modules, then ker(ϕ) and im(ϕ) are free. The cokernel is where d divides every entry of M. We can carry out the Euclidean algorithm via our elementary row operations on m−k coker(ϕ) = Z/d1 ⊕ · · · ⊕ Z/dk ⊕ Z any two row headers to make one of them the gcd of the m two. Therefore, we can arrive at a matrix with a1,1 | ai,1 Lemma 29.3. If W ⊂ Z is any submodule, then W is for all i (that is, a1,1 divides all the row headers). We can free. then subtract a multiple of the first row from every other row to arrive at a matrix of the form Proof. We will assume for now (and prove later) that W is finitely generated. So suppose that w1, . . . , wn are gen-   m n m a1,1 a1,2 ··· a1,n erators of W ⊂ Z . This gives us a map Z → Z which   sends the ith basis vector to a generator w . The image  0  i  . .  of this map is W , and hence W is free as desired.  . ..   .  0 Corollary 29.4. Any finitely-generated abelian group G is of the form If a1,1 does not divide all column headers, we can make a a1,1 the gcd of all the column headers using column op- G = Z ⊕ Z/d1 ⊕ · · · ⊕ Z/dk di |di+1, ∀i erations. Of course, this undoes all the work we have performed on the row headers, and after zeroing out the Proof. Equivalently, we know that G is a finitely- column headers, we must repeat the (very inefficient) al- generated Z-module. Choosing a set of generators m gorithm from the beginning. Since a either gets strictly v1, . . . , vm, we get a map Z − G sending the ei 7→ vi. 1,1 n smaller or divides all headers at the beginning of each rep- The kernel, being free, is isomorphic to some Z . Hence, n m etition of the algorithm, the algorithm terminates. G is the cokernel of the inclusion Z → Z , which yields This is almost what we want; we still want the sub- our desired result. matrix M to contain only entries which are multiples of a1,1. Suppose M has an entry ai,j which a1,1 does not Lecture 30 — 4/9/12 divide. We can use row or column operations to bring ai,j to the first column or row (recall that all the headers We return briefly to a claim we made in a previous lec- are 0). By repeating our algorithm, since a1,1 must get ture, that over an arbitrary ring R, matrices A, B ∈ MnR smaller each time (or becomes equal to 1), this process satisfy eventually stops with our desired matrix. det(A) · det(B) = det(AB)

42 Math 123—Algebra II Max Wang

Let us begin by replacing these matrices with variable In non-Euclidean domains, however, this does not hold. matrices X = (xij) and Y = (ykl). What we wish to do In the case of R = F [x, y], for example, R contains the is prove that the identity ideal I = (x, y), which is not a free module. Then the map ϕ : R − R/(x, y) cannot be diagonalized as a matrix, det(X) · det(Y ) = det(XY ) for then I = ker(ϕ) would be free. holds; we can then substitute elements of any ring R for Definition 30.3. Let R be a ring, ϕ : Rn → Rm a ho- the entries of X and Y . momorphism of finitely-generated free R-modules. Given n m We define this substitution relative to Z[{xij}, {ykl}], some choice of basis for R and R , we can represent ϕ the polynomial ring over the integers in 2n2 variables. by a matrix There is a unique homomorphism → R for any ring Z   R, which yields, given some matrices A, B ∈ MnR, the a1,1 ··· a1,n substitution homomorphism  . . .  A =  . .. .   . .  Z[{xij}, {ykl}] −→ R am,1 ··· am,n sending xij 7→ aij and ykl 7→ bkl. Consider the module But since the determinant function respects ring ho- m momorphism, if our identity holds in Z, it holds in any M = coker(ϕ) = R / im(A) ring R. Now consider the inclusion with quotient map Z[{xij}, {ykl}] ⊂ C[{xij}, {ykl}] ψ : Rm −→ M and the polynomial ei 7−→ vi f(x , y ) = det(X) · det(Y ) − det(XY ) ij ij for the corresponding residues vi ∈ M. Clearly, M is generated by the v ; moreover, each column of A gives a We already know that, in , the polynomial function f i C relation satisfies a1,jv1 + am,jvm = 0 f(xij, yij) = 0, ∀xij, yij ∈ C Conversely, every linear relation among the v is a linear But then f = 0 as a polynomial over , and hence also i C combination of these. So M is fully determined by the v as a polynomial over . i Z along with these relations; we call A the presentation of Observation 30.1. Recall now our corollary, that any M. finitely-generated abelian group G is of the form To obtain a presentation for our module M, we ex- a plicitly defined it as the cokernel of a map between free G = Z ⊕ Z/d1 ⊕ · · · ⊕ Z/dk di |di+1, ∀i modules. Suppose that we instead took any arbitrary R- It is also true for this decomposition that the a and di are module M—what conditions must we impose to obtain a all unique. Note that presentation? Obviously, we must restrict our attention to finitely- a = dimQ(G ⊗Z Q) generated modules. In this case, we can choose generators v , . . . , v for M; this gives us a map ψ : Rm − M. Since, if α ∈ G, we have d α = 0, and so 1 m i Then the relations on M are precisely those elements of 1 the kernel ker(ψ). However, if we want to obtain a pre- α ⊗ 1 = diα ⊗ = 0 di sentation, this kernel must be finitely generated as well. If it is, we can choose generators u1, . . . , un for ker(ψ); Observation 30.2. Note also that our proof of the di- this gives us a map ϕ : Rn → Rm with coker(ϕ) = M. agonalization of matrices in Z applies equally well to any Note that the ker(ψ) itself need not be free, but it must Euclidean domain. That is, for any Euclidean domain R, be finitely generated for us to determine Rn. any matrix in MnR can be diagonalized as How, then, can we tell if ker(ψ) is finitely generated?    This is a hard question to answer, since the map ψ already d1 depends on the choice of generators for M. Instead, we    ..   impose further restrictions on the base ring R.  .       dk  Definition 30.4. A ring R is Noetherian if any ideal   0 I ⊂ R is finitely-generated.

43 Math 123—Algebra II Max Wang

Proposition 30.5. Let R be a Noetherian ring, M a By our induction hypothesis, the image module ϕ(N) ⊂ n−1 finitely-generated R module. Then any submodule of M R is finitely-generated. Choose generators v1,..., vk, is again finitely-generated. and choose also representatives vi ∈ N such that ϕ(vi) = vi. Then, ∀v ∈ N, we can write Lecture 31 — 4/11/12 X ϕ(v) = civi, ci ∈ R Theorem 31.1. Let R be a ring, M a module over R. Then the following are equivalent: Consider the corresponding linear combination of the vi; we can see that 1. Every submodule of M is finitely-generated. X 2. Any infinite sequence v − civi ∈ ker(ϕ)

M1 ⊂ M2 ⊂ · · · ⊂ M But ker(ϕ) ⊂ R. Since R is Noetherian, ker(ϕ) is finitely- generated; choose generators w , . . . , w . Then we can of submodules of M eventually stabilizes; that is, 1 l write any element N as a linear combination of the v ∃n : ∀n, m ≥ n , M = M . This second condi- i 0 0 n m and the w . tion is called the ascending chain condition. j Now let M be any arbitrary finitely-generated R- n Definition 31.2. A Noetherian module is an R-module module, not necessarily free. Let ϕ : R − M be satisfying the above condition. the canonical map taking ei to some n generators of M, and let N ⊂ M be any submodule. By the above, Proof. The reverse direction is easy. Suppose N ⊂ M is ϕ−1(N) ⊂ Rn is finitely-generated; choose generators not finitely-generated. Then we claim that we can con- v1, . . . , vm. Then N is generated by ϕ(v1), . . . , ϕ(vm), struct an infinite, non-stabilizing sequence which completes our proof.

M1 ( M2 ( M3 ( ··· Note that “finitely-generated” has a different mean- Start with any v1 ∈ N, and take M1 = hv1i. Then choose ing depending on whether we are speaking about rings or v2 ∈ N − M1, and take M2 = hv2i. Because N is not modules. For instance, the ring finitely-generated, we can always find such an element vi; we achieve our desired sequence by induction. 1 a Z[ ] = : a ∈ Z, n ∈ N ⊂ Q Suppose now that we have an infinite sequence M1 ⊂ 2 2n M2 ⊂ · · · ⊂ M. Set [ is finitely-generated over Z as a ring, but not as a module. N = Mi ⊂ M As a module, multiplication is only defined with respect to the base ring Z, so no finite number of generators will This is a submodule of M, and hence is finitely-generated. yield all the inverse powers of 2. Choose generators v1, . . . , vk for M. For each i, there must exist some index ni such that ∀n ≥ ni, Mn 3 vi. Lemma 31.4. Let R be a Noetherian ring. Then for any Then there is some module Mn0 in the chain (with ideal I ⊂ R, the quotient R/I is Noetherian. n0 = max{ni}) such that every Mi ⊃ Mn0 in the chain contains all the vi. But then we have Mi = N for every Proof. Let ϕ : R → R/I be the quotient map. Let i > n0. Thus, the chain stabilizies, which completes our J¯ ⊂ R/I be any ideal in R/I. Then J = ϕ−1(J¯) is an proof. ideal in R. By assumption, J is finitely-generated. Choos- ing generators v , . . . , v , we see that J¯ is generated by Theorem 31.3. If R is a Noetherian ring, then every 1 k ϕ(v ), . . . , ϕ(v ). finitely-generated R-module M is Noetherian. 1 k

n Proof. We will start with the case M = R for some Note that we can obtain presentations for any ﬁnitely- n. Our base case for n = 1 is true by assumption; we generated module M over a Noetherian ring R. That is, proceed by induction. Let N ⊂ M be any submodule; we every R-module where R is Noetherian is the cokernel of want to show that N is ﬁnitely-generated. Consider the a map ϕ : Rn → Rm. projection map Theorem 31.5 (Hilbert Basis Theorem). Let R be a ϕ : Rn −→ Rn−1 Noetherian ring. Then the ring R[x1, . . . , xn] is Noethe- (x1, . . . , xn−1, xn) 7−→ (x1, . . . , xn−1) rian.

44 Math 123—Algebra II Max Wang

Proof. Since we obtain R[x1, . . . , xn] by adjoining xn to Definition 31.6. We say that a ring R is R[x1, . . . , xn−1], it suffices to prove the theorem for R[x]. finitely-generated over a field K if K,−→ R and if Let I ⊂ R[x] be any ideal. For every f ∈ R[x] of the form ∃v1, . . . , vk ∈ R such that n f = anx + ··· + a0, with an 6= 0, we define K[x1, . . . , xk] − R lc(f) = an by the evaluation homomorphism. This is notation for the leading coefficient of f. Now consider the set Corollary 31.7. Any finitely-generated ring over a field or Z is Noetherian. A = {lc(f) : f ∈ I} ∪ {0}

We claim that A is an ideal in R. This is fairly easy to Lecture 32 — 4/13/12 see. Let α, β ∈ A, with α = lc(f) and β = lc(g). Take any c ∈ R; if cα 6= 0, then Deﬁnition 32.1. A sequence of R-module homomorphisms cα = lc(cf)

ϕn ϕn−1 ϕ1 and hence A absorbs multiplication in R. To show Mn −−→ Mn−1 −−−−→ Mn−1 −→ · · · −→ M1 −−→ M0 that A is closed under addition, suppose WLOG that deg f ≥ deg g. Then we also have β = lc(xdeg f−deg g · g), is called exact if and hence ∀k < n, ker ϕk−1 = im ϕk α + β = lc(f + xdeg f−deg g · g) We say a sequence is a complex if (unless, of course, α + β = 0). Since A ⊂ R, it is finitely-generated. So let us ∀k < n, ϕk−1 ◦ ϕk = 0 choose generators α1, . . . , αk for A along with polynomials f1, . . . , fk ∈ I with lc(fi) = αi. Take n = that is, if im ϕi ⊂ ker ϕi−1. In this case, the quotients max{deg fi}. WLOG, we can assume the fi all have the (ker ϕk−1)/(im ϕk) are called the cohomology modules. n−deg fi same degree by multiplying each fi by x . Now set Definition 32.2. A short exact sequence is an exact sequence of the form P0 = {f ∈ R[x] : deg f ≤ n} 0 −→ M −→ N −→ P −→ 0 This is a free module over R, isomorphic to Rn+1 (it is, however, clearly not a ring). Let also As a result M,−→ N is an injection, and P = N/M. P = P ∩ I = {f ∈ I : deg f ≤ n} 0 Example. Let R = Z. Let M be a finitely-generated -module; hence, we get a map ψ : m → M onto M’s This is a submodule of P ; hence, P is finitely-generated Z Z 0 generators. We know that ker(ψ) =∼ n for some n is free. as an R-module. Let us choose generators g , . . . , g for Z 1 l This yields a short exact sequence P . We claim now that {fi} ∪ {gj} together generate I. We prove this by induction on degree. Let f ∈ I. If 0 −→ n −→ m −→ M −→ 0 deg f ≤ n, then f ∈ P and we are done; this is our base Z Z case. Suppose deg f > n. We can write lc(f) as a linear Similarly, take R = F [t]. Any finitely-generated mod- combination X ule R-module M is the cokernel of some map lc(f) = ciαi n ϕ m which corresponds to a linear combination of the fi. 0 −→ R −→ R −→ M −→ 0 Then, if we take Since F [t] is a Euclidean domain, we can obtain a diago- X deg f−n g = f − ci(x fi) nalized presentation

∼ a since deg g < deg f, by the induction hypothesis, g is ex- M = R ⊕ R/(d1) ⊕ R/(d2) ⊕ · · · ⊕ R/(dk) pressible as a linear combination of the fi and gj, and therefore so is f. where di ∈ F [t] and ∀i, di |di+1.

45 Math 123—Algebra II Max Wang

Example. Let R = F [x, y] for some ﬁeld F and let The R-module given by M = coker(A), that is, presented M = R/(x, y) =∼ F . Consider the quotient map by A, is generated by two elements v, w with 2 3 π : F [x, y] −→ F (t − 3t + 1)v + (t − 1) w = 0 f 7−→ f(0, 0) (t − 2)v + (t2 − 3t + 2)w = 0 which has ker(π) = (x, y). (x, y) is generated by x and Unfortunately, this doesn’t give us much insight into the y, so we have a natural map ϕ : R2 → (x, y) mapping module’s structure. the standard basis onto the generators. Thus, we have an However, we can diagonalize A as exact sequence 1 0 A = 3 2 ϕ 0 t − 3t + 2t R2 − (x, y) ,−→ R −→π F from which we have The kernel of ϕ is given by M = coker(A) = F [t]/(t3 − 3t2 + 2t) ker(ϕ) = {(f, g) : xf + yg = 0} We can factor t3 − 3t2 + 2t = t(t − 1)(t − 2), which means we can further decompose Since xf = −yg, y | f and likewise x | g. We can write f = yf 0 and g = xg0. But then substituting, we get that M = F [t]/(t) ⊕ F [t]/(t − 1) ⊕ F [t]/(t − 2) xyf 0 = −xyg0. If we set h = g0, then we have f = −yh Each of these direct summands is congruent to F . and g = xh, which means that Observation 33.1. Note that, in general for modules ker(ϕ) = {(−yh, xh)} over a Euclidean domain, we can arrange for the di in the

2 diagonalization to be prime powers rather than dividing This gives a natural map ψ : R → R given by the matrix ai one another in sequence. If R = F [t], can require di = fi −y where the fi ∈ F [t] are irreducible. In the special case ψ = ai x F = C, can take each di = (t − ci) We will now turn brieﬂy to the study of modules in It is easy to see that ker(ψ) = 0, and hence we have an relation to vector spaces and linear operators. Let V be exact sequence an n-dimensional vector space over F , T : V → V a lin-

ψ 2 ϕ π ear map. We can give the group V the structure of an 0 → R −→ R −→ R −→ F → 0 F [t]-module by deﬁning

Now let R = F [x1, . . . , xr], M a finitely-generated R- t · v = T v module. We know that there exists a surjection ϕ0 : ∼ m0 Then we have f(t) · v = [f(T )]v. Since V is finite- M0 → M where M0 = R . Since M0 is Noetherian, dimensional, it is finitely-generated over F [t]. ker(ϕ0) is finitely-generated. Choosing a set of m1 generators of this kernel, we get In the case F = C, we can decompose ∼ a1 an ϕ1 ϕ0 V = C[t]/(t − c1) ⊕ · · · ⊕ C[t]/(t − cn) M1 −−→ M0 −−→ M −→ 0 Each of the direct summands Vi is a submodule of V over ∼ m1 where M1 = R . F [t] as well as a vector space invariant under T . Theorem 32.3 (Hilbert Syzygy Theorem). The process Definition 33.2. A module which is generated by one described above terminates after at most r steps. In gen- element is called cyclic. eral, this is called a free resolution of M. In the case of F [t], modules of the form F [t]/(f) are all cyclic. Lecture 33 — 4/16/12 Consider a single direct summand V := C[t]/(t − c)a, which is also an a-dimensional vector space over C. We In general, if M is an R-module over a Noetherian ring can choose a basis 1, t, t2, . . . , ta−1 for V . The linear map R, it can be realized as the cokernel of a homomorphism T is defined by left-multiplication by t. For each basis vec- a−1 of free modules. Moreover, M has a free resolution. tor vi, we simply have tvi = vi+1 except for va−1 = t , where we have Example. Let R = F [t] for F a field. Consider the ma- a−1 trix a X a a−i i t2 − 3t + 1 t − 2 t = − c t A = i (t − 1)3 t2 − 3t + 2 i=0

46 Math 123—Algebra II Max Wang

Definition 33.3. With respect to the choice of basis Then there exists a polynomial {1, t, t2, . . . , ta−1}, the matrix for T is given in rational canonical form, P (a0, . . . , am, b0, . . . , bn)   0 −a0 such that P (a, b) = 0 iff f and g have a common zero. 1 0 −a   1   .  Proof. Note that f and g have a common zero iff  1 .. −a  T =  2   . .  (f, g) = {af + bg : a, b ∈ [t]} [t]  .. .  C (C  0 .     1 0 −an−2 Let 1 −an−1 ∼ k+1 Sk = {f ∈ C[t] : deg(f) ≤ k} = C

Alternatively, we can choose our basis {vi} to be Consider the map 1, t − c, (t − c)2,..., (t − c)a−1. Multiplication by t, or equivalently, application of T , is given by ϕ : Sn−1 × Sm−1 −→ Sm+n−1 (a, b) 7−→ af + bg v0 7−→ v1 + cv0

v1 7−→ v2 + cv1 If f and g have a common zero, then ϕ is not surjective. Suppose they do not have common zeros. Then if . . af + bg = 0, every zero of f must also be a zero of b, and va−2 7−→ va−1 + cva−2 likewise for g and a. Then a = b = 0, meaning that ϕ is injective and hence an isomorphism. va−1 7−→ cva−1 2 m+n−1 Now choose a basis 1, t, t , . . . , t for Sm+n−1, Definition 33.4. With respect to the choice of basis (1, 0), (t, 0),..., (tm+n−1, 0), (0, 1), (0, t),..., (0, tm+n−1) 2 a−1 {1, t−c, (t−c) ,..., (t−c) }, the matrix for T is given for Sn−1 × Sm−1. We can define in Jordan normal form,   P (a0, . . . , am, b0, . . . , bn) = det(A) c   where A is the matrix representing ϕ of the form 1 c   1 c  T =      . .  a0 0 b0 0  .. ..     . .   a a . ..  1 c  1 0   . . .   . .. .  We are able to achieve the rational canonical form  . a1 . b0   . . . .  for arbitrary fields F , in which case the coefficients in A =  . .. . .  am . a0 . .  the rightmost column will be in F . However, we cannot  .   .. .  achieve Jordan normal form for vector spaces over arbi-  am . a1 bn .   . .  trary F . Note also that R = [t] does not exhibit the  ......  Z  . . . .  same behavior as R = F [t], so this discussion does not 0 am 0 bn hold. Then P (a, b) = 0 iff ϕ is nonisomorphic, which is the case Lecture 34 — 4/18/12 iff f and g have common zeros, as desired.

Theorem 34.1 (Bezout’s Theorem). Let f, g ∈ C[x, y] Proof. (of Bezout’s Theorem) We can write be relatively prime, and say m = deg(f), n = deg(g). m Deﬁne f(x, y) = am(x)y + ··· + a0(x) n Γ = {(x, y) : g(x, y) = f(x, y) = 0} g(x, y) = bn(x)y + ··· + b0(x) Then #Γ ≤ mn. For how many values of x is it the case that f and g have Claim 34.2. Let f, g ∈ C[t] of degrees m and n respec- common zeros? Consider our matrix A now where all the tively, with entries are viewed as polynomials in x. Then the set of x such that f(x, y) and g(x, y) have common zeros is just m f(t) = amt + ··· + a0 the zeros of det(A). Then since deg(ai(x)) ≤ m − 1 and n g(t) = bnt + ··· + b0 deg(bj(x)) ≤ n − 1, we have deg(det(A)) ≤ mn.

47 Math 123—Algebra II Max Wang

Lecture 35 — 4/20/12 Claim 35.6. If f ∈ F [x] is irreducible and non-constant, then f fails to have distinct roots iff f 0 = 0 ∈ F [x]. In the final lecture, we will revisit field theory by dis- cussing the separability of fields. Proof. To say that f does not have distinct roots means ∃K/F , α ∈ K : (x−α)2 |f. Then (x−α)|f 0. Since f and Definition 35.1. Let F be a field, f ∈ F [x] with f 0 have a common factor in K[x], but f is irreducible in deg(f) = n. We say that f has distinct roots if f F [x], then f |f 0 ∈ F [x]. So we must have f 0 = 0 ∈ F [x]. has n distinct roots in its splitting field. Equivalently, ∀F,−→ K, α ∈ K,(x − α)2 -f ∈ K[x]. Definition 35.7. Let F be a field, char(F ) = p > 0. We Observation 35.2. Note that the property of having can define a field homomorphism given by distinct roots is independent of the ground field F ; that ϕ : F −→ F is, for any extension F,−→ K, if f ∈ F [x] has distinct p roots, then f ∈ K[x] does as well. Also, note that if f α 7−→ α has distinct roots and g |f, then g also has distinct roots. since (α + β)p = αp + βp. We say that F is perfect if ϕ is Definition 35.3. We say a polynomial f ∈ F [x] is an isomorphism (i.e., is surjective). We say that all fields separable over F if every irreducible factor of f ∈ F [x] of characteristic zero are perfect. has distinct roots. In particular, note that if F is finite, then F is perfect. Note that this definition, unlike the previous defini- Next, we claim that tion, is dependent on the ground field F . It is still the Claim 35.8. If F is perfect, then every polynomial f ∈ case that if f ∈ F [x] is separable, then f ∈ K[x] is sepa- F [x] is separable. rable, since f’s irreducibles can only decompose more in the extension field K/F . However, the converse is not Proof. Suppose that ∃f ∈ F [x] inseparable. Then ∃f ∈ true in general. F [x] irreducible but without distinct roots. Suppose first that char(f) = 0. Then all nonconstant polynomials have Example. Let F = Fp(t), the field of rational functions p nonzero derivatives. By the previous claim, f must have over Fp. Let f = x − t ∈ F [x]. This is irreducible by the Eisenstein criterion; does it have distinct roots? distinct roots. We claim it does not. Let K/F be any extension, Now suppose char(f) = p > 0. Since F is perfect, we α ∈ K a root of f ∈ K[x]. Then αp = t in K. So in K[x], can write f(x) = g(xp) f(x) = xp − t n p p for some g = anx + ··· + a0 ∈ F [x]. We can also choose = x − α p bi ∈ F such that b = ai. Then p i since Fp = (x − α) f(x) = g(xp) So f is not separable. X = a xpi Theorem 35.4. Let F be any field. If ∃f ∈ F [x] insep- i X i p arable, then char(F ) = p > 0 and #F = ∞. = (bix )

Proof. Recall the derivative, defined for X i p = ( bix ) n f = anx + ··· + a0 so f is not irreducible, a contradiction. as 0 n−1 f = nanx + ··· + a1 This claim, along with the definition of perfect field, yields our desired result. We claim first that Claim 35.5. f ∈ F [x] has distinct roots iff f and f 0 have Definition 35.9. Let F,−→ K be a field extension. no common roots in any extension K/F . α ∈ K is separable over F if its minimal irreducible polynomial f ∈ F [x] is separable. We say that K/F is Proof. Suppose that f(α) = f 0(α) = 0 in K[x], for separable if every element α ∈ K is separable. α ∈ K. Then f(x) = (x − α)g(x) for some g ∈ K[x]. Then Definition 35.10. A field extension K/F is normal if f 0(x) = g(x) + (x − α)g0(x) ∀f ∈ F [x] irreducible, f has a root in K iff f splits completely in K. by the product rule. Then if f 0(α) = 0, we must have g(α) = 0, so (x − α) | g. Then (x − α)2 | f, a contradic- Theorem 35.11 (Fundamental Theorem of Galois The- tion. ory). K/F is Galois iff K/F is normal and separable.