Download Worksheets from Math 594 and more Summaries Differential Equations in PDF only on Docsity!
These are the worksheets from a graduate algebra course focusing on rings and modules, taught at the
University of Michigan in Fall 2021. The worksheets were written by David E Speyer and are released under a Creative Commons By-NC-SA 4.0 International License. If you wish to use them for teaching, contact David E Speyer ([email protected]) for the LATEX source. Many thanks to the students: Emilee Cardin, Heitor Anginski Cotosky, Zach Deiman, Ram Ekstrom, Taey- oung Em, Yuqin Kewang, Sandra Nair, Mia Smith and Ying Wang, for their work and suggestions.
- WORKSHEETS FROM MATH 594: ALGEBRA II, UNIVERSITY OF MICHIGAN, WINTER
- Symmetries of polynomials CONTENTS
- Characters of the symmetric and alternating groups
- A weak version of unsolvability of the quintic
- Groups
- Group actions
- Normal subgroups, quotient groups, short exact sequences
- Simple groups
- Subnormal series, composition series
- The Jordan-Holder theorem
- Solvable groups
- Direct products
- Semidirect products
- Abelian extensions
- The Sylow Theorems
- Some problems with Sylow groups
- Review of polynomial rings
- Degrees of field extensions, and constructible numbers
- Splitting fields and maps between them
- Introduction to field automorphisms
- Galois extensions
- Towers of field extensions and Galois groups
- Artin’s lemma
- Kummer’s theorem and Galois’s criterion for radical extensions
- A. Left and right splittings
- B. Center, central series and nilpotent groups
- C. Finite nilpotent groups are products of p-groups.
- D. Schur-Zassenhaus, the abelian case
- E. The Schur-Zassenhaus theorem, general case
- F. Solvable extensions and unsolvability of the quintic
- G. The Galois correspondence
1. SYMMETRIES OF POLYNOMIALS
Let Sn be the group of permutations of 1 , 2 ,... , n. For two permutations σ and τ , we will write either σ ◦ τ or στ for the composition: (στ )(j) := σ(τ (j)). We will often write permutations using cycle notation: (i 1 i 2 · · · ik) means the permutation which cycles i 1 7 → i 2 7 → · · · 7 → ik 7 → i 1 and fixes everything not in {i 1 , i 2 ,... , ik}. We will let Sn act on the ring of polynomials C[r 1 ,... , rn] in the obvious way.
Set ∆ =
i
3. A WEAK VERSION OF UNSOLVABILITY OF THE QUINTIC
One of the highlights of this course will be the proof of the unsolvability of the quintic. This worksheet proves a weaker version of this result.
Let L be the field of rational functions C(r 1 , r 2 ,... , rn). Define e 1 , e 2 ,... , en as the coefficients of the polynomial:
(x − r 1 )(x − r 2 ) · · · (x − rn) = xn^ − e 1 xn−^1 + e 2 xn−^2 − e 3 xn−^3 + · · · ± en.
Theorem (Ruffini): Starting from e 1 , e 2 ,... , en, it is impossible to obtain the elements r 1 , r 2 ,... , rn of L by the operations +, −, ×, ÷, √n^ , under the condition that, every time we take an n-th root, we must stay in L.
At any point in the computation, there will be some list of elements of L which we have computed so far. Call them θ 1 , θ 2 , θ 3 ,... where each θk is either
(1) An element of C(e 1 ,... , en). (2) Of the form θi + θj , θi − θj , θi × θj or θi/θj , for i, j < k. (3) Of the form n
θj for j < k.
Let Gj be the subgroup of Sn fixing θ 1 , θ 2 ,... , θj.
Problem 3.1. (1) If θk ∈ C(e 1 ,... , en), show that Gk = Gk− 1. (2) If θk is of the form θi + θj , θi − θj , θi × θj or θi/θj , for i, j < k, show that Gk = Gk− 1. (3) If θk is of the form n
θj for j < k, show that there is a character χ : Gk− 1 → C∗^ with kernel Gk.
Deleting the duplicate groups, we obtain a chain of subgroups
Sn = G 0 ) G 1 ) G 2 ) · · ·
such that, for each k ≥ 1 , there is a character χ : Gk− 1 → C∗^ with kernel Gk.
Problem 3.2. Let n ≥ 2. Show that the first step of the chain must be Sn ) An.
Problem 3.3. Let n ≥ 5. Show that the chain ends at An.
Problem 3.4. Prove Ruffini’s Theorem!
4. GROUPS
Definition: A group G is a set with a binary operation ∗ : G × G → G obeying the properties (1) There is an element 1 of G such that 1 ∗ g = g ∗ 1 = g for all g ∈ G. (2) For all g ∈ G, there is an element g−^1 obeying g ∗ g−^1 = g−^1 ∗ g = 1. (3) For all g 1 , g 2 , g 3 ∈ G, we have (g 1 ∗ g 2 ) ∗ g 3 = g 1 ∗ (g 2 ∗ g 3 ). Given a group G, a subgroup of G is a subset containing 1 and closed under ∗ and g 7 → g−^1.
Depending on context, we may denote ∗ by ∗, ×, · or no symbol at all, and we may denote 1 as 1 , e or Id.
Problem 4.1. Show that a group G only has one element 1 obeying the condition (1).
Problem 4.2. Let G be a group and let g ∈ G. Show that G only has one element obeying the condition (2).
Definition: Given two groups G and H, a group homomorphism is a map φ : G → H obeying φ(g 1 ∗ g 2 ) = φ(g 1 ) ∗ φ(g 2 ). A bijective group homomorphism is called an isomorphism and two groups are called isomorphic if there is an isomorphism between them.
A group homomorphism can also be called a “map of groups” or a “group map”.
Problem 4.3. Let φ : G → H be a group homomorphism. Show that φ(1) = 1 and φ(g−^1 ) = φ(g)−^1.
Problem 4.4. Let φ : G → H be a group homomorphism.
(1) The image of φ is Im(φ) := {φ(g) : g ∈ G}. Show that Im(φ) is a subgroup of G. (2) The kernel of φ is Ker(φ) := {g ∈ G : φ(g) = 1}. Show that Ker(φ) is a subgroup of G.
Definition: Given two groups G and H, the product group is the group whose underlying set is G × H, with multiplication structure (g 1 , h 1 ) ∗ (g 2 , h 2 ) = (g 1 g 2 , h 1 h 2 ).
Problem 4.5. Let G and H be two groups and let π 1 and π 2 be the projections G×H → G and G×H → H onto the first and second factor. Show that G × H obeys the universal property of products, meaning that, for any group F with maps φ 1 : F → G and φ 2 : F → H, there is a unique map (φ 1 , φ 2 ) : F → G × H such that the diagram below commutes:
F (φ 1 ,φ 2 )
φ 1
!!
φ 2
$ $
G × H
π (^1) //
π 2
G
H
We close with a few more definitions:
Definition: For g ∈ G, the conjugacy class of g is the set Conj(g) := {hgh−^1 : h ∈ G}.
Definition: A group G is called abelian if g 1 ∗ g 2 = g 2 ∗ g 1 for all g 1 , g 2 ∈ G.
If G is abelian, we often denote ∗ by + and 1 by 0. We never use these notations for a non-abelian group.
6. NORMAL SUBGROUPS, QUOTIENT GROUPS, SHORT EXACT SEQUENCES
Problem 6.1. Let G be a group and let N be a subgroup. Show that the following are equivalent:
(1) For all g ∈ G, we have gN g−^1 = N. (2) N is a union of (some of the) conjugacy classes of G. (3) All elements of G/N have the same stabilizer, for the left action of G on G/N. (4) Every left coset of N in G is also a right coset. (5) If g 1 N = g′ 1 N and g 2 N = g 2 ′N , then g 1 g 2 N = g′ 1 g 2 ′N.
Definition: A subgroup N obeying the equivalent conditions of Problem 6.1 is called a normal sub- group of G. We write N E G to indicate that N is a normal subgroup of G.
Problem 6.2. Let G be S 3. Which of the following subgroups are normal?
(1) The subgroup generated by (12). (2) The subgroup generated by (123).
Problem 6.3. Let G be a group and let N be a normal subgroup of G.
(1) Prove or disprove: Let α : F → G be a group homomorphism. Then α−^1 (N ) is normal in F. (2) Prove of disprove: Let β : G → H be a group homomorphism. Then β(N ) is normal in H. (3) At least one of the statements above is false. Find an additional hypothesis you could add to make it true.
Definition: Given a group G and a normal subgroup N , the quotient group G/N is the group whose underlying set is the set of cosets G/N with multiplication such that (g 1 N )(g 2 N ) = g 1 g 2 N.
This definition makes sense by Part (4) of Problem 6.1. I won’t make you check that this is a group, but do so on your own time if you have any doubt. Also, I won’t make you check this, but the groups G/N and N \G, defined in the obvious ways, are isomorphic.
Let φ : G → H be a group homomorphism. Recall that the image and kernel of φ are Ker(φ) := {g ∈ G : φ(g) = 1} and Im(φ) := {φ(g) : g ∈ G}.
Problem 6.4. Show that the kernel of φ is a normal subgroup of G.
Problem 6.5. Show that the “obvious” map from G/ Ker(φ) to Im(φ) is an isomorphism.
We often discuss quotients using the language of short exact sequences:
Definition: A short exact sequence 1 → A −→α B
β −→ C → 1 is three groups A, B and C, and two group homomorphisms α : A → B and β : B → C such that α is injective, β is surjective, and Im(α) = Ker(β).
I will occasionally write 0 instead of 1 at one end or the other of a short exact sequence. I do this when the adjacent group (meaning A or C) is abelian and it would feel bizarre to denote the identity of that abelian group as 1.
We’ll write Cn for the abelian group Z/nZ. This is called the cyclic group of order n.
Problem 6.6. Show that there is a short exact sequence 1 → Cm → Cmn → Cn → 1.
Problem 6.7. Show that there is a short exact sequence 1 → C 3 → S 3 → S 2 → 1.
Problem 6.8. Show that there is a short exact sequence 1 → C 22 → S 4 → S 3 → 1.
7. SIMPLE GROUPS
Definition: A group G is called simple if G has precisely two normal subgroups, G and { 1 }.
We remark that the trivial group is not simple, since it only has one normal subgroup.
Problem 7.1. Prove or disprove: Let G be simple and let H be any group. For every group homomorphism φ : G → H, either φ is injective or else φ is trivial.
Problem 7.2. Prove or disprove: Let G be any group and let H be simple. For any group homomorphism φ : G → H, either φ is surjective or else φ is trivial.
Problem 7.3. Let p be a prime. Show that Cp (the cyclic group of order p) is simple.
Problem 7.4. In this problem we will show that An is simple, for n ≥ 5. Let N be a nontrivial normal subgroup of An. Let g be a non-trivial element of N.
(1) Show that there is some 3 -cycle (ijk) in An which does not commute with g.
We set h = g(ijk)g−^1 (ijk)−^1.
(2) Show that h ∈ N. (3) Show that h has one of the following cycle structures: (abc)(def ), (abcde), (ab)(cd), (abc). (4) Show that N contains a 3 -cycle. In the case where h has cycle type (ab)(cd), you’ll need to use that n ≥ 5. This part is a nuisance, and you may want to skip ahead and come back to it. (5) Show that N = An.
After Cp and An, the most important simple groups are the projective special linear groups. Let F be a field. The group SLn(F ) is the group of n × n matrices with entries in F and determinant 1. Let Z ⊂ SLn(F ) be {ζ Idn : ζ ∈ F with ζn^ = 1}. The projective special linear group PSLn(F ) is defined to be SLn(F )/Z. The group PSLn(F ) is simple, except in the cases of PSL 2 (F 2 ) (which is isomorphic to S 3 ) and PSL 2 (F 3 ) (which is isomorphic to A 4 ). The proof that PSLn(F ) has a lot of good ideas in it, but it is too long to make a worksheet problem; it might appear as a bonus lecture.
9. THE JORDAN-HOLDER THEOREM
We recall the definitions from last time:
Definition: A subnormal series of a group G is a chain of subgroups G 0 / G 1 / G 2 / G 3 / · · · / GN ⊆ G where Gj− 1 is normal in Gj. A composition series is a subnormal series where G 0 = {e}, GN = G and each subquotient Gj /Gj− 1 is simple. A quasi-composition series is a composition series where G 0 = {e}, GN = G and each subquotient is either simple or trivial.
Let G be a group with a composition series {e} = G 0. G 1. · · ·. GN = G. We define N to be the length of the composition series and write N = `(G). For a simple group Γ and a composition series G•, we define m(G•, Γ) to be the number of quotients Gj /Gj− 1 which are isomorphic to Γ. Our aim today is to prove
Theorem (Jordan-Holder): Let G be a group and let G• and G′• be two composition series for G. Then (G•) =(G′•) and, for any simple group Γ, we have m(G•, Γ) = m(G′•, Γ).
Let 1 → A −→α B →
β −→ C → 1 be a short exact sequence of groups. Let B• be a composition series for B. Recall that we proved on the previous worksheet that { 1 } = α−^1 (B 0 ) ⊆ α−^1 (B 1 ) ⊆ · · · ⊆ α−^1 (Bb) = A is a quasi-composition series for A and { 1 } = β(B 0 ) ⊆ β(B 1 ) ⊆ · · · ⊆ β(Bb) = C is a quasi-composition series for C.
Problem 9.1. With the above notations, let A• and C• be the composition series obtained from deleting duplicate entries from the quasi-composition series above.
(1) Show that (B•) =(A•) + `(C•). (2) For any simple group Γ, show that m(B•, Γ) = m(A•, Γ) + m(C•, Γ).
At this point, you have enough to prove the Jordan-Holder theorem for finite groups, by induction on #(G).
Problem 9.2. Check the base case: Jordan-Holder holds for the trivial group.
Problem 9.3. Check also that Jordan-Holder holds for simple groups.
Problem 9.4. Suppose that G is a finite group which is neither simple nor trivial, and suppose that Jordan- Holder holds for all groups of size less than #(G). Show that Jordan-Holder holds for G. This completes the induction, for #(G) < ∞.
The Jordan-Holder theorem is also true for infinite groups that have composition series! Proving this requires no big new ideas, but a little more finesse. Define L(G) = min `(G•), where the minimum is over all composition series for G. Note L(G) = 0 if and only if G is trivial, and L(G) > 0 for any nontrivial G.
Problem 9.5. Check that L(G) = 1 if and only if G is simple.
Problem 9.6. Let 1 → A → B → C → 1 be a short exact sequence of groups.
(1) Show that L(B) ≥ L(A) + L(C).^1 (2) If A and C are nontrivial, show that L(B) > L(A) and L(B) > L(C).
Problem 9.7. Prove the Jordan-Holder theorem by induction on L(G).
(^1) In fact, equality holds and you have the tools to show it, but you don’t need this.
10. SOLVABLE GROUPS
Now that we have the Jordan-Holder theorem, we can start to classify groups according to what kind of factors appear in their composition/subnormal series. A basic example of this is the solvable groups:
Definition: A group G is called solvable if it has a subnormal series 1 = G 0 E G 1 E G 2 E · · · E GN = G such that Gj /Gj− 1 is abelian.
Problem 10.1. Show that S 3 and S 4 are solvable.
Problem 10.2. Show that a subgroup of a solvable group is solvable.
Problem 10.3. Show that a quotient group of a solvable group is solvable.
Problem 10.4. Show that, if 1 → A −→α B
β −→ C → 1 is a short exact sequence, and A and C are solvable, then B is solvable.
Problem 10.5. Show that a finite group is solvable if and only if all its Jordan-Holder factors are cyclic of prime order.
There is a standard algorithm to test whether a group is solvable, using the derived series.
Definition: Let G be a group. The commutator subgroup, also called the derived subgroup, is the group generated by all products ghg−^1 h−^1 for g and h ∈ G. It can be denoted D(G) or [G, G].
Problem 10.6. Show that D(G) is normal in G.
Problem 10.7. Show that G/D(G) is abelian.
Definition: The quotient G/D(G) is called the abelianization of G and denoted Gab.
Problem 10.8. Prove the universal property of the abelianization: If G is a group, A is an abelian group and χ : G → A is a group homomorphism, then there is a unique homomorphism φ : Gab^ → A such that the diagram below commutes: G χ !!
Gab φ //A
Definition: The derived series of G is the chain of subgroups G D D(G) D D(D(G)) D · · ·. We’ll denote the k-th group in this chain as Dk(G).
Problem 10.9. Show that G is solvable if and only if there is some N for which DN (G) = {e}.
Problem 10.10. (1) For n ≥ 2 , show that Sab n ∼= {± 1 }. (2) For n ≥ 5 , show that Aab n is trivial. (3) For n ≥ 5 , show that Sn is not solvable.
12. SEMIDIRECT PRODUCTS
Once again, we ask how we can stick groups A and C together into a short exact sequence 1 → A → B → C → 1. After direct products, the next most basic way is semidirect products. This time, we’ll do the internal version first. Again, I’ll underline the internal version for this worksheet, but the standard notation is to use the same symbol for both. We recall the definition:
Definition: Let B be a group and let A and C be subgroups. Then AC is the set {ac : a ∈ A, c ∈ C}.
We proved last time that, if A ∩ C = {e}, then the map (a, c) 7 → ac is a bijection from A × C to AC.
Problem 12.1. Let B be a group, let A be a normal subgroup of B and let C be any subgroup of B. Show that AC is a subgroup of B.
Definition: Let B be a group, let A be a normal subgroup of B and let C be any subgroup of B. Suppose that A ∩ C = {e} and that B = AC. Then we say that B is the internal semidirect product of A and C and write B = AoC.
Problem 12.2. Show that S 3 = A 3 oS 2 , with S 2 embedded as the permutations that fix 3.
Problem 12.3. Let B = AoC. Define a map φ : C → Aut(A) by φ(c)(a) = cac−^1.
(1) Show that φ(c) is, as promised, an automorphism of A. (2) Show that φ : C → Aut(A) is a group homomorphism. (3) Show that (a 1 c 1 )(a 2 c 2 ) =
a 1 φ(c 1 )(a 2 )
(c 1 c 2 ).
We use the formula in the last problem to define the external semidirect product:
Definition: Let A and C be groups and let φ : C → Aut(A) be a group homomorphism. We define A oφ C to be the group whose underlying set is A × C, with multiplication (a 1 , c 1 )(a 2 , c 2 ) = (a 1 φ(c 1 )(a 2 ), c 1 c 2 ). We sometimes omit φ when it is clear from context.
Problem 12.4. Check that A oφ C is a group.
So Problem 12.3 says that, if B = AoC, then B ∼= A oφ C for the action φ(c)(a) = cac−^1.
Problem 12.5. Give two actions of C 2 on C 3 such that S 3 ∼= C 3 o C 2 for one action and C 6 ∼= C 3 o C 2 for the other.
Problem 12.6. Let p be prime. Show that Cp 2 6 ∼= Cp o Cp for any action of Cp on Cp.
Problem 12.7. Let 1 → A −→α B −→β C → 1 be a short exact sequence and suppose that there is a group
homomorphism σ : C → B with β ◦ σ = Id. In this case, we will say that the sequence 1 → A −→α B
β −→ C → 1 is right split.
(1) Show that B = α(A)oσ(C). (2) Show that α(A) ∼= A and σ(C) ∼= C, so B ∼= A o C.
Problem 12.8. For any groups A and C, and any action of C on A, show that there is a right split short exact sequence 1 → A → A oφ C → C → 1.
13. ABELIAN EXTENSIONS
Here is a lemma from the homework; check that everyone in your group solved it.
Problem 13.1. Let 1 → A −→α B
β −→ C → 1 be short exact. Let C˜ be any subset of G such that β : C˜ → C is bijective. Then every b ∈ B can be uniquely written in the form α(a)˜c for a ∈ A and c˜ ∈ C˜.
In this worksheet, we will study short exact sequences 1 → A → G → H → 1 with A abelian; when such a short exact sequence exists, we say that G is an abelian extension of H. A special case is when A is central in G, in this case, we say that G is a central extension of H.
Problem 13.2. Let 1 → A → G → H → 1 be an abelian extension. Since A is normal in G, we get an action of G on A by g : a 7 → gag−^1. Show that the map G → Aut(A) factors through H.
We’ll write φ : H → Aut(A) for the resulting action.
Problem 13.3. Show that the action φ is trivial (meaning φ(h)(a) = a for all h ∈ H and a ∈ A) if and only if the extension 1 → A → G → H → 1 is central.
Classifying abelian extensions with fixed (A, H) thus comes down to two parts (1) classifying all actions of H on A and (2) for each action φ, classifying all abelian extensions that result. We know there is always at least one such extension: the semidirect product A oφ H.
Problem 13.4. Let p be a prime number, let H be a group of order pk^ and let 1 → Cp → G → H be an abelian extension. Show that it must be a central extension.
Problem 13.5. Let n be a positive integer and let 1 → Z → G → Cn → 1 be a central extension. Show that G is abelian. (Hint: Let g ∈ G map to a generator of Cn. Use Problem 13.1 with S = { 1 , g, g^2 ,... , gn−^1 }.)
Problem 13.6. Let p be a prime number and let G be a group of order pk. Show that G lies in a central extension 1 → Cp → G → H → 1 for some H of order pk−^1.
Problem 13.7. Let p be prime. Show that every group of order p^2 is isomorphic to C p^2 or Cp 2.
Problem 13.8. Let p and q be distinct prime numbers, let A ∼= Cp, H ∼= Cq and let 1 → A → G → H → 1 be an abelian extension.
(1) If p 6 ≡ 1 mod q, show that the action of H on A is trivial. (2) If the action of H on A is trivial, show that G ∼= Cpq ∼= Cp × Cq. (3) If the action φ of H on A is nontrivial, show that G ∼= Cp oφ Cq.
Problem 13.9. Let p be an odd prime, let A ∼= Cp, H ∼= C p^2. In this problem, we will classify abelian extensions 1 → A → G → H → 1. We write z for a generator of A and ˜x and y˜ for lifts of x and y to G.
(1) Show that z is central in G. (Hint: What can φ be?) (2) Show that every element of G is uniquely of the form x˜a^ y˜bzc^ for a, b, c ∈ { 0 , 1 ,... , p − 1 }. (3) Show that ˜xp, y˜p^ and ˜yx˜˜y−^1 ˜x−^1 are of the form zi, zj^ and zk^ for some i, j and k ∈ Z/pZ. (4) Suppose that k = 0. Show that G is abelian and is isomorphic to either C p^3 or Cp 2 × Cp. (5) Suppose that k 6 = 0 and (i, j) = (0, 0). Show that (˜xa^1 y˜b^1 zc^1 )(˜xa^2 y˜b^2 zc^2 ) = ˜xa^1 +a^2 y˜b^1 +b^2 zc^1 +c^2 +kb^1 a^2. Show that G is isomorphic to the group of matrices of the form
[ 1 ∗ ∗
1 ∗ 1
]
with entries in Z/pZ. (6) Suppose that (i, j) 6 = (0, 0). Show that there are a and b not both 0 mod p such that (˜xa^ y˜b)p^ = 1. This is where you will need that p is odd. (7) Suppose that (i, j) 6 = (0, 0) Show that G ∼= Cp 2 oφ Cp and describe the action of Cp on Cp 2.
Problem 13.10. Let p be an odd prime. Show that every group of order p^3 is isomorphic to one of
C p^3 , Cp 2 × Cp, Cp 3 , Cp 2 o Cp,
{[ (^1) a c 0 1 0 0 1 b
]
: a, b, c ∈ Z/pZ
15. SOME PROBLEMS WITH SYLOW GROUPS
Problem 15.1. Let G be a group of order pkm where p does not divide m. Show that the number of p-Sylow subgroups of G divides m.
Problem 15.2. Let G and H be finite groups and p a prime number. Let P and Q be p-Sylow subgroups of G and H.
(1) Show that P × Q is a p-Sylow subgroup of G × H. (2) Show that every p-Sylow subgroup of G×H is of the form P ′^ ×Q′^ for P ′^ and Q′^ p-Sylow subgroups of G and H.
Problem 15.3. Let 1 → A −→α B
β −→ C → 1 be a short exact sequence of finite groups and let Q be a p-Sylow subgroup of B. Show that α−^1 (Q) and β(Q) are p-Sylow subgroups of A and C respectively.
Problem 15.4. Let p < q be primes and let G be a group of order pq.
(1) Show that the q-Sylow subgroup of G is normal. (2) Conclude that there is a short exact sequence 1 → Cq → G → Cp → 1. (3) Show that G ∼= Cq o Cp for some action of Cp on Cq.
Problem 15.5. Show that there are no simple groups of order 40. (Hint: Look at 5 -Sylows.)
Problem 15.6. In this problem, we will show that there is no simple group G of order 80.
(1) Show that, if G were such a group, then G would have five 2 -Sylow subgroups. (2) Consider the map G → S 5 to get a contradiction.
Problem 15.7. A standard rite of passage is to check that there are no non-abelian simple groups of order < 60 , so let’s do that. Let G be a non-abelian simple group.
(1) Show that the order of G is not a prime power. (2) Show that, for every prime p dividing #(G), there must be some np dividing #(G) with np > 1 and np ≡ 1 mod p.
At this point, we have ruled out all cases except 12 , 24 , 30 , 36 , 48 and 56.
(3) In the notation of the previous problem, show that furthermore we must have #(G)| n 2 p !.
This rules out 12 , 24 , 48 (take p = 2) and 36 (take p = 3).
(4) Suppose that G were a simple group of order 30. Show that G would contain 24 elements of order 5 and 20 elements of order 3 ; deduce a contradiction. (5) Suppose that G were a simple group of order 56. Show that G would contain 48 elements of order 7 and > 8 elements whose order is a power of 2 ; deduce a contradiction.
16. REVIEW OF POLYNOMIAL RINGS
Throughout this worksheet, let k be a field. Let k[x] be the ring of polynomials with coefficients in k.
Here are some things that you hopefully know, and may use without proof.
- Let b(x) ∈ k[x] be a nonzero polynomial of degree d. Let a(x) be any polynomial in k[x]. Show that there are unique polynomials q(x) and r(x), with deg r < d, such that a(x) = b(x)q(x) + r(x).
- The ring k[x] is Euclidean, is a PID and a UFD.
- If p(x) is an irreducible polynomial, then p(x)k[x] is a maximal ideal, and k[x]/p(x)k[x] is a field.
Problem 16.1. Let b(x) ∈ k[x] be a nonzero polynomial of degree d. Show that the ring k[x]/b(x)k[x] is a k-vector space of dimension d.
Let K be a larger field containing k. For θ ∈ K, we say that θ is algebraic over k if there is a nonzero polynomial f (t) in k[t] with f (θ) = 0.
Problem 16.2. Let θ ∈ K be algebraic over k. Let I ⊂ k[t] be {f (t) ∈ k[t] : f (θ) = 0}.
(1) Show that I = m(t)k[t] for some irreducible polynomial m. (2) Show that k[θ], meaning the subring of K generated by k and θ, is isomorphic to k[t]/m(t)k[t].
The polynomial m(t) is called the minimal polynomial of θ.
Problem 16.3. Let K be a larger field containing k. Let α and β be two algebraic elements of K which have the same minimal polynomial. Show that there is an isomorphism k[α] → k[β] taking α to β.
Problem 16.4. Show that θ is algebraic over k if and only if dimk k[θ] < ∞.
Problem 16.5. Show that the set of elements of K which are algebraic over k is a subfield of K.
18. SPLITTING FIELDS AND MAPS BETWEEN THEM
Definition: Let k be a field, let f (x) be a polynomial in k[x] and let K be an extension field of f. We will say that f splits in K if f factors as a product of linear polynomials in K[x]. We say that K is a splitting field of f if f splits as a product c
(x − θj ) in K[x] and the field K is generated by k and by the θj.
For example, if k = Q and θ 1 , θ 2 ,... , θn are the roots of f (x) in C, then Q[θ 1 ,... , θn] is a splitting field of f (x).
Problem 18.1. Let k be a field and let f (x) be a polynomial in k[x]. Show that f has a splitting field.
Problem 18.2. Let L be a splitting field for x^3 − 2 over Q. Show that [L : Q] = 6. (Hint: At one point, it will be very useful to use the fact that Q[ 3
2] is a subfield of R.)
Problem 18.3. Let L = C(x 1 , x 2 ,... , xn). Let ek be the k-th elementary symmetric polynomial and let K = C(e 1 , e 2 ,... , en) ⊂ L. Show that L is a splitting field for xn^ − e 1 xn−^1 + e 2 xn−^2 − · · · ± en over K.
Problem 18.4. Let
f (x) =
x − cos 27 π
x − cos 47 π
x − cos 87 π
8 x^3 + 4x^2 − 4 x − 1
I promise, and you may trust me, that f (x) is irreducible. Let K = Q(cos 27 π ).
(1) Show that [K : Q] = 3. (2) Show that f (x) splits in K. Hint: Use the double angle formula. (3) Show that there is an automorphism σ : K → K with σ(cos 27 π ) = cos 47 π.
Problem 18.5. Let k be a field and let f (x) be a polynomial in k[x]. Let K be a splitting field of f in which f splits as
(x − αj ). Let σ : k → L be a field homomorphism and let σ(f ) :=
σ(fj )xj^ split in L. Show that there is an injection φ : K → L making the diagram
k
σ
K φ^ &^ &//L
commute. Hint: Think about k ⊆ k[α 1 ] ⊆ k[α 1 , α 2 ] ⊆ · · · ⊆ k[α 1 , α 2 ,... , αn] = K.
Problem 18.6. Let k 1 and k 2 be two fields and let σ : k 1 → k 2 be an isomorphism. Let (x) =
fj xj^ be a polynomial in k 1 [x] and let σ(f )(x) :=
σ(fj )xj^. Let K 1 be a splitting field of f and let K 2 be a splitting field of σ(f ). Show that there is an isomorphism K 1 ∼= K 2 making the diagram
k 1
σ (^) //k 2
K 1
∼= (^) // K 2
commute.
Problem 18.7. Let k be a field and let f (x) be a polynomial in k[x]. Let K 1 and K 2 be two splitting fields of f. Show that there is an isomorphism K 1 ∼= K 2 making the diagram
k
' ' K 1
∼= (^) // K 2
commute. So splitting fields are unique.
19. INTRODUCTION TO FIELD AUTOMORPHISMS
Definition: Let K ⊆ L be fields. An automorphism of L is a bijection σ : L → L with σ(x + y) = σ(x) + σ(y) and σ(xy) = σ(x)σ(y). An automorphism of L fixing K is an automorphism of L obeying σ(a) = a for all a ∈ K. We write Aut(L) for the automorphisms of L and Aut(L/K) for the automorphisms of L fixing K.
Problem 19.1. Let K ⊆ L be fields. Let f (x) be a polynomial in K[x]; let {θ 1 , θ 2 ,... , θr} be the roots of f in L.
(1) Show that Aut(L/K) maps {θ 1 , θ 2 ,... , θr} to itself. (2) Show that stabilizer of θj in Aut(L/K) is Aut(L/K(θj )). (3) Let L = Q( 4
- and let f (x) = x^2 − 2. Show that the roots of f (x) in L are {±
2 } and show that Aut(L/Q) fixes both of them.
Problem 19.2.. Let K be a field, let f be a polynomial in K[x], let L be a splitting field for f and let {θ 1 , θ 2 ,... , θn} be the roots of f in L. Assume {θ 1 , θ 2 ,... , θn} are distinct.^1
(1) Show that the action of Aut(L/K) takes {θ 1 , θ 2 ,... , θn} to itself. (2) Show that this action of Aut(L/K) gives an injection Aut(L/K) ↪→ Sn.
Problem 19.3. Let K, f , L and {θ 1 , θ 2 ,... , θn} be as in Problem 19.2. Let g(x) be an irreducible fac- tor of f (x) in K[x] and renumber the θ’s so that {θ 1 , θ 2 ,... , θm} are the roots of g in L. Show that {θ 1 , θ 2 ,... , θm} is the Aut(L/K)-orbit of θ 1 in L. Hint: Apply Problem 18.6 to the diagram
K[θi]
K[x]/g(x)K[x] oo^ ∼=^ ∼=^ // K[θj ]
L //L
Problem 19.4. Let L be the splitting field of x^3 − 2 over Q. Show that Aut(L/Q) ∼= S 3.
Problem 19.5. Let L = Q(cos 27 π ). Show that Aut(L/Q) ∼= C 3.
(^1) This happens if and only if GCD(f (x), f ′(x)) = 1, see the homework.
