Algebra I & II

Question 1

1. a) How many monic polynomial of degree 2 with coefficients in Fp are irreducible over Fp?
   b) How many monic polynomials of degree 3 with coefficients in Fp are irreducible over Fp?

Answer 1

1.#irr=#all-#red
a) #all=p^2
f red => f=gh where g,h monic linear. p + pC2 linear possibilities (p for equal factors, pC2 for distinct factors) => #red=p+pC2
#irr = p^2-p-pC2=p(p-1)/2
(Couting the factors works because Fp is field so Fp[x] is PID and so also UFD.)
b) #all=p^3
If f factors into irr then f=(x^2+ax+b)(x+c) or f=(x+a)(x+b)(x+c).
By a) there are p(p(p-1)/2) for first possibility
There are p+2pC2+pC3 possibilities for the second (all the same + 2 diff + 3 diff)
#irr=p^3-p(p^2-p-pC2)-p-2pC2-pC3
#irr=p^3-p(p(p-1)/2)-p-2pC2-pC3=p(p^2-1)/3

Question 2

2. Let G=F*17 be the multiplicative group of the field F17. Does there exist a transitive group action of G?
   a) on a set with 8 elements?
   b) on a set with an odd number of elements?

Answer 2

2. a) If H is a subgroup G acts on the quotient group G/H (G mod H) by multiplication. All cosets xH are in the orbit of H.
   Let H={1,16}. Then G mod H has |G/H|=|G|/|H|=16/2=8 elements by Lagrange’s theorem.
   b) Unless |T|=1, this is not possible. Since action is transitive Orb(t0)=T for any t0 in T. But by the Orbit-Stabilizer Thm we need that |T|=|Orb(t0)| divides |G|=16=2^4.

Question 3

3. Let G be a group of order 12.
   a) Must G be abelian?
   b) List all possible abelian G of order 12 (up to isomorphism)

Answer 3

3. a) No, A4, the symmetry group of the tetrahedron (4-base/seat) is not abelian: (23)(34)=(234)!=(243)=(34)(23)
   Dihedral group D12 the symmetry group of the regular hexagon (six corners) is not abelian. (Just compose a rotation and reflection)
   b) Z2 x Z6 and Z12
   [From Prop B-3.25 (call it Basis Theorem of finitely generated abelian groups / fundamental theorem of finite abelian groups) it follows that:
   G = C(d1) x … x C(dr) s.t. d1|…|dr.
   So now either r=1 and d1=12
   or r=2 and d1=2 and d2=2*3=6.]

Question 4

4. Is the ideal generated by x^7+1 and x^5+1 a principal ideal in F2[x]? If not, why? If so, find a generator.

Answer 4

4. (F2 is a field and F2[x] is a Euclidean ring)
   Using the Euclidean algorithm:
   x^7+1=x^2(x^5+1)+x^2+1
   x^5+1=x^3(x^2+1)+x^3+1
   x^3+1=x(x^2+1)+x+1
   x^2+1=(x+1)(x+1)
   So by backsubstituting, we can see that we can write x^7+1 and x^5+1 in terms of x+1. So (x^7+1,x^5+1) s.s.o. (x+1).
   By polynomial division (or by using above equations):
   (x^7+1)=(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)
   (x^5+1)=(x+1)(x^4-x^3+x^2-x+1)
   We get (x+1) s.s.o. (x^7+1,x^5+1), and so (x+1)=(x^7+1,x^5+1).

Question 5

5. How many elements in S5 have order exactly 6?

Answer 5

5. Let sigma be in S5 of order 6 with complete factorization sigma=alpha_1 ** alpha_n, where alpha_i is a cycle of length ri. Then 6=lcm(r1,…,rn), so sigma must have a cycle structure of (2,3) since there are no cycles of length 6. Hence there are 5C2*(3-1)!=20 elements of order 6.

Question 6

6. Let d be the smallest index of a proper subgroup of S8. What is d? Do there exist two distinct subgroups of S8 of index d?

Answer 6

6. The index [G:H] of H is defined as |G/H| (the number of cosets of H), which by Lagrange is |G|/|H|.
   So by definition it has to be an integer. We know A8 has index 2, so d=2.
   There can’t be any others. Suppose [S8:H]=2. Then H is a normal subgroup because finite groups can only contain normal subgroups of index 2 by question (7) (ie index 2 implies normality).
   [Generally speaking, {1},An,Sn are the only normal subgroups of Sn [see pdf] but specifically:]
   By the second isomorphism theorem An n H is normal too. Since An is simple, either An n H is {e} or An. If it is An then H=An because |An|=|H|. If it is {e} then besides e, H only contains odd permutations (and more than one, since |An|=|H|). So let sigma and tao be odd permutations, then by uniqueness of their inverses, either sigma*tao != e or sigma^2 !=0. In either case, H would contain an even permutation, giving us a contradiction.

Question 7

7. Let G be a finite group. Let H be a subgroup of G which is not normal. Can the index of H be 2? Can the index of H be 3?

Answer 7

7. 2: No, since if g not in H then gHuH=G=HuHg, so gH=G\H=Hg, i.e. H is normal.
   3: Is possible, {(1),(12)} is not normal in S3, for (23)(12)(23)!=(13) iff (132)=(12)(23)!=(23)(13)=(123).

Question 8

8. Are the following rings UFDs? Are the rings PIDs?
   a) Z[x]
   b) (Z/2Z)[[x]]
   c) R[x]/(x^2-1)
   d) R[x]/(x^2+1)

Answer 8

8.
a) Z is PID (since for a,b in Z there are s,r s.t. gcd(a,b)=sa+rb), so Z UFD. By Gauss Z[x] is now also UFD. But not PID, since e.g. (2,x) is not principle.
b) It is a PID: For any given nonzero power series p, we can write it as a product of X^nq for some integer n and an invertible power series q (recall that a power series is invertible iff its constant term is a unit)
Thus, (p)=(X^nq)=(X^n). Now, for any two non zero power series p,q there are integers m,n s.t. (p,q)=(X^n,X^m)=(X^min(n,m).
[Exercise: A-3.29, (ii)
https://math.stackexchange.com/questions/1264615/inverse-rule-for-formal-power-series]
Every PID is a UFD, so it is also UFD.
c) x+1 and x-1 are zero divisors, for (x+1)(x-1)=x^2-1=0 in R[x]/(x^2-1).
d) As we know R[x]/(x^2+1)~C, so it is a field, hence PID and UFD.

Question 9

9. Which of the following rings are fields? Explain why or why not?
   a) Z[x]/(3,x^2+x+1)
   b) Z[x]/(4,x^2+x+1)
   c) Z[x]/(2,x^2+x+1)

Answer 9

9. a) Z[x]/(3,x^2+x+1)=Z3[x]/(x^2+x+1). It is not a domain: (x+2)(x+2)=x^2+4x+4=x^2+x+1=0
   Additionally: Z3=F3=k is a field, and by prop (3.83) k[x]/(f) is a field iff its a domain iff f !=0 is irreducible.
   b) Not a domain, since 2*2=4=0.
   c) Z[x]/(2,x^2+x+1)~Z2[x]/(x^2+x+1) is a field iff x^2+x+1 is irreducible (prop 3.83), which is its since x^2+x+1=gh => g,h in {x,x+1} but no combination of these yields x^2+x+1.

Question 10

10. What is the automorphism group Aut(G) of the following groups G?
    a) Z/2Z
    b) Z/8Z
    c) Z
    d) S3

Answer 10

10. a) Let sigma be an automorphism, then it is a homomorphism, so sigma(0)=0. Then sigma(1)=1. So Aut(Z2)={id}
    b) sigma in Aut(Z8). Then sigma(0) and sigma is uniquely determined by sigma(1). Since sigma is an isomorphism, sigma(1) must be sent to a generator of Z8. These are 1,3,5 and 7. So, the Aut(Z8)={(0,1,2,3,4,5,6,7), (0,3,6,1,4,7,2,5),(0,5,2,7,4,1,6,3),(0,7,6,5,4,3,2,1)}.
    [By the classification theorem of abelian groups every finite abelian group G is a direct sum of cyclic groups G=C(d1)+…+C(dr) where d1|…|dr. So Aut(Z8) is isomorphic to Z4 or Z2xZ2. We can see that every element is its own inverse, so Aut(Z8)=Z2xZ2.
    [Lem: If a in G, phi hom, then ord(phi(a))|ord(a). So if phi is hom ord(phi(a))=ord(a).]
    c) The general homomorphism is phi(n)=n*phi(1) in Z. Since we need surjectivity, phi(1) is 1 or -1. And so Aut(Z)={-id,id}
    d) Again phi((1))=(1).
    [(12) and (23) generate S3; since (12)(23)(12)=(13) and we can decompose any 3 cycle into these three elements.]
    Since we can generate all elements with transpositions, we only need to assign the transpositions to each other to uniquely determine an automorphism. Assigning 3 elements to 3 elements is S3. So Aut(S3)~S3.

Question 11

11. In the ring C[x,y], do the following inclusions of ideals hold? Prove your answer:
    a) (x-1,y-2) s.s.o. (y^2-4x)
    b) (y^2-4x) s.s.o. (x-1,y-2)

Answer 11

11. a) No, since y^2-4x as a poly in y has degree 2 and y-2 has degree 1.
    b) Yes, since y^2-4x=(y+2)(y-2)-4(x-1)

Question 12

12. Let a in Z s.t. [a]=[3] mod 15, [a]=[1] mod 17.

Does this information suffice to compute a mod 255?What is [a] mod 255?

Answer 12

12. Yes, since gcd(15,17)=1 and 15*17=255 so the existence follows form the chinese remainder theorem.
    Def. a=b mod m iff m|(a-b).
    => 15|a-3 and 17|a-1 => 15p=a-3 and 17q=a-1 => there exist p and q s.t. 15p+3=a=17q+1. (p,q)=(1,1) are solutions and then a=18. So 18 solves both equivalences and a=18 mod 255 also by the chinese remainder theorem.

Question 13

13. Do all groups of order 4 occur as subgroups of S4? Do all groups of order n occur as subgroups of Sn?

Answer 13

13. n=4: There are only two groups of order 4 [see lem below]. By the classification theorem of finite abelian groups these are Z2xZ2 and Z4. Z4 is isomorphic to span({(1234)}) and Z2xZ2 is isomorphic to span({(12),(34)}) both are subgroups of S4.

n=general: Let |G|=n. Put elements of G into bijective correspondence with {1,…,} by phi:G–>{1,…,n}. Now G acts on {1,…,n} by gi=phi(gphi^-1(i)) and every g corresponds to a bijective function on {1,…,n} i.e. a permutation. Thus, G is isomorphic to a subgroup of Sn.

“Let G act on itself by left-mult, this is a faithful group action and thus induces an injective homomorphism G–>Sym(G)~Sn.”

Lem: There are only 2 groups of order 4.
Pf: Because of uniqueness of inverses, at least one of a,b,c must be its own inverse. W.l.o.g. say c (w.l.o.g. because we can just switch names). Then either ab=e or not. If so then the other elements in the Cayley table follow. If not then a and b are both their own inverses. Each of a,b,c multiplied with another will then yield the third. Thus there are only 2 ways of filling out the Cayley table.

Question 14

14. Let G be the group of rotational symmetries of a regular solid cube.
    a) How many elements does G have?
    b) Does there exist a surjective homomorphism from G to S3?

Answer 14

14. a) 1 vertex can go to 8 corners of a cube. The cube is then uniquely determined by the orientation of the 3 adjoining vertices/edges. Therefore there are 3*8 possibilities to permute the edges without breaking the symmetry.
    b) Yes, each rotation takes the faces of the cube and therefore permutes the three axes (crossing through two opposite faces).

Question 15

15. a) Does the alternating group A17 have any subgroup of index 2 or 3?
    b) Does the alternating group A17 have any elements of order 12?

Answer 15

15. a) 2: Since A17 is simple and any group with index 2 is normal (gH=G\H=Hg) the only subgroups of A17 with index 2 are itself and {(1)}.
    3: Suppose there was a subgroup H with index 3. Let A17 act on A17/H, this gives us a homomorphism phi from A17 to S3, 3 since |A17/H|=3. ker(phi) is normal in A17, which is simple and phi is non-trivial so ker(phi) is trivial. So injective. Thus, A17 is isomorphic to im(phi) some subgroup of S3. Now Lagrange’s theorem delivers the contradiction: 17!/2 | 3!.
    https: //math.stackexchange.com/questions/3281671/does-a-17-have-a-subgroup-of-index-3
    b) An cycle of length r has order r. We can write a cycle of length r larger than 1 as r-1 transpositions by: (1…r)=(23)…(r-1 r)(r 1). So we can write a 12 cycle as 11 transpositions and we can add a disjoint transposition to make the permutation even and in A17 (this does not change the arder since 2|12).

Question 16

16. Describe all possible homomorphisms:
    a) f:S3–>Z6
    b) g: S5–>Z2

Answer 16

16. a) Since S3 is generated by (12) and (23) [(12)(23)(12)=(13)]. We only need to determine them. [Make a table of ordes of Z6]. Since they have order 2 and the order of their images must divide their orders and only 0,3 have orders that 2, 1 and 2 respectively, they must be sent to 0 and/or 3. Moreover, since (123) has order 3, its image must be 0,2 or 4. So 2 divides f(123)=f(12)+f(23), so f(12)=f(23). Arguing analagously gives f(132) is 0, 2 or 4 and f(12)=f(13)=f(23)= 0 or 3. So there are two possible homomorphisms: f=const=0 or f(12)=f(13)=f(23)=3 and f=0 else.
    b) Since the function only takes on two values, we can determine it by its kernel. We know that ker(f) is normal in S5. The only normal subgroups of S5 are {(1)}, A5 and S5. However the function arising from ker(f)={(1)} is not a homomorphism (it isn’t injective but has trivial kernel or 1=f(123)!=f(12)+f(23)=0). ker(f)=A5 gives a type of sign homomorphism and ker(f)=S5 gives us the trivial homomorphism.

Question 17

17. Which of the following statements are true for all groups G of order 56?
    a) They contain a subgroup of order 2.
    b) They contain a subgroup of order 3.
    c) The number of different subgroups of order 8 contained in G is exactly 2.

Answer 17

17. a) Every element of a group has a unique inverse. Since the neutral element is its own inverse, there are 55 elements left. This implies that there is at least one element g in the group that is its own inverse. So {1,g} is a subgroup.
    b) Suppose H is a subgroup with |H|. By Lagrange, |G|=56=2^3*7 is divisible by |H|=3. Contradiction.
    c) No, consider following counterexamples:
    Z7xZ8: Since 7 is prime, all elements of Z7 have order 7, except 0, which has order 1 and is the only element whose order divides 8. So every element in Z7xZ8 with order 8 must be of the form (0,x). But the subgroup {0}xZ8 already has order 8. So this is the only subgroup with order 8.
    Z56: Z56 is cyclic and so all its eubgroups are cyclic too. If g has order 8 in Z56, then 56|8g iff 7|g. So g occurs is one of 7, 14, 21, 28, 35, 42, 49. But of these, we see that only 7, 21, 35 and 49 are of order 8.

Question 18

18. In the ring R[x], are the following subsets ideals of R? Prove your answer.
    a) {f(x) in R[x] : f(pi)=0}
    b) {f(x) in R[x] : f’(pi)=0}
    c) {f(x) in R[x] : f(i)=0}
    d) {f(x) in R[x] : f(0)=1}

Answer 18

18. a) yes, check, std. [equal to (x-pi), since all poly with root pi are div by (x-pi)]
    b) Not an ideal since for f(x)=(x-pi)^2+1 or g(x)=x^2-2pix, f’(pi)0=g’(pi) but multiplying p(x)=x does not stay in set for (pf)’(pi)=p’(pi)f(pi)+p(pi)f’(pi)=f(pi)=1!=0 and (pg)’(pi)=p’(pi)g(pi)+p(pi)g’(pi)=f(pi)=-pi^2!=0
    c) Yes, check, std. Furthermore, x^2+1 is the minimal polynomial, so the ideal is (x^2+1).
    d) No, for 1 the set but multiplying by x we get (x1)(0)=0, so x1 is not in set.

Question 19

19. Compute the conjugate of the group element a=(3524) in S5 by the element (23). How many elements of S5 are conjugate to a?

Answer 19

19.(23)(3524)(23)=(2534).
By Theorem A-4.7 [two elements of Sn are conjugates iff they have equal cycle structure (i.e. their complete factorizations (factorization into disjoint cycles) have the same number of r-cycles for each r in N)] all elements that are conjugate to a have the same cycle structe. So it is enough to count how many 4 cycles there are in S5. There are 5C4 choices of 4 elements of 5 and 4! ways of ordering these in a row. Since we are ordering them in a circle, we counted every ordering 4 times to many, so there are 4!*5C4/4=30.

Question 20

20. Let F2 be the field with two elements.
    a) Show that, for every degree n g/e 1, there exists an irreducible polynomial of degree n in F2[x].
    b) Let bar(F2) be the algebraic closure of F2. Show that bar(F2) is infinite.

Answer 20

20. a) Corollary to Galois: Let E/Fp be an extension field with p^n elements, which exists by Galois' thm (and proof, constructed field was splitting field of x^p^n-x in Fp[x]). Let alpha in E be primitive, then Fp(alpha)=E.
Let g(x)=irr(alpha,Fp) in Fp[x]. It is a irr. poly with alpha as a root [and exists by Thm A-3.87i]. If deg(g)=d, then [Fp[x]/(g):Fp]=d [by Prop A-3.84v] and Fp[x]/(g)~Fp(alpha)=E [by thm A-3.87i].
Also [E:Fp]=n [since if {x1,...,xm} basis, there are p^m lincombs]. Hence, n=d, and so g is an irr. poly of degree n.
b) A field is algebraically closed if every nonconstant f in K[x] has a root in K.
However, if bar(F2)={a1,...,an} were finite, then f(x):=1+(x-a1)*...*(x-an) is a nonconstant polynomial in bar(F2)[x] with no roots in bar(F2). Contradiction.

Question 21

21. Let A be a module over the ring C[x]. Since C s.s.o. C[x], A is a C-vector space.
    a) List all C[x]-modules A with dim_C(A)=1
    a) List all C[x]-modules A with dim_C(A)=2
    a) List all C[x]-modules A with dim_C(A)=3

Answer 21

21.

Question 22

22. What is the degree over Q of the splitting field over Q of the polynomial x^3-2?

Answer 22

22. https://math.stackexchange.com/questions/400660/determine-splitting-field-k-over-mathbbq-of-the-polynomial-x3-2

Let K be a splitting field, a1, a2, a3 three roots of p(X)=x^3-2. Then a1/a1, a1/a2, a1/a3 are three different third roots of unity in K. As there can be at most three third roots of unity, this means that we have found all. Moreover, we know that the third roots of unity form a group. Since any finite subgroup of the multiplicative group of any field is cyclic, so is the group of third roots of unity. So let w be a generator. Now, consider the subextension E=Q(w,a1). It contains a1,a1w,a1w^2, which are three different roots of p, so E is already a splitting field of p, so K=E. Now we can determine [K:Q]: w satisfies w^2+w+1=0 (wiki: Cyclotomic poly), a1 satisfies a1^3-2=0, and these are their minimal polynomials, so [K:Q] has to be divisible by 2 and 3, hence is 6. On the other hand, the splitting field has at most dimension 3!, so indeed we have [K:Q]=6.

https://math.stackexchange.com/questions/1326078/how-should-i-find-splitting-field-of-x3-2-over-mathbb-q?rq=1

Question 23

23. Is the field Q[1+i] s.s.o. C the splitting field in C over Q of some polynomial over Q? If so, give an example of such a polynomial f.

Answer 23

23. 1) Show: Q[i+1]=Q[i]=Q(i) of which f(x)=x^2+1 is a splitting field.
    2) It is the splitting field of the polynomial (x-(1+i))(x-(1-i))=x^2-2x+2, which is the minimal polynomial over 1+i over Q.

Question 24

24. Let R be a comm. ring with unit, and let R* s.s.o. R be the group of units. Consider elements x in R* s.t. x=x^-1.
    a) Show that if R=k is a field, then there are at most two such elements. When is there exactly one such element?
    b) What if R is a domain? Are there still only at most two elements x in R* with x=x^-1? If so, give a proof, if not a counterexample.
    c) What if we let R be a noncommutative ring with a unit? Are there still at most two such elements?

Answer 24

24. a) i) x=x^-1 iff x^2=1 iff 0=x^2-1=(x-1)(x+1). So x=+/- 1.
    a) ii) follows that exactly two sols iff +1=-1 iff char(k)=2.
    b)Suppose there is a 3rd element a: s.t. a=a^-1 but different so a != 1 so a-1 !=0 and a != -1 so a+1!=0.
    It then follows that (a+1)(a-1)=a^2-1=1-1=0. Which contradicts something (domain or assumptions on a).
    c) Observe the ring M_n(IR). There are at least 4 such elements: ((a,0),(0,b)), where a,b in {-1,1}.

Question 25

25. Let GL_2(F_2) act on F_2^2{(0,0)} in the usual way. Is the induced homomorphism an isomorphism?

Answer 25

25.”injectivity”: iff ker phi = {1}. Let A be in G - {1}. List the elements of GL_2(F_2). For each find a vector a in F_2^2 that s.t. A*a != a.
“surjectivity”: (pigeon hole principle) follows with injectivity and |G|=|S_3|.

Question 26

26. Let G be a group of order p^n (p prime) acting on a finite set X. Let X^G s.s.o. X be the set of fixed points. Is it true that |X|=|X^G| mod p?

Answer 26

26. Yes. Pf:
    The orbits of G in T form a partition (t1 ~ t2 iff Ex. g s.t. g*t1=t2). So let Y be a set of representatives, then X = U orb(x) (disjoint union over x in Y). Then |X|=Sum(orb(x)) (over x in Y). By the orbit-stabilizer theorem, |orb(x)| = |G|/|Stab(x)|=p^k for some integer k=

Question 27

27. Give an example (with proof) of a group G which is not abelian and not isomorphic to any of the groups Sn (for n>=Sn).

Answer 27

27. • A general linear group over an infinite field like Q or R is obviously not isomorphic to Sn.
      * Quaternion group Q8={+/-1,+/-i,+/-j,+/-k}. Define: i^2=j^2=k^2= -1, ij=k=-ji, jk=i=-kj, ki=j=-ik but then ji=-k
      * Dihedral group D4 (symmetries of a square)
      * Fundamental group of 2-rose
      * A4 (alternating group n=4), has cardinality 12 but 1!=1, 2!=2, 3!=6, 4!=24. So |Sn| != |A4|

Question 28

28. Calculate gcd(k^2+k+1,3k^2+4k+5) for k in Z

Answer 28

28. Use Euclidean algorithm (Lem A-218: (i) gcd(a,b)=gcd(a,r) if b=qa+r).
    gcd=3 if k~1mod3, gcd=1 else

Question 29

29. Let z be the real third root of 5. Is the field Q(i,sqrt(3),z) the splitting field in C over Q of some polynomial f in Q[x]? If so give an example of such a polynomial.

Answer 29

29. The splitting field of x^3-5 is Q(i,z).
    The splitting field of x^2-3 is Q(sqrt(3)).
    So since Q[x] UFD the splitting field of (x^3-5)(x^2-3) is Q(i,sqrt(3),z)……maybbe this is wrong…fabian says (x^2+1)(x^3-5)

Question 30

30. Is sqrt(2)-sqrt(3) algebraic over Q? If so, find the minimal polynomial over Q.

Answer 30

30. Yes, find poly by turning x=sqrt(2)-sqrt(3) into:
    f(x):=x^4-10x^2+1=0. Claim that f is minimal poly.
    Pf (Example A-3.89): Factor f in R[x]. As all roots and constant terms are in the form +/-sqrt(2)+/-sqrt(3). So if f is reducible in Q[x] then its factors must be quadratic. Let g|f then g(x):=(x-asqrt(2)-bsqrt(3))(x-csqrt(2)-dsqrt(3))
    =x^2-((a+c)sqrt(2)+(b+d)sqrt(3))x+2ac+3bd+(ad+bc)sqrt(6). So g is in Q[x] iff linear coefficient is rational iff a+c=0=b+d but this would imply that ad+bc non zero iff constant term not in Q. So g isn’t in Q[x], f is irreducible in Q[x]. f is obviously also monic and so f is the minimal poly.

Question 31

31. Let Fq be the field with q:=p^n elements where p is an odd prime. For how many elements alpha in Fq does there exist a solution in Fq of the quadratic equation x^2-alpha=0?

Answer 31

31. Claim: there are (q+1)/2 such equations.
    Pf1: alpha=0 has one solution, namely alpha=0.
    Define the group homomorphism phi:Fq–>Fq by phi(x):=x^2 (pf hom uses abelian). Since F field there are 2 sols to x^2=1 iff x=x^-1 by Q-24 and so ker(phi)={+/-1}. By the FIT Fq/ker(phi)~im(phi) and so (by “index formula”) |im(phi)|=|Fq|/2=(q-1)/2. Recalling that 0 is a solution we conclude.
    Pf2: If beta is a solution to x^2-alpha=0, then so is -beta. Since the poly then factors, there are no more solitions and Fq[x] is UFD for each alpha there are at most two solutions. There is one solution iff beta=-beta iff beta=0. But we can choose any element in Fq as beta, so but choosing beta or -beta yields the same equation, so there are (q+1)/2 solutions.

Question 32

32. Let R be an integral fomain with finitely many elements. Must R be a field? Answer with a proof.

Answer 32

32. Yes
    Pf1: ince |R|R, def by f(x):=ax. This is injective since ax=ay => x=y (Cancellation Law). Since R is finite, it is even a bijection (pigeonhole principle) and so a has an inverse.

Question 33

33. Let Fq be the field with q=2^n elements. For how many alpha in Fq does there exist a solution in Fq to the quadratic equation x^2-alpha=0?

Answer 33

33. Pf1: Define phi:Fq–>Fq by phi(x):=x^2. Since

Question 34

34. Let phi:R–>S be a homomorphism of comm rings.
    a) If I s.s.o R is an ideal, must phi(I) be an ideal?
    b) If J s.s.o. S is an ideal, must phi^-1(J) be an ideal?

Answer 34

34. a) No, needs to be surjective. Counterexample: phiZ–>Z(sqrt(2)) embedding. Then sqrt(2)=sqrt(2)*phi(1) not in phi(Z).
    b) Yes, axioms: i) o in I; ii) a,b in I => a+b in I; c) a in R, b in I => ab in R.

Question 35

35. Let R be a non trivial commutative ring whose only ideals are {0} and R. Prove R is a field.

Answer 35

35. Pf1: Let r != 0, then (r)=R by assumption and since 1 is in (r), r must have an inverse. So r is a unit.
    Pf2: As (0) is the only proper ideal, it is maximal and R~=R/(0).

Question 36

36. a) Consider p=x^3-5/2x+3/2x and q=2x^2-x-3 in Q[x]. Compute the Euclidean division of p by q.
    b) Fnd a single generator of the principal ideal (p,q)Q[x] s.s.o. Q[x].

Answer 36

36. a) Interpret polydiv/Euclidean algorithm:
    p/q=(0.5*x-1)+(2x-3)/(2x^2-x-3).
    b)Euclidean algorithm (Lem A-2.18: (i) If b=qa+r, then gcd(a,b)=gcdd(r,a)) gives gcd=2x-3 which is the generator.

Question 37

37. Let R be a UFD. Let a,b in R be coprime elements (that is, gcd(a,b) in R*) and c in R. Suppose that a | c and b | c. Prove that ab | c.

Answer 37

37. Ex 1. Sh. 6: By assumption, there exist r,s in R s.t. ar=c=bs. It suffices to show a | s (since then ab | sb=c). Since R is a UFD, there exist u,v,w,z in R, na,nb,ns,nr in N0 andd irreducible elements a(1),…,a(na), r(1),…,r(nr),b(1),…,b(nb) and s(1),…,s(ns) in R s.t.
    a=ua(1)…a(na), r= vr(1)…r(nr), b=wb(1)…b(nb), s=zs(1)…s(ns).
    Then ar=bs reads
    a(1)…a(na)(uvr(1))r(2)…r(nr)=b(1)…b(nb)(wzs(1))s(2)…s(ns)
    (domains are comm rings) and by uniqueness of factorization each ai is associated to some bi or some si. Suppose that, for some i, ai is associated to bj. Then ai | gcd(a,b) in R, a contradiction (a,b coprime but ai irreducible ie not unit!). Hence each ai divides one of the elements sj, and since in the factorizations of c the associated elements on the two sides can be taken in a bijective correspondence, we can choose for each i a distint ji s.t. s(ji)=lambda(i)a(i), so that
    lambda(1)…lambda(na)a=lambda(1)…lambda(na)a(1)…a(na)
    =s(j1)…s(jna) | s, so that a | s and ab | c as desired.

Question 38

38. Find the finite order (if finite) of the following elements:
    a) i, e^(isqrt(3)pi) and e^(2pii/17) in the group C.
    b) 1/sqrt(2)[[1,1];[-1,1]] and [[2,3];[1,4]]

c) 1, 2 and 3 in F17*.

Answer 38

38. a) Ex. 2.a) Sh 7: i^4=1 but not i^k for 0 a^2=3b^2 but 3 on lhs even times, odd times on rhs]
    b) First: We could multiply (but 8 multiplications is a lot) or see that there is an group isomorphism between C* and {[[a,-b],[b,a]] | a,b in R}. Since isomorphisms are order preserving and our matrix gets mapped to 1/sqrt(2)(1-i)=e^(i3pi/8) so it has order 8.
    Second: The determinant is 5 and since the determinants are multiplicative the order is infinite (i.e. detA^n=(detA)^n =5^n !=1 for any n.
    c) Ex. 2.c) Sh 7: i) Since 1 is the neutral element of F17, it has order 1 by definition. For the other two elements, we consider some of their powers modulo 17.
    ii) 2^2=4,2^3=8,2^4=16= -1, 2^8=(-1)^2. Notice that for k in {5,6,7}, we can say for sure that 2^k != 1, because else 2^(8-k)=2^8(2^k)^-1=1 => 2^k=1, which contradicts the lower powers. This implies that ord(2)=8.
    iii)@3^2=9, 3^3=27=10,3^4=30=13= -4 => 3^8=16= -1 => 3^16=1.
    @If k in {12,13,14,15} => 16-k in {1,2,3,4} => (3^k)^-1=3^(-k)=3^(16-k) != +/- 1 => 3^k != +/- 1.
    @If k in {5,6,7} then k+8 in {13,14,15} => 3^k= -3^(k+8)!=-1 (1st @)
    @ 3^9=33^8=-3=14!=1,
    3^10=33^9=-9=8!=1
    3^11=3*3^10=24=7!=1

Question 39

39. Ak:={ A in Gl2(Fp) | (detA)^k=1}
    a) Show Ak subgroup of G.
    b) Is Ak a normal subgroup of Gl2(Fp)?
    c) What is the index of A{p-1} in GL2(Fp)?

Answer 39

39. Prop: H subgroup iff H non-empty and x(y^1) in H whenever x,y in H.
    a) Ak non empty since (det id )^k=1. (det g(h^-1))^k=1 by multiplicativity of determinant.
    b) g in GL2(Fp), A in Ak then (det(gAg^-1))^k=detA^k=1
    a’,b’) Let phi:GL2(Fp)–>Fp* by phi(B)=det(B)^k. This is a group hom. phi(BC)=phi(BC)^k=(detBdetC)^k=phi(B)phi(C) and its kernel is Ak.
    c) Let B be in GL2(Fp) then det(B)!=0. By Fermat’s Little Theorem, det(B)^{p-1}=1 and so A_{p-1}=GL2(Fp) and its index is 1.

Question 40

40. Let G be a group of order 3^17, and let H be a subgroup of index 3. Must H be normal?

Answer 40

40. p=3: Let G act on G/H by x*(gH)=xgH. This induces a hom phi:G–>S3. ker(phi) s.s.o. H since only elements of H fix its cosets [Pf]. Let K:=ker(phi). By FIT G/K~im(phi) s.s.o. S3 and so |G/K| | 3! and |G/K| has no divisors larger than p (so 2 or 3). Also |G/K| | |G|=3^17, so |G/K|=3.

{G finite, p smallest prime divisor of |G|. H subgroup of indey p. Then H is normal:
Let G act on G/H by x*(gH)=xgH. This induces a hom phi:G–>Sp. ker(phi) s.s.o. H since only elements of H fix its cosets [Pf]. Let K:=ker(phi). By FIT G/K~im(phi) s.s.o. S3 and so |G/K| | p! and |G/K| has no divisors larger than p (|G/K| = for some d | p and d=<p> [H:K]=1 => H=K=ker(phi) and kernels are normal.</p>

Question 41

41. i) Show that the center of S17 is trivial.
    ii) What is [S17,S17]? Recall: for a group G, the commutator [G,G] is defined as the subgroup of G generated by {ab(a^-1)(b^-1) : a,b in G}.

Answer 41

41. i) * Z(S17) is normal: z in Z, g in G s.t. zg=gz by def.
    * Z(S17) doesn’t contain A17: ( 1 2 3 )=(2 3)(1 3) doesn’t commute with ( 1 2) but are in A17.
    * Only {e},A17,S17 normal in S17: Supp. H normal in S17 then by SIT, A17 normal in S17 and simple => H n A17 normal in A17 so H in {e,A17,S17} [Pf?]
    * Conclusion: Z(S17)={e}
    ii) *[G:G] is normal: u in [G,G], g in G then gug^-1=gug^-1u^-1u=[g,u]u in [G,G] since [g,u],u in [G,G] (group)(S8E6)
    * [S17,S17] in A17: Every element in [S17,S17] has even sign, since we can decompose any element of S17 into transpositions and the inverse has the same amount of transpositions.
    * [S17,S17] non-trivial: (1 2 3)=(1 2)(1 3)(1 2)(1 3) in [S17,S17]
    * As a non-trivial [S17,S17] subgroup of A17 it must be A17 since A17 is simple.

Question 42

42. Let Sn act on {1,…,n} via the standard action. Define an action of Sn on {1,…,n}^2 via g*(i,j):=(g(i),g(j)). Find and describe all orbits of this action.

Answer 42

42. If i=j then g(i)=g(j), so its orbit is {(i,i)| 1=<i></i>

Question 43

43. Is there a transitive action of S5 on a set with 10 elements? Is there a transitive action of S5 on a set with 20 elements?

Answer 43

43. Every group acts transitively on its quotient groups which has the order of the index of the respective subgroup. So we need only find subgroups of Sn with index 10 and 20 in other words subgroups of orders 5!/10=0.5*4! and 5!/20=3!. So S5 acts transitively on the left cosets of A4 and S3.

Question 44

44. Let a in Z{-1,0,1} be a square-free integer, that is, an integer which is not divisible by any perfect square except 1. Prove that for each n in N, the polynomial x^n-a in Q[x] is irreducible.

Answer 44

44. Solution follows from Eisenstein. There must be some prime p that divides a since it isnt unit or 0. Since p doesnt divide 1=an and p^2 doesnt divide a (for a is square-free). Now f is irreducible.
    Eisenstein: p prime. Let pi_p:Z[x]–>Fp[x] be the canonical projection (surjective ring hom). f in Z[x] s.t. p divides all coefficients but the leading one and p^2 does not divide the constant term. Then f is irreducible.
    Pf S11E2.b): Suppose not, g0,h0 non-const. in Q[x] s.t. f=g0h0 then by Gauss’ Lemma (Thm A-3.65: f in Z[x], f=GH in Q[x] s.t. degG,degH R[x] UFD)
    [Lem: Same asumptions implies that pi_p(f) is a monomial of degree n in Fp[x].
    Pf S11E2.a): Rewrite ak’=p*ak for k in {1,…,n-1}. Then one applies pi_p and sees this.]

Question 45

45. Let K be a field of characteristic !=2 and L/K a field extension of degree 2. Show that there exists alpha in L s.t. alpha^2 in K and L=K(alpha).

Answer 45

45. Let alpha be in K\L. Since L is a 2-dim VS over K and {1,beta} lin indep. it is a basis. beta^2 is in L since L is a field and there is a lin. combination beta^2= -bbeta-c as it is a VS. Hence there is a poly f(x)=x^2+bx+c s.t. b, c in K and beta is a root. So f is irreducible over K.
    Now: beta^2+bbeta+c=0 iff
    beta^2+2betaa/2+(a/2)^2-(a/2)^2+b=0 iff
    (beta+a/2)^2-(a/2)^2+b=0 iff
    alpha^2=(a/2)^2-b in K and {1,alpha} is a basis of L too.
    <1,alpha>=K(alpha).
    https://math.stackexchange.com/questions/3212079/doubts-regarding-quadratic-field-can-be-obtained-by-adjoining-square-root

Question 46

46. Let L=K(alpha)/K be a field extension such that [L:K] is odd. Prove that L=K(alpha^2).

Answer 46

46. Suppose not, i.e. K(alpha) not in K(alpha^2), then alpha isn’t in K(alpha^2). Then K(alpha) isomorphic to K(alpha^2)[X]/(X^2-alpha^2) (by Thm. A-3.87, pf with eval map at alpha) is a extension of 2nd degree (by Prop. A-3.89). But now [L:K]=[L:K(alpha^2)][K(alpha^2):K]=2[K(alpha^2):K] is even.

(Alt: [L:K]=[K(alpha):K(alpha^2)]*[K(alpha^2):K] is odd and so also [K(alpha):K(alpha^2)] is odd. Furthermore alpha is a root of f(x)=x^2-alpha^2 with f in K(alpha^2)[x]. Since the splitting field of f(x) is K(alpha) we also get
1=…)

Question 47

47. Is the algebraic closure of Q countable? Are there real numbers which are not algebraic over Q?

Answer 47

47. The algebraic closure is the set of algebraic numbers. It is countable because there are only finitely many roots for each polynomial. There are countably many polynomials because there are countably many degrees and for each degree there are finitely many coefficients and countable many ways to choose them. Since the reals are uncountable, the existence follows from the fact that the difference is nonempty. We also know that pi and e are transcendental.

Question 48

48. Let M be an nxn-matrix with complex coefficients. Let J={f in C[x] | f(M)=0} be the set of polynomials which vanish when evaluated at M.
    a) Is J an ideal of C[x]?
    b) Let g be a non zero poly in J of minimal degree. Can the degree of g be equal to n+1?

Answer 48

48. a) Yes, 0 in J; f,g in J => f+g in J; h in C[x], f in J => hf in J.
    b) Remember the Cayley-Hamilton theorem from linear algebra that says that if p(x):=char_M(x) then p(M) (i.e. p is in J). The characteristic polynomial is however always of degree n and not n+1. So no.

Question 49

49. Let R=Z[x,y]. Can you find a tower of ideals I0 in I1 in I2 in I3 in R, where each of the Ii is a prime idea?

Answer 49

49. Recal I is prime iff R/I is domain.
    {0} in (x) in (x,2) in (x,y,2) in R.
    - {0} trivial ideal is prime since Z[x,y] is a domain (since Z is a domain)
    - (x) is prime since Z[x,y]/(x)=Z[y] domain (Z domain)
    - (x,2) is prime since Z[x,y]/(x,2)=F2[y] domain (F2 domain)
    - (x,y,2) is prime since Z[x,y]/(x,y,2)=F2 domain (F2 domain)
    Last three by FIT.

Question 50

50. Consider the cubic polynomial f(x)=(x^2+17x+1)(x-17) in Z[x]. For each prime p, let f^bar(x) i Fp[x] be the polynomial defined by taking the coefficients of f mod p. For which primes does f^bar(x) have repeated roots in the algebraic closure of Fp?

Answer 50

50. (.^bar:Z[x]–>Fp[x] is a hom (phi(0)=0; phi(f+g)=phi(f)+phi(g); phi(fg)=phi(f)phi(g))
    f has repeated roots if (x-17)|(x^2+17x+1) or if (x^2+17x+1)=(x-alpha)^2 for some alpha.
    This is also true in Fp[x] since .^bar is a hom.
    In the first case [17]_p is the root of x^2+17x+1 and so it must be true that 579=17^2+1717+1=0 mod p and 579=3193=mp iff p is 3 or 193.
    In the second case, we know that x^2+17x+1=x^2+x+1 is irreducible in F2 and so p !=2.
    So we can find solutions to x^2+17x+1 with repeated roots by using the quadratic formula with the discriminant =0. And 0=D=17^2-4=285=557=3519 and so p is in {3,5,19}.
    So in conclusion p is in {3,5,19,193}.

Question 51

51. Let K be a splitting field of x^7-5 over Q. Prove that K contains a primitive 7th root of unity. Determine the degree of K/Q.

Answer 51

51. It holds that K=Q(sqrt[7]{5},sqrt[7]{5}w,…,sqrt[7]{5}w^6), where w:=exp(2pi*i/7), since x^7-5=(x-sqrt[7]{5})…(x-sqrt[7]{5}w^6). We have w=sqrt[7]{5}w/sqrt[7]{5}. So K=Q(sqrt[7]{5},w).
    To determine the the degree [K:Q] consider the subextension Q(sqrt[7]{5}). As this is a real extension, it does not contain any of the 7th roots of unity (except for 1 iteself). We have [K:Q]=[K:Q(sqrt[7]{5})][sqrt[7]{5}:Q]. We now determine the degrees of the subextensions:
    +Q(sqrt[7]{5})~Q[X]/(X^7-5) (by Prop 3.84, Thm 3.87) is a degree 7 extension over Q(x^7-5) is irreducible by Eisenstein with p=5)
    +K~Q(sqrt[7]{5})[X]/(Phi_7(X)) is a degree 6 extension, as deg(Phi_7)=phi(7)=6, where Phi_n is the nth cyclotomic polynomial and phi(n) is the Euler totient function.
    To show that the last two are isomorphic:
    Argue that the intermediate field extension Q(w)/Q is of degree 6 (minimal polynomial of w had degree 6) (Phi_7(X) is irreducible over Q[X]) and therefore [K:Q] is divisible by 6 and thus at least 42. On the other hand, the degree of the minimal polynomial of w “is certainly not larger” over the field Q(sqrt[7]{5}) than over Q, so we get at most 42.

Question 52

52. Let K=Q(sqrt5,zeta), where zeta is a primitive 15th root of unity. (Me: def of separable, Galois Group & Extension)
    a) Prove that K is Galois over Q and find the order of the Galois group G.
    b) Show that Gal(K/Q(zeta)) is a normal subgroup of G.

Answer 52

52. Def. An irreducible poly p(x) is separable if it has no repeated roots. An arbitrary poly f(x) is separable if each of its irreducible factors has no repeated roots; otherwise, it is inseparable.
    Def. A Galois Group of an extension field E/k, denoted by Gal(E/k), is the set of all those automoprhisms of E that fix k. If f(x) in k[x] and E=k(z1,…,zn) is a splitting field of f over k, then the Galois group of f over k is defined to be Gal(E/k).
    Def. If E/k is a finite extension field with Galois G=Gal(E/k), then E/k is called a Galois extension if one of the following equivalent (Thm A-5.42) statements hold:
    i) E is a splitting field of some separable polynomial f(x) in k[x].
    ii) k=E^G (:={a in E | sig(a)=a for all sig in G}; fixed field of G)
    iii) If a monic irreducible p(x) in k[x] has a root in E, then it is separable and splits in E[x].
    a) Notice it is the splitting field of x^15-27=(x^5-3)(x^10+3x^5+9). Clearly, this has distinct roots, thus is separable and hence K is Galois. Note that [K:Q]=[K:Q(sqrt5)]:[Q(sqrt5):Q].
    +Q(sqrt5)~Q[x]/(x^5-3) is a degree 5 extension by Eisenstein (p=3) x^5-3 is irreducible and by Prop 3.84, Thm 3.87.
    +K~Q(sqrt5)/Phi_15(x) is an extension 8 extension since deg(Phi_15(x))=phi(15)=phi(5)phi(3)=4*2=8.
    Hence: [K:Q]=40 (Me: and I think [K:Q]=[Gal(K/Q):Gal(K/K)]=|Gal(K/Q)| from Galois’ Thm A-5.51 (iv) )
    b) By FTGT (A-5.51 (v)), it is normal iff Q(zeta)/Q is Galois.
    I) another way to see Q(zeta)/Q is Galois is by noting that Q(zeta) is the splitting field of x^15-1 over Q (Def GF (i) ).
    II) This is the case, as it is a degree 8 extension and there are 8 possible choices for sig(zeta), where sig is an automorphism of the extension, since:
    sig(zeta) must satisfy Phi_15(x) (why???) i.e. be another primitive root of 15th root. Further, as zeta generates the group of 15th roots of unity, the choice for sig(zeta) fully determines sig. There are 8 possinle choices for sig(zeta).

Question 53

53. Let G be a finite group. Does there always exist a finite dimensional extension of fields E/F which is Galois with Galois group G?

Answer 53

53. Yes. Let n=|G| and consider the extension C(x1,…,xn)/C(e1,…,en), which is Galois with Galois group Sn (Mentioned as result of FTGT in lecture 9). By Q13 (All groups of order n are symmetric to a subgroup of Sn) G is isomorphic to a subgroup G’ of Sn.
    Then, C(x1,…,xn)/C(x1,…,xn)^G (??????) is a Galois extension with Galois group G (F^G represents the fixed subfield of F).

Question 54

54. Let Q(zeta)/Q be the extension of the rational numbers determined by zeta=exp(2pi*i/29) in C. Let alpha in Q(zeta) be a number which is constructible over Q. What are the possible values of [Q(alpha):Q]?

Answer 54

54.

Question 55

55. Let E be the splitting field of the polynomial f(x):=x^4-2x^2-3 over Q.
    a) What is the degree of E/Q?
    b) Find all proper subextensions of E/Q

Answer 55

55. a) Substituting x^2=:y, leads to a quadratic formula and the 4 roots +/-sqrt(3), +/-i. The splitting field is therefore E=Q(i,sqrt(3)). Since sqrt(3) not in Q(i), we conclude that x^2-3 is irreducible in Q(i) and therefore:
    [E:Q]=[Q(sqrt(3),i):Q]=[Q(sqrt(sqrt(3),i):Q(i)][Q(i):Q]=22=4.
    b) E is the splitting field of a separable poly, since the irreducible factors of f have no repeated roots, thus E/Q is Galois. Gal(E/Q) has order 4, i.e. Gal(E/Q)=Z4 or Z2xZ2. We conclude that its Z2xZ2 because it has no elements of order 4.
    The proper subextensions of E/Q can be found using the Galois correspondence of non trivial subgroups of Gal(E/Q) which are span(g) for g in {(0,1),(1,0),(1,1)}. With the correspondence:
    sigma(k,j):=((-1)^ki,(-1)^jsqrt(3)), we get that:
    FixedField( span(1,0) )=Q(sqrt(3))
    FixedField( span(0,1) ) = Q(i)
    FixedField ( span(1,1) ) = Q(isqrt(3)).

Question 56

56. Solve the equation x^3-6x-9=0 using Cardano’s formula.

Answer 56

56. Thm (Cardano’s formula). The roots of f(x)=x^3+qx+r are:
    x1:=g+h, x2:=wg+w^2h and x3:=w^2g+wh,
    where g^3=0.5(-r+sqrt(R)), h= -q/3g, R=r^2+4q^3/27 and w= -0.5+i*sqrt(3)/2 is a primitive cube root of unity.

R=49, g=2, h=1, x1=3, x2,3=-3/2 +/- i*sqrt(3)/2.

“possible topics for further discussion: Solvable groups, Abel theorem”

Question 57

57. Let F2 be the field with 2 elements.
    a) Factor the polynomial f(x)=x^4+x^2+x+1 into irreducible factors in the ring F2[x].
    b) Let E be the splitting field of f over F2. How many elements foes E have?

Answer 57

57. a) We see that f(1)=0 but f(0)=0. So by polynomial division f(x)=(x+1)(x^3+x^2+1). The second factor is a degree 3 polynomial, so if it were to split further, it would have to have a root. However, neither 0 nor 1 are roots of x^3+x^2+1, and so it must be irreducible.
    b) An exercise (S13Ex2b) tells us that if Fq with q=p^n and f in Fq[x] is irr of positive degree d and x a root in the closure of Fq (bar(Fq)). Then x,x^q,…,x^q^{d-1} are roots too.
    So let alpha be a root of x^3+x^2+1 in the closure. Then F2(alpha)~F2[x]/(x^3+x^2+1) is a degree 3 extension. By the exercise alpha^2 and alpha^4 are also roots of f and x^3+x^2+1=(x-alpha)(x-alpha^2)(x-alpha^4) splits.
    So, |E|=|F2|^3=2^3=8 [2 options for coefficients of 3 basis vectors].

Question 58

58. Is the regular 272-gon constructible by straightedge and compass?

Answer 58

58. Yes.
    Gauss-Wantzel: n-gon constructible iff n can be written as a product of a power of 2 and a (possibly empty) product of Fermat primes (i.e. p=2^2^k+1 for some k).
    272=2^4*17 and 17 is a Fermat prime because 2^2^2+1=17.

Question 59

59. Let C(e1,e2,e3) s.s.o. C(x1,x2,x3) be the subfield of symmetric functions. What is the degree of the extension of C(e1,e2,e3) generated by the element z=x1+2x2+3x3?

Answer 59

59. Using the Fundamental Theorem of Galois Theory we need to find out which subgroup of the Galois group of the extension C(x1,x2,x3)/C(e1,e2,e3) (which is isomorphic to S3 by proof of Abel-Ruffini), which fixes C(e1,e2,e3)(z). Only the identity fixes z, so the subgroup which fixes the base field is the trivial group, so the extension is of degree 3!=6.

Question 60

60. Let C(e1,e2,e3) s.s.o. C(x1,x2,x3) be the subfield of symmetric functions. What is the degree of the extension of C(e1,e2,e3) generated by the element x1^2x2+x2^2x3+x3^2*x1?

Answer 60

60. We know that the extension C(x1,x2,x3) over C(e1,e2,e3) has Galois Group S3 (from proof of Abel-Ruffini theorem), so by Galois-Correspondence, the extension generated by x1^2x2+x2^2x3+x3^2x1 sorresponds to a subgroup H of S3, which fixes the extension. Now notice that only all 3-cycles and the identity fix x1^2x2+x2^2x3+x3^2x1, therefore H is generated by 3 cycles, so H=A3 and the degree of the extension is 2.
    [Since A3 is normal, this is even Galois]

Question 61

61. Let C(e1,e2,e3) s.s.o. C(x1,x2,x3) be the subfield of the symmetric functions. What is the minimal polynomial over C(e1,e2,e3) of the element x1+x2 in C(x1,x2,x3)?

Answer 61

61. f(z)=(z-(x1+x2))(z-(x2+x3))(z-(x1+x3))
    f(z)=z^3-2e1z^2+(e1^2+e2)z-e1*e2+e3.
    Now we need to argue, why this is irreducible. We will do this in an indirect way by showing the degree of the minimal polynomial must be 3. By considering all permutations of S3 we see that {id,(x1 x2)} fix x1+x2. Its index is 3, so the minimal polynomials of x1+x2 must have degree 3, which was to be shown…..

Question 62

62. Factorize x^8-1 and x^9-1 into irreducible factors in the ring Q[x].

Answer 62

62. x^8-1=(x^4+1)(x^2+1)(x+1)(x-1)
    The first factor is irreducible over Q because all its roots are complex and they cant be grouped to make two rational quadratic polynomials.
    x^9-1=(x^3)^3-1=(x^3-1)(x^6+x^3+1)=(x-1)(x^2+x+1)(x^6+x^3+1).
    The second factor has a negative discriminant and so has only complex roots.
    The last factor is irreducible, since: it has roots in Q iff:
    (x+1)^6+(x+1)^3+1=x^6+6x^5+15x^4+21x^3+18x^2+9x+3 has roots in Q, which it doesn’t by Eisenstein with p=3.
    Alternatively: We know that x^n-1=Prod(Phi_d(x) : d|n). Moreover, all cyclotomic polys are irreducible over Q.

Question 63

63. a) Does there exist an irreducible polynomial f(x) in Q[x] of degree 4 with Galois group over Q isomorphic to S3xS3?
    b) Does there exist an irreducible polynomial g(x) in Q[x] of degree 5 with Galois group over Q isomorphic to S3xS3?

Answer 63

63.

Question 64

64. a) What is the automorphism group of the field Q?

b) What is the automorphism group of the field R?

Answer 64

64. a) Let f in Aut(Q). Then f(0)=0 and f(1)=1 since f hom. By induction f(n)=n for all N. Not since 0=f(1)+f(-1)=1+f(-1) and uniqueness of inverses, it follows that f(-1)=-1 and so f(-n)=-n for all n in N.
    Now f(p/q)=f(p)f(1/q)=f(p)1/f(q)=p/q, so f fixes Q and Aut(Q) is trivial.
    b) Let x be real and positive. Then sqrt(x)^2=x and so
    f(sqrt(x)^2)=f(sqrt(x))^2 is greater than zero.
    So for x greater than y,
    x-y is greater than 0 and
    f(x-y) is greater than zero too.
    So f(x) is greater than f(y).
    So f is monotonous. Let c be irrational.
    Let A={x str smaller than c} and B={x str larger than c}.
    Then for all a in A: a=f(a) smaller than f(c).
    for all b in B: b=f(b) larger than f(c).
    So f(c) lies in neither set and is therefore c.
    So Aut(R) is trivial too.

Question 65

65. Let F243 be the field with 243 elements.
    a) Let tao:F243–F243 be defined by tao(x)x^80. Is tao an automorphism of F243? If so, calculate the number of elements of the fixed field of tao.

Answer 65

65. a) No, since 243=3^5, the characteristic is 3 () and 2^3=(3+1)2=2, then if tao is a hom then tao(1)=1 but tao(2)=2^80=2^81/2=(((2^3)^3)^3)^3/2=1. So tao is not bijective.
    b)
    (*) Arg for char: https://math.stackexchange.com/questions/130064/determining-the-characteristic-of-a-field.

Question 66

66.

Answer 66

66

Question 67

67. Consider the real number alpha=sqrt(17)+2/sqrt(17). Is alpha algebraic over Q? If so, find the minimal polynomial of alpha over Q.

Answer 67

67. Yes, its minimal poly is x^2-alpha^2.
    alpha=19/sqrt(17)=>alpha^2 is in Q, so x^2-alpha^2 is in Q[x], however it is irreducible in Q since it factors in R[x] into (x-alpha)(x+alpha). The only lower degree poly with alpha as a root is x-alpha which is not in Q, so x^2-alpha^2 is minimal.

Question 68

68

Answer 68

68

Question 69

69

Answer 69

69

Question 70

70

Answer 70

70

Question 71

71

Answer 71

71

Question 72

72

Answer 72

72

Question 73

73

Answer 73

73

Question 74

74

Answer 74

74

Question 75

75

Answer 75

75

Question 76

Define minimal polynomial.

Answer 76

If K/k is an extension field and alpha in K is algebraic over k, then the unique monic irreducible polynomial p(x) in k[x] having alpha as a root is called the minimal polynomial of a over k; it is denoted by irr(alpha,k)=p(x).

Question 77

State the isomorphism theorems.

Answer 77

2nd (groups): 164

Question 78

Sn contains only {1}, An and Sn as normal subgroups.

Answer 78

See pdf on this.