Suggested solutions to selected exercises in sections 26, 27 and 36
1.
Section 26
Exercise 20:
Let R be a commutative ring with unity of prime characteristic p. Show that the map φ p : R → R given by φ p (a) = a p is a homomorphism. Solution:
We have to show the following: (1) φ p (a + b) = φ p(a) + φ p (b) for all a, b ∈ R. ab) = φ p(a)φ p (b) for all a, b ∈ R. (2) φ p (ab) Ad (1): p
p p p φ (a+b) = (a+b) = a b = a+ b = a +b = φ (a)+φ )+φ (b) i 0 p ≡ 0 mod p for 0 < i < p.p. (Check that!) because p
p
p−i i
p
p
p
p
p
i=0
p i
ab) = (ab) ab) p = a p b p = φ p (a)φ p (b) where we used that R is Ad (2): φ p (ab) commutative. Exercise 30:
An element a of a ring R is called nilpotent if an = 0 for some n ∈ N. Show that the collection of all nilpotent elements in a commutative ring R is an ideal, called the nilradical of R of R. Solution:
Let N = {a ∈ R|a is nilpotent}. We have to show two things: +). (1) (N, +) is a subgroup of ( of (R, +). (2) ra ∈ N for all a ∈ N , r ∈ R (Note that since R is commutative we do not have to check that ar ∈ N as well). Ad (1): (i) N is closed under addition: addition: Let a and b be nilpotent elements of R, then there exist n, m ∈ N such that an = 0, bm = 0. Set l = n + m − 1. Then l ∈ N and l
n−1
l
l l l (a + b) = ab = ab + ab l
i l−i
i=0
i
i l−i
i=0
i
i l−i
i=n
i
=0
where the first sum is zero because of l of l − i ≥ m (so bl−i = 0) and the second sum is zero because of i ≥ n (so ai = 0 in this case). (ii) It is obvious that the identityelement 0 is in N . 1
p
2
(iii) Let a be in N . Then the additive inverse −a is also in N since if an = 0, then (−a)n = −an = 0. Ad (2): Let a ∈ N , r ∈ R and let n ∈ N such that an = 0. Then (ra)n = rnan = r n 0 = 0 (Here we used that R is commutative). 2.
Section 27
Exercise 3:
Find all prime ideals and all maximal ideals of Z2 × Z2 . Solution:
The only additive subgroups of Z2 × Z2 are {0}, Z2 × Z2 , < (1, 0) >= {(0, 0), (1, 0)}, < (0, 1) >= {(0, 0), (0, 1)} and < (1, 1) >= {(0, 0), (1, 1)}. Among these all but the last one are ideals (Check that!). By Theorems 27.9 and 27.15 we are looking for ideals N such that Z2 × Z2 /N is an integral domain/a field. By definition Z2 × Z2 is neither a prime ideal nor a maximal ideal and we have that Z2 × Z2 /{0} ∼ = Z2 × Z2 , so {0} is neither a prime ideal nor a maximal ideal. (If you are not sure, try to figure out why Z2 × Z2 is not an integral domain.) Convince yourself that Z2 × Z2 / < (1, 0) >∼ = Z2 and Z2 × Z2 / < (0, 1) >∼ = Z2 , so (since Z2 is an integral domain and a field) < (1, 0) > and < (0, 1) > are prime ideals and maximal ideals. Exercise 6:
Find all c ∈ Z3 such that
Z3 [x]/
< x3 + x2 + c > is a field.
Solution:
By Theorem 27.9 we have to find c ∈ Z3 such that < x3 + x2 + c > is a maximal ideal. By Theorem 27.25 we have to find c ∈ Z3 such that x3 + x2 + c is irreducible over Z3 . Since in this case we can choose c only among three elements, we can check them one by one: (1) c = 0: x3 + x2 = x(x2 + x), so the polynomial is reducible over Z3 and Z3 [x]/ < x3 + x2 > is not a field. (2) c = 1: We observe that 1 is a zero of the polynomial x3 + x2 + 1 over Z3 , so we can write x3 + x2 + 1 = (x − 1)q (x) for some q (x) ∈ Z3 [x] which means that x3 + x2 + 1 is reducible over Z3 and 3 2 Z3 [x]/ < x + x + 1 > is not a field. (3) c = 2: You should check that x3 + x2 + 2 has no zeros in Z3 . This means that this polynomial is irreducible over Z3 since it is of degree 3. (If a polynomial of degree 3 is reducible, then at least one factor must have degree 1, which yields a zero of the polynomial). So Z3 [x]/ < x3 + x2 + 2 > is a field.
3
3.
Section 36
Exercise 13:
Show that every group of order 45 has a normal subgroup of order 9. Solution:
Write 45 = 32 5. By the first Sylow theorem we know that a group of order 45 has at least one subgroup of order 9. Note that a subgroup of order 9 in a group of order 45 is a Sylow 3-subgroup. By the third Sylow Theorem we see that the number of Sylow 3subgroups divides 45 and is congruent to 1 modulo 3. The only possibility here is that this number is equal to 1. So there is only one Sylow 3-subgroup and by the second Sylow Theorem every conjugate of this subgroup over the whole group is the subgroup itself, which means that the subgroup is normal in the whole group. Exercise 14:
Prove the following: Let G be a finite group. Then G is a p-group if and only if |G| is a power of p. Solution:
Recall the definition of p-group ( p is a prime number): A group G is a p-group if every element in G has order a power of p. First we observe that if G is the trivial group, that is G = {e}, then the statement is trivially true. So in the following we assume that |G| ≥ 2. Here we have to show an equivalence. First we show the "if part", that is we have to show that if |G| is a power of p, then G is a p-group: We pick an (arbitrary) element a ∈ G. By Lagrange’s Theorem we know that the order of a has to divide the order of G which is a power of p. So the order of a has to be a power of p, that is G is a p-group. Now we prove the "only if" part, that is we have to show that if G is a p-group, then |G| is a power of p. Equivalently, we can show that if |G| is not a power of p, then G is not a p-group: = p divides |G|. Then If |G| is not a power of p, then some prime q by Cauchy’s Theorem G has an element of order q . Thus G is not a p-group. Exercise 19:
Show that there are no simple groups of order pr m, where p is a prime, r is a positive integer and m < p. Solution:
4
We have to distinguish between two cases: (1) m = 1: Let G be a group of order pr . Then by the first Sylow Theorem G has a normal subgroup of order pr−1 , that is G has a proper nontrivial normal subgroup, hence G is not simple. (2) m > 1: Let G be a group of order pr m. By the first Sylow Theorem there exists at least one subgroup of G of order pr , that is a Sylow p-subgroup (since m < p). By the third Sylow Theorem, the number of Sylow p-subgroups divides pr m and is congruent to 1 modulo p. Now the only divisors of pr m are 1, p , . . . , pr , m , p m , . . . , p r m and of these only 1 is congruent to 1 modulo p. So there exists only one Sylow p-subgroup and by the second Sylow Theorem every conjugate of this subgroup over G is the subgroup itself, that is it is normal in G. Therefore G has a proper nontrivial normal subgroup, so it is not simple. Exercise 22:
Let G be a finite group and let P be a normal p-subgroup of G. Show that P is contained in every Sylow p-subgroup of G. Solution:
First we show that if P is contained in one Sylow p-subgroup of G, then P is contained in every Sylow p-subgroup of G: Let S 1 , . . . , Sn be all Sylow p-subgroups of G and let P ⊆ S i for one i ∈ {1, . . . , n}. Now let j ∈ {1, . . . , n}, j = i. Then by the second Sylow Theorem, S i and S j are conjugate over G, that is there exists a g ∈ G such that S j = gS i g −1 . Moreover, P is normal in G, so gP g −1 = P . We conclude that P = gP g −1 ⊆ gS i g −1 = S j . That is (since S j was arbitrary) P is contained in every Sylow p-subgroup of G. It remains thus to show that P is contained in (at least) one Sylow p-subgroup of G: Write |G| = pr m, r an integer, p does not divide m, and |P | = pl , l an integer, l ≤ r. If l = r, then P itself is a Sylow p-subgroup, so we are done. Assume therefore that l < r. By the first Sylow Theorem (part 2) P is contained in a subgroup Q1 of G of order pl+1, and Q1 is contained in a subgroup Q2 of G of order pl+2 and so on. So we obtain a chain of subgroups P ≤ Q1 ≤ Q2 ≤ Q3 ≤ · · · ≤ Qr−l , where all the Qi are subgroups of G and |Qi | = pl+i . That is, Qr−l is a Sylow p-subgroup of G and P is contained in Qr−l .
