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Math 541 Problem Set 8
4.5.8. The order or S 5 is 5 · 4 · 3 · 2 · 1 = 2^3 · 3 · 5, so a Sylow 2-subgroup will have order 2^3 = 8. The dihedral group D 8 has order 8 and acts naturally on the 4 corners of the square, so it can be made to act on 5 elements by leaving the fifth alone. The only group element that does nothing in this action is the identity, so the map D 8 → S 5 given by r 7 → (1 2 3 4), s 7 → (2 4) is injective. Thus 〈(1 2 3 4), (2 4)〉 is a subgroup of S 5 of order 8. If we conjugate this by (1 2), we have (1 2)(1 2 3 4)(1 2) = (1 3 4 2) and (1 2)(2 4)(1 2) = (1 4), so 〈(1 3 4 2), (1 4)〉 is another Sylow 2-subgroup.
7.1.5. (a) If a/b and c/d have odd denominators then so too does a/b − c/d = (ad − bc)/bd, and this will still be true after we write it in lowest terms; thus the set is closed under subtraction. Similarly, it is closed under multiplication since a/b × c/d = ac/bd. It is not empty since 1/3 is an element. Thus it is a subring of Q. (b) This is not closed under addition, since 1/6 + 1/6 = 1/3. (c) This is does not contain additive inverses, hence is not a group under addition. (d) This is not closed under addition, since 1/4 + 1/4 = 1/2. (e) This is not closed under addition, since 1 + 1 = 2. (f) If a/b is written in lowest terms and a is even then b is odd. If a/b and c/d have even numerators and odd denominators then so too does a/b − c/d = (ad − bc)/bd, and this will still be true after we write it in lowest terms, so the set is closed under subtraction. Similarly, it is closed under multiplication since a/b × c/d = ac/bd. It is not empty since 2/3 is an element. Thus it is a subring of Q.
7.1.6. Throughout, let S denote the subset.
(a) Let f, g ∈ S, so f (q) = g(q) = 0 for all q ∈ Q ∩ [0, 1]. S is closed under subtraction since (f − g)(q) = f (q) − g(q) = 0 − 0 = 0, so f − g ∈ S. S is closed under multiplication since (f g)(q) = f (q)g(q) = 0 · 0 = 0. S is not empty since 0 ∈ S. Thus S is a subring. (b) It is well-known that this is closed under subtraction and multiplication and is not empty. (c) This is not closed under subtraction. Let
f (x) =
0 if x = (^13) 1 otherwise
g(x) =
0 if x = (^23) 1 otherwise
Then f ∈ S and g ∈ S, but f − g vanishes on an infinite set. (d) This is not closed under subtraction. Let
f (x) =
0 if x < (^12) 1 otherwise
g(x) =
0 if x > (^12) 1 otherwise
Then f, g ∈ S, but f − g vanishes only at 12. (e) Let f, g ∈ S. S is closed under subtraction since
lim x→ 1 −
[f (x) − g(x)] = lim x→ 1 −
f (x) − lim x→ 1 −
g(x) = 0 − 0 = 0.
S is closed under multiplication since
lim x→ 1 −^
f (x)g(x) = lim x→ 1 −^
f (x) · lim x→ 1 −^
g(x) = 0 · 0 = 0.
S is not empty since 0 ∈ S. Thus S is a subring.
(f) Clearly S is nonempty and closed under subtraction. To show that it is closed under multiplication, it suffices to observe that
sin mx cos nx = 12 [sin(m + n)x + sin(m − n)x] cos mx cos nx = 12 [cos(m + n)x + cos(m − n)x] sin mx sin nx = 12 [− cos(m + n)x + cos(m − n)x].
If m − n is negative, we can rewrite sin(m − n)x as − sin(n − m)x and cos(m − n)x as cos(n − m)x. Thus S is a subring.
7.2.2. Let p(x) = anxn^ + · · · + a 0 ∈ R[x], where an 6 = 0. If there is a nonzero b ∈ R such that bp(x) = 0 then p(x) is a zero divisor. Conversely, suppose that p(x) is a zero divisor and let g(x) = bmxm^ + · · · + b 0 , where bm 6 = 0, be of minimal degree such that g(x)p(x) = 0. The leading term of g(x)p(x) is bmanxm+n, so bman = 0. Since R is commutative, anbm = 0 as well. Now
ang(x) = anbmxm^ + anbm− 1 xm−^1 + · · · = 0 + anbm− 1 xm−^1 + · · ·
has lower degree than g(x) and (ang(x))p(x) = an(g(x)p(x)) = 0, but we assumed that g(x) had minimal degree among such polynomials, so ang(x) = 0. Now
0 = g(x)p(x) = g(x)(anxn^ + an− 1 xn^ + · · · ) = g(x)anxn^ + g(x)(an− 1 xn−^1 + · · · ) = 0 + g(x)(an− 1 xn−^1 + · · · )
so bman− 1 = 0, so an− 1 g(x) = 0 by the same argument as above. Similarly bman− 2 = 0, bman− 3 = 0, and so on, so bmg(x) = 0, as desired.
7.2.3. (a) Since addition in R[[x]] is done componentwise, it is an abelian group under + because R is; in fact, R[[x]] ≡ R × R × R × · · · as abelian groups. For the rest, let a(x) = a 0 + a 1 x + a 2 x^2 + · · · be an element of R[[x]] and similarly b(x) and c(x). The nth^ coefficient of a(x)b(x) is
∑^ n
i=
aibn−i =
0 ≤i,j≤n i+j=n
aibj
and since multiplication in R is commutative, this is the same as the nth^ coefficient of b(x)a(x). The nth^ coefficient of (a(x)b(x))c(x) is
∑
0 ≤m,k≤n m+k=n
0 ≤i,j≤m i+j=m
aibj
ck =
0 ≤i,j,k≤n i+j+k=n
(aibj )ck
and since multiplication in R is associative, this is the same as the nth^ coefficient of a(x)(b(x)c(x)). The nth^ coefficient of (a(x) + b(x))c(x) is ∑
0 ≤i,j≤n i+j=n
(ai + bi)cj =
0 ≤i,j≤n i+j=n
(aicj + bicj ) =
0 ≤i,j≤n i+j=n
aicj +
0 ≤i,j≤n i+j=n
bicj
which is the nth^ coefficient of a(x)c(x) + b(x)c(x). The proof of distributivity on the other side is similar. The identity is 1 + 0x + 0x^2 + 0x^3 + · · · , since the nth^ coefficient of 1a(x) is 1 an + 0an− 1 + 0an− 2 + · · · + 0a 1 + 0a 0 = an and similarly a(x)1 = a(x). (b)
(1 − x)(1 + x + x^2 + · · · ) = (1 + x + x^2 + · · · ) − x(1 + x + x^2 + · · · ) = 1 + x + x^2 + · · · − x − x^2 − x^3 − · · · = 1.
