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Homework 4
1.9.1. Claim: For any x, y, and n ∈ N,
n
(x + y) =
n X
n
k=0
k
xk y n−k .
Proof: We will prove
the above equation by induction on n. Suppose first that n = 1.
Then (x + y)1 = 10 x + 11 y, so the equation holds. Now suppose the equation holds for
n = l. For n = l + 1, we know that (x + y)l+1 = (x + y)(x + y)l . So by assumption,
l+1
(x + y)
l X
l k l−k
= (x + y)
x y .
k
k=0
Using the distributive property, we have
(x + y)
l+1
l l X
l k+1 l−k X l k l−k+1
x y
.
x y
+
=
k
k
k=0
k=0
If we now group together the monomials of the same bidegree, we have
l+1
(x + y)
=x
l+1
+y
l+1
l X
l
l
(
+
)xk y l+1−k .
+
k
k−1
k=1
Then by Lemma 1.9.3, this is
l+1
(x + y)
=
l+1 X
l+1
k
k=0
xk y l+1−k .
1.10.2. Claim: The set C4 = {1, i, −1, −i} is a group under complex multiplication.
Proof: First, we check that the set C4 is closed under multiplication. We must check that
for every a, b, ∈ C4 , ab ∈ C4 . We see 12 = (−1)2 = 1 ∈ C4 and i2 = (−i)2 = −1 ∈ C4 .
Also, 1b = b and −1b = −b ∈ C4 for each b ∈ C4 . Finally, i(−i) = 1 ∈ C4 . Hence C4 is
closed under complex multiplication.
Complex multiplication is always associative, so we must check for an identity element
and inverses to prove that C4 is a group. 1 ∈ C4 is the multiplicative identity, since 1a = a
1
for all a ∈ C4 . The inverse of −1 is −1 since (−1)2 = 1. The inverse of i is −i, and
the inverse of −i is i, since i(−i) = 1. This shows that C4 is a group under complex
multiplication.
1.10.4. Claim: The group Z4 is isomorphic to C4 .
Proof: Consider the bijection of sets [0] ↔ 1, [1] ↔ i, [2] ↔ −1, [3] ↔ −i. We check the
multiplication tables for C4 and Z4 and see that this bijection is an isomorphism of the
groups C4 and Z4 .
Z4
[0]
[1]
[2]
[3]
C4
1
i
−1
−i
[0]
[0]
[1]
[2]
[3]
[1]
[1]
[2]
[3]
[0]
[2]
[2]
[3]
[0]
[1]
1
1
i
−1
−i
i
i
−1
−i
1
−1
−1
−i
1
i
[3]
[3]
[0]
[1]
[2]
−i
−i
1
i
−1
1.10.9. Claim: The set of affine transformations of Rn is a group under composition of
maps.
Proof: Consider the set of affine maps T (~x) = S(~x) + ~b, where T is an invertible linear
transformation from Rn to itself, and ~b is a vector. First, we show this set is closed under
composition. If T1 = S1 +~b1 and T2 = S2 +~b2 , then T1 ◦T2 (~x) = T1 (S2 (~x)+ b~2 ) = S1 (S2 (~x)+
~b2 ) +~b1 . Using the properties of linear transformations, this is S1 S2 (~x) + (S1~b2 +~b1 ). Since
S1 and S2 are invertible, S1 S2 is invertible, and T1 ◦ T2 is an affine linear transformation.
Function composition is always associative, so we must check that there is an identity
and inverse functions for each affine map. First, if we take S = 1n the n × n identity
matrix, and ~b = 0, then we claim this is the identity under composition. To check, for
any other affine transformation T , 1n T (~x) = 1n S(~x) + 1n~b = S(~x) = T (~x), and T 1n (~x) =
T (1n ~x) + ~b = T (~x). Hence 1n is the identity affine transformation.
Now we show that T (~x) = S(~x) + ~b has an inverse transformation. Let T −1 (~x) =
S −1 (~x) − S −1~b. Then T ◦ T −1 (~x) = T (S −1 (~x) − S −1~b) = S(S −1 (~x) − S −1~b) +~b = ~x. This is
the same as 1n (~x). Now we check T −1 ◦ T (~x) = T −1 (S(~x) +~b) = S −1 (S(~x) +~b) − S −1~b = ~x.
This is again the same as 1n (~x). Therefore T ◦ T −1 = 1n and T −1 ◦ T = 1n , so T has
2
an inverse under composition. This completes the proof that affine transformations are a
group under function composition.
1.11.2. Claim: Let K be any field. Then K[x] is a commutative ring under polynomial
addition and multiplication. Further, 1 is the multiplicative identity and the only units are
constant polynomials.
Proof: First, we show that K[x] is a group under addition. Let f (x) = an xn + an−1 xn−1 +
· · · + a1 x + a0 and g(x) = bm xm + am−1 xm−1 + · · · + b1 x + b0 be polynomials in K[x]. Say
n ≥ m. Then
f (x) + g(x) = an xn + · · · am+1 xm+1 + (am + bm )xm + · · · + (a1 + b1 )x + (a0 + b0 ).
This is another polynomial in K[x], since K is closed under addition. Further, polynomial
addition is associative since addition in K is associative. The identity polynomial is 0,
and 0 + f (x) = f (x) = f (x) + 0. Each polynomial f (x) has an additive inverse −f (x) =
−an xn + · · · + (−a1 )x + a0 .
Now that we see K[x] is a group under addition, we check that multiplication is associative in K[x]. But this follows from the facts that multiplication and addition in K are
associative. So now we check the distributive property. Let h(x) = cl xl + · · · c1 x + c0 . Say
n ≥ m ≥ l. Then
f (x)(g(x) + h(x)) = f (x)
=
m
X
!
(bi + ci )xi
i=0
n
i
XX
aj (bi−j + ci−j )xi
i=0 j=0
=
n X
i
X
aj bi−j xi + aj ci−j xi
i=0 j=0
= f (x)g(x) + f (x)h(x).
(Note that in this formula we pretend n = m = l, or you can take bi = 0 and cj = 0 for
i > m or j > l.)
This shows that K[x] is a ring. It is commutative since multiplication in K is commutative. The element 1 ∈ K is the identity in K[x] as well. Suppose now that f (x) ∈ K[x]
is a unit, and g(x) is its inverse. Then g(x)f (x) = 1. This means that the degree of g(x)
added to the degree of f (x) is 0. Hence f (x) and g(x) are both constant polynomials in
K.
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