Homework 6: Math 3360 Applicable Algebra
Chapter 7: Rings and Fields
44.
Let
F
be a ﬁeld of characteristic 2. Show that for any
a,b
∈
F
(i)
a
=

a
(ii) (
a
+
b
)
2
=
a
2
+
b
2
Solution:
(i) Since
F
has characteristic 2, then we know 1+1 = 2 = 0. So by the distributive
property 0 =
a
(0) =
a
(1 + 1) =
a
+
a
, and then
a
=

a
.
(ii) We use the distributive property again to get (
a
+
b
)(
a
+
b
) =
a
2
+
ab
+
ba
+
b
2
.
Since
F
is a ﬁeld, then it is commutative under multiplication, so we have
ab
=
ba
. Then
(
a
+
b
)
2
=
a
2
+
ab
+
ab
+
b
2
=
a
2
+ 2
ab
+
b
2
=
a
2
+ 0(
ab
) +
b
2
=
a
2
+
b
2
.
Chapter 9: Fermat’s and Euler’s Theorems
20.
Find the order of [3] in
Z
/
23
Z
.
Solution:
Now 23 is prime so 3 is a unit and then we have 3
22
≡
1
mod
23. Thus the order of
[3] must divide 22, i.e. it is either 2
,
11, or 22. Now
3
2
≡
9
mod
23
6≡
1
mod
23
and
3
11
≡
(3
3
)
3
(3
2
)
mod
23
≡
(27)
3
(9)
mod
23
≡
(4)
3
(9)
mod
23
≡
64(9)
mod
23
≡
(

5)(9)
mod
23
≡ 
45
mod
23
≡
1
mod
23
33.
Show that for prime
p
, then [(
p

1)!]
p
= [

1]
p
.
Solution:
Note that since
p
is prime, then every element in
Z
/p
Z
is a unit, so every every ele
ment has an inverse. First we will show that the only elements that are their own inverses are
[1] and [
p

1]. Suppose that [
a
] is its own inverse modulo
p
. Then we have that
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 Spring '08
 BILLERA
 Algebra, Prime number, Identity element

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