Algebra , Assignment 12
Solutions
I wasn’t sure anymore and I will tell you, it is a strange process
to feel one’s mind changing, allowing ideas into your brain which
it had once considered unthinkable. I cannot say it’s painful, or
particularly pleasurable, but that it requires a certain relaxation
of the hold one keeps over oneself, and is to that degree both a
thrill and a horror.
– from The Chess Garden, by Brooks Hansen
1. These problems deal with the field
K
=
Q
[
x
]
/
(
x
3

2).
(You can
assume, without proof, that
x
3

2 is irreducible over
Q
.) All elements
should be written in the form
a
+
bx
+
cx
2
with
a, b, c
∈
Q
. (There
may be intermediate steps in other forms.)
(a) Find (
x
2
+ 3
x
+ 7)(2
x
2
+
x
+ 3).
Solution:
2
x
4
+7
x
3
+20
x
2
+16
x
+21 but now we reduce by
x
3
= 2
and
x
4
= 2
x
to get 4
x
+ 14 + 20
x
2
+ 16
x
+ 21 or 20
x
2
+ 20
x
+ 35.
(b) Find
x
4
.
Solution:
As
x
3
= 2,
x
4
= 2
x
.
(c) Find
x
30
.
Solution:
As
x
3
= 2,
x
30
= (
x
3
)
10
= 2
10
= 1024.
(d) Find the multiplicative inverse of
x
+ 3.
Solution:
(Method 1) First we divide
x
+ 3 into
x
3

2:
x
3

2 = (
x
2

3
x
+ 9)(
x
+ 3) + (

29)
and we are done as

29 is a unit. Reversing

29 = (
x
3

2) + (

x
2
+ 3
x

9)(
x
+ 3)
so that
1 =

1
29
(
x
3

2) +

1
29
(

x
2
+ 3
x

9)(
x
+ 3)
and so the multiplictative inverse of
x
+ 3 is

1
29
(

x
2
+ 3
x

9)
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
(Method 2)
(
α
+
xβ
+
x
2
γ
)(
x
+ 3) = 3
α
+ 3
xβ
+ 3
x
2
γ
+
xα
+
x
2
β
+ 2
γ
as the
x
times
x
2
γ
term becomes
x
3
γ
which is 2
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 JOSEPHS
 Algebra, Prime number, Greatest common divisor, Multiplicative inverse, Inverse element

Click to edit the document details