Math 210
Answers for Homework 10
1. (a) We have
Z2 [x]/(x3 + x + 1) = cfw_[rx2 + sx + t] | r, s, t Z2
= cfw_[0], [1], [x], [x + 1], [x2 ], [x2 + 1], [x2 + x], [x2 + x + 1] with x3 + x + 1 = 0
Roots of x3 + x + 1 are [x], [x2 ] and [x2 + x]. Indeed, we ca
Math 210
Answers for Homework 2
1.
(a) Suppose that there are only nitely many primes, say r < , of the form 3n + 2
with n N. Label them p1 , p2 , . . . , pr . Square these primes : p2 , p2 , . . . , p2 . Then
1
2
r
take the product of all these numbers a
Math 210
Answers for Homework 3
1.
(a) Since ay bx = k and ay0 bx0 = k, by subtracting, we get
a(y y0 ) = b(x x0 )
a (y y0 ) = b (x x0 )
Since gcd(a , b ) = 1, it follows that
a |x x0
Write
x x0 = a t with t Z
then
y y0 = b t
Therefore,
x = x0 + a t
y = y
Math 210
Answers for Homework 1
1. For each pair of numbers a, b we repeatedly apply division with remainder to nd
the gcd, and then start at the bottom and substitute back up to obtain an expression
for gcd(a, b) as an integer linear combination au + bv
Rings and Fields: Final Handout
Goals of the Course:
0. Develop mathematical maturity.
1. Understand important examples of rings:
Z, Q, R, C, M2x2 (R), R S, R[x], Zn , F [x]/p(x), R/I
where R, S are arbitrary rings, F a eld, I an ideal of R. Also extensio
M2P4
Rings and Fields
Mathematics
Imperial College London
ii
As lectured by Professor Alexei Skorobogatov
and humbly typed by as1005@ic.ac.uk.
CONTENTS
iii
Contents
1 Basic Properties Of Rings
1
2 Factorizing In Integral Domains
5
3 Euclidean domains and
Groups, Rings and Fields
Karl-Heinz Fieseler
Uppsala 2010
1
Preface
These notes give an introduction to the basic notions of abstract algebra,
groups, rings (so far as they are necessary for the construction of eld extensions) and Galois theory. Each sect
Page 1 of 15
Math 210
Final Exam
April 17, 2012
Queens University
Faculty of Arts and Science
Department of Mathematics and Statistics
Instructor: Michael Dewar
Instructions: This exam has 7 questions, for a total of 200 points.
The exam is three hours in
Homework and Tutorials
I downloaded these from the exambank a year ago: 2006, 2007, 2008, 2009, 2010, 2011. Bear
in mind that different instructors using different textbooks will place slightly different emphasis on
material from year to year.
Homework 1:
Math 210
Answers for Homework 8
1. To obtain solutions modulo m(x) = m1 (x)m2 (x), we may use the same algorithm
developed for Z earlier. To nd all polynomials f (x) F [x] satisfying
f (x) a(x) (mod m1 (x) and f (x) b(x) (mod m2 (x)
Step 1: Find polynomia
MATH 210
Midterm Examination
March 3, 2014
Instructions: The exam has six questions labeled 1 through 6. Each question has
subquestions. The total mark is 100.
The exam paper has seven pages.
The exam is two hours in length.
To receive full credit you mus
Math 210
Answers for Homework 11
1.
(a) Find all roots of x2 x + 18 in Z/5Z.
(b) Find all roots of x2 x + 18 in Z/8Z.
(c) Find all roots of x2 x + 18 in Z/40Z.
Solutions: Put f (x) = x2 x + 18.
(a) Z/5Z = cfw_0, 1, 2, 3, 4. Then f (x) = x2 x + 3 (Z/5Z)[x]
Math 210
Answers for Homework 5
1.
(a) Dene addition and multiplication on K by matrix addition and matrix mul0 0
tiplication. Then K has the additive identity
and the multiplicative identity
0 0
1 0
a b
. Any non-zero element
is a unit as its multiplicat
Math 210
Answers for Homework 7
1.
(a) f (x) and g(x) are associates if and only if f (x) = ug(x) for some unit u F ,
if and only if g(x) = u1f (x). Hence f (x) and g(x) are associates if and only if
g(x)|f (x) and f (x)|g(x).
(b) Let p(x) F [x], and g(x)
Math 210
Answers for Homework 6
1.
(a) If R is not an integral domain, then there exist a, b R such that a = 0, b = 0 but
ab = 0. Consider the polynomials f (x) = a and g(x) = b. Both are nonzero polynomials
in R[x] but the product f (x)g(x) = ab = 0. So
Math 210
Answers for Homework 9
1. (a) Nonunits in Z9 are 0, 3, 6. For any r Z9 , r cfw_0, 3, 6 cfw_0, 3, 6. So the set of
nonunits in Z9 is an ideal of Z9 .
(b) 2Z, 3Z, 4Z, 6Z and 12Z.
(c) Let I = cfw_h(x) Z[x] | 3|h(0) . For f (x), g(x) I, 3|f (0) g(0),
Math 210
Answers for Homework 4
1. (a) Since 325 = 52 13 the single congruence is equivalent to the system of three
congruences
7x 8 (mod 52 )
(1)
7x 8 (mod 13)
(2)
Step 1: The congruence (1) is transformed to the congruence
7 7 x 8 7 (mod 52 )
which give