Homework 1 (Week 2)
Please write all answers in complete sentences.
Arithmetic in F2
1. Show that the equation x2 + x + 1 = 0 has no solutions in F2 .
2. Show that the equation x3 + x + 1 = 0 has no s
Spring 2013
Math 370: Applied Algebra
Prof. Assaf
Solutions to Problem Set 2
1. Using Euclids Algorithm takes a while, but gets the job done with 18 steps. Solving for the
remainder in each equation a
Spring 2013
Prof. Assaf
Math 370: Applied Algebra
Solutions to Problem Set 1
3(1)2 1
2
.
2=
2
3n2 n
2 . Then
1. We proceed by induction. The Base Case is for n = 1. In this case we have 1 =
For the In
Spring 2013
Math 370: Applied Algebra
Prof. Assaf
Solutions to Midterm Exam 1
1. Let p N be a prime number. Prove that p is irrational.
Solution : Suppose p is prime and that p is rational. Then p = n
Spring 2013
Math 370: Applied Algebra
Prof. Assaf
Solutions to Problem Set 4
1. First suppose that cfw_r1 , r2 , . . . , rm is a complete set of representatives. Then since ri
ri (mod m), the unique
Spring 2013
Math 370: Applied Algebra
Prof. Assaf
Solutions to Problem Set 6
1. By problem 3 on problem set 5, there exists a prime p such that p a = 0 in F. In particular,
the Freshmans Dream applies
Spring 2013
Math 370: Applied Algebra
Prof. Assaf
Solutions to Problem Set 10
1. Compute xn (mod x2 + x + 1) in Z/3Z until a pattern appears:
x0 1 (mod x2 + x + 1)
x1 x (mod x2 + x + 1)
x2 x 1 2x + 2
Spring 2013
Math 370: Applied Algebra
Prof. Assaf
Solutions to Problem Set 9
1. We rst prove by induction that f (x) can be factored into a product of irreducible polynomials.
If deg(f (x) = 1, then f
Spring 2013
Prof. Assaf
Math 370: Applied Algebra
Solutions to Problem Set 8
1. Let p(x) = x5 and q (x) = x. Then p(x) = q (x) since they have dierent degrees, but, by
Fermats Little Theorem, for any
Spring 2013
Math 370: Applied Algebra
Prof. Assaf
Solutions to Problem Set 5
1. Since f is a homomorphism, f (0) = 0 and f (1) = 1. For n N, f (n) = f (1) + + f (1) =
1 + + 1 = n, and f (n) = f (n) =
Spring 2013
Prof. Assaf
Math 370: Applied Algebra
Solutions to Problem Set 3
1. Since m is not prime, we may write m = ab where 2 a, b m/2. If both a and b are m/2,
then m = m2 /4, so m = 4. Since m =
Spring 2013
Prof. Assaf
Math 370: Applied Algebra
Solutions to Problem Set 11
1. The exponent of U16 is (24 ) = 22 = 4, so every element of U16 is a solution to x4 1. Since
(24 ) = 23 , there are 8 so
Spring 2013
Math 370: Applied Algebra
Prof. Assaf
Solutions to Problem Set 12
1. Since 1001 = 7 11 13, we know U1001 (1000) = U7 (1000) U11 (1000) U13 (1000). Therefore
#
U1001 (1000) = gcd(6, 1000) g
Homework Week 3
Irreducibility
1. Write down all the irreducible polynomials of degree 2 with coefcients in F2 .
2. Write down all the irreducible polynomials of degree 3 with coefcients in F2 .
3. Wr
Homework Week 4
Computing with the (15, 7)-code from class
1. Given the data bit (1, 1, 0, 1, 0, 0, 1), construct the sent word s(x) to which it encodes.
2. Suppose we know that at most 2 errors have
Homework Week 5
Polynomial arithmetic
1. Using the algorithm given in class, determine the greatest common divisor of x3 + 1 and
x4 + 1 over F2 .
More nite elds
2. Suppose n is an integer 2. Consider
Homework Week 8
Orders of elements and Eulers totient function
1. Consider the group Z/mZ of units in the commutative ring Z/mZ discussed on the last
homework.
i) In class, I dened the order of an ele
Homework Week 9
Groups
1. We know that the exponent of a group divides the order of the group. Consider the group
Z/nZ . The order of Z/nZ is given by Eulers -function, as we studied last week.
i) Com
Homework Week 11
Applications of orders
1. Suppose p > 2 is a prime number, and consider the polynomial x2 + 1 = 0 in Fp .
i) Show that x2 + 1 = 0 has a solution in Fp if and only if there is an eleme
Homework Week 13
On the Chinese remainder theorem
1. Find the smallest positive solution, if any, of
x9
x 17
mod 16
mod 28.
2. Find the smallest positive solution, if any, of
x 10
mod 15
x 17
mod 28.
M idtermL-Math370
Instructions
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s
o
The following examconsists f several roblems, omeof which havemultiple parts' There are125
points availableon the exam. To get full credit, you must do enoughprobl
M idterm2-Math370
Instructions
There are 130 points
The following exam consistsof threeproblems,each with multiple parts'
completeEnglish sentences'You are not
s
availableon the exam. Your explanation
Last 1
First Last
Mrs. St. Amant
LA 1, Period 4
17 December 2013
My Interesting Title
Indent your first line. Here is where you will write your essay. Be sure to double
space. In order to do that for