Department of Mathematics
University of British Columbia
MATH 342 Midterm 2
March 17, 2016, 9:30 AM - 10:50 AM
Family Name:
Initials:
I.D. Number:
Problem
Signature:
Mark
Out of
1
20
2
20
3
20
4
20
5
20
Total
100
CALCULATORS, NOTES OR BOOKS ARE NOT PERMITTED.
THERE ARE 5 PROBLEMS ON THIS EXAM.
JUSTIFY YOUR ANSWERS.
MATH 342
Math 342 Midterm 2
Below are the addition and multiplication tables for GF (4).
(GF (4), +):
+
0
1
a
b
0
0
1
a
b
1
1
0
b
a
a
a
b
0
1
b
b
a
1
0
·
0
1
a
b
0
0
0
0
0
1
0
1
a
b
a
0
a
b
1
b
0
b
1
a
(GF (4), ·):
2
3
1. (a) Find the multiplicative inverse of 53 in Z157 . Give your answer as a principal remainder.
Note: 157 is prime. You do not need to verify this fact.
MATH 342
Math 342 Midterm 2
(b) Find the principal remainder of (739)33 mod 11.
Solution:
a. EA:
157
53
51
2
= 2 · 53 + 51
= 1 · 51 + 2
= 25 · 2 + 1
=
1·2
Working backwards through the EA:
1 = 51 − 25 · 2 = 51 − 25 · (53 − 1 · 51) = 26 · 51 − 25 · 53
= 26 · (157 − 2 · 53) − 25 · 53 = 26 · 157 − 77 · 53
Thus, 53−1 = −77 = 80 in Z157 .
b. 739 = 2 mod 11. Thus (739)33 = 233 mod 11. By Fermat’s little theorem, 211 = 2 mod 11.
Thus, (739)33 = (211 )3 = 23 = 8 mod 11.
It is also possible to do part b without using Fermat’s little theorem.
MATH 342
Math 342 Midterm 2
4
MATH 342
5
Math 342 Midterm 2
2. Let C be the span of {2430, 2443, 4323} in Z5 . Find the following.
(a) A generator matrix in standard form for C (or for a code equivalent to C by a permutation
of codeword positions).
(b) The dimension of C.
(c) |C|.
(d) The number of cosets of C (as a subgroup under vector addition) in V (4, 5).
Solution:
a. A generator matrix in standard form is obtained by deleting the zero rows of the RREF and
permuting columns if necessary.


2 4 3 0
 2 4 4 3 
4 3 2 3


2 4 3 0
 0 0 1 3 
0 0 1 3


2 4 3 0
 0 0 1 3 
0 0 0 0


1 2 4 0
 0 0 1 3 
0 0 0 0


1 2 0 3
 0 0 1 3  = RREF
0 0 0 0
A generator matrix is:
1 2 0 3
0 0 1 3
1 0 2 3
0 1 0 3
G=
A generator matrix in standard form is:
G=
b. The dimension is 2 since the generator matrix has 2 rows.
c. |C| = 52 = 25.
d. The number of cosets is 54 /52 = 25.
MATH 342
Math 342 Midterm 2
6
MATH 342
Math 342 Midterm 2
7
3.
(a) Find all solutions to the equation x3 + x2 + x + 1 = 0 in the rings Z2 , Z3 , Z4 and GF (4).
(b) Show that in a field F , the only solutions to x2 = 1 are x = 1 and x = −1.
(c) Show that in a field F , the only elements x ∈ F such that x = x−1 are x = 1 and x = −1.
(d) Find a counterexample to part c if F is merely a ring.
Solution:
a. Z2 : 1
Z3 : 2
Z4 : 1, 3
GF (4): 1
b. x2 = 1 iff (x − 1)(x + 1) = 0 iff x = ±1, the latter because in a field the product of two nonzero
elements is nonzero.
c. x = x−1 iff x2 = 1; so this reduces to part b.
d. In Z8 , 3 = 3−1 and 3 6= ±1.
MATH 342
Math 342 Midterm 2
8
9
4. Let q be a positive integer and a ∈ Zq such that a 6= 0. Show that a has a multiplicative inverse
in Zq iff gcd (a, q) = 1.
MATH 342
Math 342 Midterm 2
Solution:
If: Suppose that gcd(a, q) = 1. By Bezout, there exist integers m, n s.t. ma + nq = 1. Thus,
ma = 1 mod q, and so m = a−1 in Zq .
Only If: Let b = a−1 ∈ Zq . Then ba = 1 mod q. So, there exists an integer n s.t. ba = 1 + nq.
Then any common divisor of a and q must divide 1. So, gcd(a, q) = 1.
MATH 342
Math 342 Midterm 2
10
11
5. Recall that GF (4) = {0, 1, a, b}. Find a bijection g : {1, a, b} → Z3 such that for all x, y ∈ {1, a, b},
g(x · y) = g(x) + g(y). Here, · is multiplication in GF (4) and + is addition in Z3 .
MATH 342
Math 342 Midterm 2
Solution:
Define g as follows:
g(1) = 0, g(a) = 1, g(b) = 2
which is clearly a bijection from {1, a, b} to Z3 .
For all x ∈ {1, a, b}, g(x · 1) = g(x) = g(x) + 0 = g(x) + g(1).
Since both groups are abelian, it suffices to verify that g(a · a) = g(a) + g(a), g(b · b) = g(b) + g(b),
and g(a · b) = g(a) + g(b):
g(a · a) = g(a2 ) = g(b) = 2 = 1 + 1 = g(a) + g(a)
g(b · b) = g(b2 ) = g(a) = 1 = 2 + 2 = g(b) + g(b)
g(a · b) = g(1) = 0 = 1 + 2 = g(a) + g(b)
Alternatively, one can observe that the tables
{1, a, b} with ·
·
1
a
b
1
1
a
b
a
a
b
1
b
b
1
a
·
0
1
2
0
0
1
2
1
1
2
0
2
2
0
1
and
{0, 1, 2} with +
are identical via the 1-1 correspondence 1 ↔ 0, a ↔ 1, b ↔ 2.
Note: the only other bijection that works is
g(1) = 0, g(b) = 1, g(a) = 2
MATH 342
Math 342 Midterm 2
12