Math 342
Homework Assignment #3 (Due Thursday, March 3, in class)
All numbered problems are worth the same value.
1.
a. Verify that 727 is prime.
b. Find the multiplicative inverse of 726 in the field Z727 .
c. Find the multiplicative inverse of 373 in the field Z727 .
Solution:
a. 727 is prime because it is not divisible by any prime ≤
by 2,3,5,7,11,13,17,19,23.
√
727, namely not divisible
b. 726 = −1 mod 727 and so 726−1 = 726 in Z727 .
c. for any integer solution to Bezout’s equation 373x + 727y = 1, x will be the
multiplicative inverse of 373 in the field Z727 . Such x, y can be found by the
Euclidean algorithm:
727 = 1 · 373 + 354
373 = 1 · 354 + 19
354 = 18 · 19 + 12
19 = 1 · 12 + 7
12 = 1 · 7 + 5
7=1·5+2
5=2·2+1
2=2·1
Then working backwards, we find
1 = 5−2·2 = 5−2·(7−1·5) = 3·5−2·7 = 3·(12−1·7)−2·7 = 3·12−5·7 = 3·12−5·(19−1·12)
= 8 · 12 − 5 · 19 = 8 · (354 − 18 · 19) − 5 · 19 = 8 · 354 − 149 · 19
= 8 · 354 − 149 · (373 − 1 · 354) = 157 · 354 − 149 · 373 = 157 · (727 − 1 · 373) − 149 · 373
= 157 · 727 − 306 · 373
Thus, 373−1 = −306 = 421 mod 727.
2. Recall the definition of the field GF (4), given by the addition and multiplication tables
in the first paragraph of page 36 in the text. Compute the following in this field.
(a) a + a
(b) (a + 1)(a + b)
(c) a1000
1
Solution:
a. a + a = 0.
b. (a + 1)(a + b) = (b)(1) = b
c. a1000 = (a3 )333 a = 1 · a = a since a3 = a2 a = ba = 1.
3. (a) Show that in a field F , for all a, b, c ∈ F such that a 6= 0, there is a unique solution
to the equation ax + b = c.
(b) Find a counter-example to part a) if F is merely a commutative ring with identity.
(c) Find all solutions to the equation x2 + x + 1 = 0 in the fields GF (2), GF (3) and
GF (4).
Solution:
a. By using the field operations, we get
ax + b = c ⇒ ax + b + (−b) = c + (−b) ⇒ ax + 0 = c − b
⇒ ax = c − b ⇒ a−1 ax = a−1 (c − b) ⇒ x = 1x = a−1 (c − b).
So, if x satisfies the equation, then it is forced to be a−1 (c − b) and this is the only
possible solution. One can see that this is indeed a solution by plugging into the
equation and using the field operations.
b. Let F = Z4 , a = 2, b = 0, c = 0. The equation 2x = 0 has two solutions x = 0 and
x = 2.
c. Solve by trial and error.
GF (2) : no solutions
GF (3) : x = 1.
GF (4) : x = a, x = b since a2 + a + 1 = b + a + 1 = 1 + 1 = 0, and likewise for b, while
02 + 0 + 1 = 1 and 12 + 1 + 1 = 1
4. Prove that if p is prime and a ∈ Zp and a 6= 0, then ap−1 = 1 mod p.
Hint: Recall from the proof that Zp is a field: {a, 2a, . . . (p − 1)a} are distinct and
non-zero mod p; thus,
{a, 2a, . . . (p − 1)a} = {1, 2, . . . (p − 1)} ( mod p)
and so the following products are equal mod p:
(a)(2a)(3a) · · · ((p − 1)a) = (1)(2) · · · (p − 1) ( mod p)
Solution: Form the preceding equation, we have
ap−1 ((p − 1)!) = (p − 1)! mod p
2
Since each i = 1, . . . , p − 1 is non zero in Zp , (p − 1)! 6= 0 in Zp . Since Zp is a field,
(p − 1)! is invertible in Zp . Thus,
ap−1 = ap−1 ((p − 1)!)((p − 1)!)−1 = (p − 1)!((p − 1)!)−1 = 1 mod p
Note: This result is known as Fermat’s little theorem.
5. Without using a computer or calculator, find the following principal remainders.
a. (99)101 mod 31
b. (263912)20111 mod 13
Solution:
99101 = 6101 = 611 mod 31, the latter equality by Fermat’s little Theorem. And 611 =
6·210 ·310 = 6·(25 )2 ·(33 )3 ·3 = 6·12 ·(−4)3 ·3 = −(18)·26 = −18·64 = −36 = 26 mod 31.
b. Since 13 divides 26 and 39, it follows that 263912 = 12 = −1 mod 13. Thus,
26391220111 = (−1)20111 = −1 = 12 mod 13.
Alternatively, observe that 13 divides 20111. Thus, by Fermat’s little theorem, it
follows that 26391220111 = 263912 = 12 mod13.
6. Let V be a vector space and u ∈ V .
(a) Show that 0u = 0
(b) Show that (−1)u = −u.
Solution:
a.
0u = (0 + 0)u = 0u + 0u
Thus
0 = 0u − 0u = (0u + 0u) − 0u = 0u
b. u + (−1)u = (1)u + (−1)u = (1 + (−1))u = 0u = 0. So, (−1)u = −u.
7. Which of the following subsets are subspaces of GF (q)n ?
(a) {0000, 1111, 1010, 0101}, q = 2, n = 4
(b) {0000, 1111, 1010, 0101}, q = 3, n = 4
(c) {000, 111, 222}, q = 3, n = 3
(d) {00, 11, aa, ab}, q = 4, n = 2
(e) {a1ab, bab1, 1b1a}, q = 4, n = 4
3
Solution:
a. Yes: an addition table shows that this set is closed under vector addition and closed
under scalar multiplication since the only scalar are 0 and 1.
b. No: it is not closed under addition: 1111 + 1111 = 2222, which is not in the subset.
c. Yes: an addition table shows that it is closed under addition; and since 2(111) =
222 and 2(222) = 111 it is closed under scalar multiplication.
d. No: b(11) = bb which is not in the subset.
e. No: 0000 does not belong to the subset.
8. For each of the following subsets S of GF (q)n , find a basis for, and the dimension of,
the span of S. Also, determine if the given S is linearly independent.
(a) S = {1100, 1010, 1001, 0101}, q = 2, n = 4
(b) S = {1234, 3142, 2413, 4321}, q = 5, n = 4
(c) S = {0140, 4322, 1233, 2141}, q = 5, n = 4
(d) S = {a1ab, bab1, 1b1a}, q = 4, n = 4
(e) S = {01ab, 1ab0, ab10, abbb}, q = 4, n = 4
Solution:
Use REF or an ad hoc method.
a.


1 1 0 0
 1 0 1 0 

A=
 1 0 0 1 
0 1 0 1


1 1 0 0
 0 1 1 0 


 0 1 0 1 
0 1 0 1


1 1 0 0
 0 1 1 0 


 0 0 1 1 
0 0 1 1


1 1 0 0
 0 1 1 0 

REF = 
 0 0 1 1 
0 0 0 0
So, the first 3 rows of the REF form a basis and the dimension is 3.
4
b.

1
 3
A=
 2
4

2
1
4
3
1
 0
REF = 
 0
0

4
2 

3 
1
3
4
1
2
2
0
0
0
3
0
0
0

4
0 

0 
0
So, a basis is the first row of the REF and the dimension is 1.
c.


0 1 4 0
 4 3 2 2 

A=
 1 2 3 3 
2 1 4 1


4 3 2 2
 0 1 4 0 


 1 2 3 3 
2 1 4 1


4 3 2 2
 0 1 4 0 


 0 0 0 0 
0 2 3 0


4 3 2 2
 0 1 4 0 

REF = 
 0 0 0 0 
0 0 0 0
So, the first two row of the REF form a basis and the dimension is 2.
d.


a 1 a b
A= b a b 1 
1 b 1 a


a 1 a b
REF =  0 0 0 0 
0 0 0 0
So, the first row forms a basis and the dimension is 1.
e.

0
 1
A=
 a
a
5
1
a
b
b
a
b
1
b

b
0 

0 
b

1
 0

 a
a

1
 0

 0
0
a
1
b
b
a
1
0
0

1
 0
REF = 
 0
0
b
a
1
b
b
a
0
a
a
1
0
0

0
b 

0 
b

0
b 

0 
b
b
a
a
0

0
b 

b 
0
So, the first 3 rows forms a basis and the dimension is 3.
In each example the REF has a zero row and so the sets are all linearly dependent.
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