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Math 431 Homework 4
2.1. Suppose that a and b are elements of a field F.
a) Show that if a · b = 0, then a = 0 or b = 0.
Either a = 0 or a 6= 0. If a = 0 we’re done. If not, multiply by a−1 to get
a−1 · (a · b) = (a−1 · a) · b) = 1 · b = b = a−1 · 0 = 0.
b) Show that (−a) · b = −(a · b).
We have
(a · b) + (−(a · b)) = 0
by definition and by the distributive law
(a · b) + (−a) · b = (a + (−a)) · b = 0 · b = 0.
Thus −(a · b) and (−a) · b are additive inverses of a · b. Let’s show that there is always
a unique additive inverse in a field. Suppose
a + b = 0,
and
a + c = 0.
Adding b we get
a + b + c = 0 + b = b,
but
(a + b) + c = 0 + c = c,
so b = c, and the additive inverse is unique. This implies −(a · b) = (−a) · b.
c)Show that −(−a) = a.
By definition of the additive inverse,
(−a) + (−(−a)) = 0,
and
(−a) + a = 0.
Since −(−a) and a are both additive inverses of −a,they are equal by the observation
in part b).
2.2. Suppose that a 6= 0 and b 6= 0 are elements of a field F.
a)Show that a−1 6= 0 and (a−1 )−1 = a.
If a−1 = 0 then
a · a−1 = a · 0 = 0,
but
a · a−1 = 1.
This contradicts the axiom
1 6= 0,
2
so
a−1 6= 0.
For the second part,
a · a−1 = 1,
so a is a multiplicative inverse of a−1 . Just as in 1.b, multiplicative inverses are unique:
a · b = 1,
a · c = 1,
gives
b = (a · c) · b = a · b · c = (a · b) · c = c.
b) Show that ab 6= 0 and (ab)−1 = a−1 b−1 .
If a · b = 0 then
a−1 · (a · b) = (a−1 · a) · b = b = 0.
This contradicts our starting assumption, so a · b 6= 0. For the second part
a−1 b−1 ab = b−1 (a−1 a)b = b−1 b = a,
so a−1 b−1 is the unique multiplicative inverse of ab.
2.4.Let Zp be the set of integers {0, 1, 2, . . . , p − 1}, and suppose that addition x ⊕ y
and multiplication x ⊗ y are carried out modulo p. That is, if xy is normal integer
multiplication and xy = pn + r, with 0 ≤ r < p, then x ⊗ y = r. Addition modulo p is
similar.
a) Construct addition and multiplication tables for Z2 and Z3 .
Here is the addition table for Z2 :
⊕ 0 1
0 0 1
1 1 0
The multiplication table for Z2 is
⊕ 0 1
0 0 0
1 0 1
The addition table for Z3 is
⊕
0
1
2
0
0
1
2
1
1
2
0
2
2
0
1
0
0
0
0
1
0
1
2
2
0
2
1
The multiplication table for Z3 is
⊕
0
1
2
3
b)Show that Z2 and Z3 are fields.
Many of the properties can be checked directly from the addition and multiplication
tables. The associative and distributive laws require a bit more work. Here is a check
of the associativity of addition for these two examples.
Z2
a
0
0
0
0
1
1
1
1
b
0
0
1
1
0
0
1
1
c a + (b + c) (a + b) + c
0
0
0
1
1
1
0
1
1
1
0
0
0
1
1
1
0
0
0
0
0
1
1
1
Z3
A full table for Z3 would have 27 rows, so to simplify the work let’s make an
observation. If a = 0 then both sums are equal, being b + c. Similarly we find
a + (b + c) = (a + b) + c if any of a, b, c is zero. After eliminating these cases the table
is more manageable, and we have
a
1
1
1
1
2
2
2
2
b
1
1
2
2
1
1
2
2
c a + (b + c) (a + b) + c
1
0
0
2
1
1
1
1
1
2
2
2
1
1
1
2
2
2
1
2
2
2
0
0
c) Is Z4 a field?
No, since 2 6= 0, but
2 · 2 = (1 + 1) · (1 + 1) = 1 + 1 + 1 + 1 = 0.
As we saw in 2.2b, the product of two nonzero elements in a field is never 0.
2.5. Show that if 2 = 1 + 1 6= 0 in a field F, then 4 = 1 + 1 + 1 + 1 6= 0 in F.
Notice that
4 = 1 + 1 + 1 + 1 = (1 + 1) · (1 + 1) = 2 · 2.
If 4 = 2 · 2 = 0, then
2−1 · 2 · 2 = 2 = 0.
The contrapositive of this last statement is if 2 6= 0 in a field F, then 4 6= 0 in F.
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2.6. Suppose that F satisfies the field and order axioms. Show that q ≥ 1 and p ≥ 0
implies pq ≥ p.
First we have q − 1 ≥ 0 by O.5. By axiom O.6
p · (q − 1) = p · q − p ≥ 0,
and by O.5 again pq ≥ p.