To prove that Z/pZ is a field, we need to show that it is a commutative ring with unity and that every nonzero element has a multiplicative inverse. We can do this by using the following facts:
Z/pZ is a set of equivalence classes of integers modulo p, where p is a prime.
Z/pZ has two binary operations, namely addition and multiplication, defined as:
[a] + [b] = [a + b] [a] * [b] = [a * b]
where [a] and [b] are equivalence classes of integers modulo p.
Z/pZ inherits the commutative, associative and distributive properties of addition and multiplication from Z.
Z/pZ has two special elements, namely zero and one, defined as:
[0] = {0, p, 2p, 3p, …} [1] = {1, p+1, 2p+1, 3p+1, …}
where [0] is the additive identity and [1] is the multiplicative identity.
Using these facts, we can show that:
Z/pZ is a commutative ring with unity. This follows from the commutative, associative and distributive properties of addition and multiplication, and the existence of zero and one as identity elements.
Every nonzero element in Z/pZ has a multiplicative inverse. This follows from the fact that p is a prime and the application of Bezout’s lemma. For any nonzero element [a] in Z/pZ, we have that gcd(a,p) = 1, since p is a prime and a is not divisible by p. By Bezout’s lemma, there exist integers x and y such that:
ax + py = 1
Taking both sides modulo p, we get:
[a] * [x] + [p] * [y] = [1]
But [p] = [0], so we have:
[a] * [x] = [1]
Therefore, [x] is the multiplicative inverse of [a]. Hence,
Z/pZ is a field.
(10) To prove that the congruence ax=b (mod m) has a solution if and only if b=0 (mod (a,m)), we need to use the following facts:
The congruence ax=b (mod m) means that m divides ax-b, or equivalently,
ax-b = km
for some integer k.
The notation b=0 (mod (a,m)) means that (a,m) divides b, or equivalently,
b = l(a,m)
for some integer l.
The notation (a,m) denotes the greatest common divisor of a and m.
Using these facts, we can show that:
If ax=b (mod m) has a solution, then b=0 (mod (a,m)). Suppose x is a solution of ax=b (mod m). Then we have:
ax-b = km
Multiplying both sides by y, where y is any integer such that
y(a,m) = 1
we get:
ayx - by = kmy
But ayx - by is divisible by (a,m), since both ayx and by are multiples of (a,m). Therefore,
by = l(a,m)
for some integer l. Hence,
b=0 (mod (a,m)).
If b=0 (mod (a,m)), then ax=b (mod m) has a solution. Suppose b=0 (mod (a,m)). Then we have:
b = l(a,m)
for some integer l. Dividing both sides by (a,m), we get:
b/(a,m) = l
Multiplying both sides by x, where x is any integer such that
x(a/m,(m/m)) = 1
we get:
bx/(a,m) = lx
But bx/(a,m) is congruent to ax/m modulo m/m, since both bx/(a,m) and ax/m are multiples of a/m. Therefore,
ax/m = lx (mod m/m)
Multiplying both sides by m/m, we get:
ax = lm/m + km
for some integer k. Hence,
ax=b (mod m).