Lecture 11:
We were in the process of proving:
Theorem: Zm is a field iff m is prime.
We proved ‘only if.’
If: We already know that Zm is a ring for all m. So it remains to
show that every nonzero element has a multiplicative inverse.
Suppose that m is prime. Let a ∈ Zm, a 6= 0
(mod m). Consider
the set
{a, 2a, 3a, . . . , (m − 1)a} (mod m)
Claim 1: None of these is 0
(mod m).
Proof: Suppose that for some 1 ≤ i ≤ m − 1, ia = 0 (mod m).
Then since m is prime, by Lemma 2, m divides i or a. But i is too
small and a 6= 0 (mod m), a contradiction.
Claim 2: {a, 2a, 3a, . . . , (m−1)a} (mod m) are distinct ( (mod m)).
Proof: Suppose two of these are the same: ia = ja (mod m),
with 1 ≤ i < j ≤ m − 1. Then (j − i)a = 0 (mod m). Thus,
either m|(j − i) or m|a, neither of which is possible. Thus, all of
{a, 2a, 3a, . . . , (m − 1)a} are distinct (mod m).
Thus,
{a, 2a, 3a, . . . , (m − 1)a} = {1, 2, 3, . . . , (m − 1)}
Thus, for some 1 ≤ i ≤ m − 1, ia = 1
exists and thus Zm is a field. (mod m). So, a−1 = i
Theorem: For every finite field F , |F | = pn where p is a prime
and n is a positive integer.
Proof (postponed) uses concept of the characteristic of a field.
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Theorem: For every prime p and positive integer n, there exists a
field F s.t. |F | = pn. In fact, the field is unique in some sense.
Proof (postponed) uses polynomials with coefficients in Zp.
The unique field of this size is denoted GF (pn) and called a Galois
field.
For n = 1, GF (p) = Zp.
Q: What is GF (4)? (it is not Z4, because that is not a field).
GF (4) is defined by the following addition and multiplication tables.
+ 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
Theorem: GF (4) as defined above is a field.
Proof: Verify that (GF (4), +) and (GF (4) \ {0}, ·) are abelian
groups; for this, deduce the group axioms and commutativity from
the addition and multiplication tables; verification of associativity
can be done on a case by case basis.
So, it remains to check distributivity. Can be done on a case-bycase basis Example of distributivity: a · (a + b) = a · 1 = a
a · a + a · b = b + 1 = a.
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Later, we will have a more systematic way of verifying associativity
and distributivity.
You can use the addition and multiplication tables above to compute in GF (4). For instance,
ab + a2 + b = 1 + b + b = 1
Compare with GF (4) with Z4:
∀x ∈ GF (4), x + x = 0; but in Z4, 1 + 1 = 2 6= 0.
Algorithm for finding inverses of elements in a GF (p) = Zp,
p prime.
Defn: Let a and b be positive integers. The greatest common
divisor of a and b (written gcd(a, b), or sometimes (a, b)) is the
largest integer which is a divisor of a and b.
Example: (30, 45) = 15, (30, 49) = 1.
The Euclidean Algorithm is an efficient method for finding the
gcd. This method is based on the Division Algorithm:
Theorem (Division Algorithm): Let a and b be nonnegative integers. There exist nonnegative integers q and 0 ≤ r < a such that
b = aq + r
Here, q is the quotient and r is the remainder.
Example: b = 306, a = 45,
306 = 45 · 6 + 36.
Euclidean Algorithm (EA): Let a and b be positive integers.
Iteratively apply the Division Algorithm until the remainder is 0: Set
r−1 = b, r0 = a
3
r−1 = r0q0 + r1,
r0 = r1q1 + r2,
r1 = r2q2 + r3,
...
ri−1 = riqi + ri+1,
...
rj−2 = rj−1qj−1 + rj ,
rj−1 = rj qj
0 ≤ r1 < r0
0 ≤ r2 < r1
0 ≤ r3 < r2
...
0 ≤ ri+1 < ri
...
0 ≤ rj < rj−1
Theorem: gcd(a, b) = rj .
Proof: will be posted.
Example: Let a = 657 and b = 963.
963
657
306
45
36
=
=
=
=
=
657 · 1 + 306
306 · 2 + 45
45 · 6 + 36
36 · 1 + 9
9·4
Thus, gcd(657, 963) = 9.
Why?
Step 1: Proof that 9 is a divisor of 657 and 963:
9 divides 9 and 36
9 divides 36 and 45
9 divides 45 and 306
9 divides 306 and 657
9 divides 657 and 963.
Step 2: Proof that any common divisor, d, of 657 and 963 divides
9:
d divides 657 and 963
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d divides 306 and 657
d divides 45 and 306
d divides 36 and 45
d divides 9 and 36
Bezout’s Identity: Let a and b be positive integers. Then there
exist integers m, n (not necessarily positive) such that
am + bn = gcd(a, b)
Proof: will be posted.
You can find m and n by working backwards through the Euclidean Algorithm:
9 =
=
=
=
=
=
=
=
45 − 36
45 − (306 − 45 · 6)
−306 + 45 · 7
−306 + (657 − 306 · 2) · 7
657 · 7 − 306 · 15
657 · 7 − (963 − 657) · 15
−963 · 15 + 657 · 22
657 · 22 − 963 · 15
So, m = 22 and n = −15.
Note that the m and n in Bezout’s Identity are by no means
unique. For example, let a = 2 and b = 3, with gcd = 1. Then
1 = 2 · (−1) + 3 · 1 = 2 · 2 − 3 · 1.
Finally, we apply EA and Bezout to find inverses in Zp when p is
prime. First, observe that if p is prime and a ∈ Zp and a 6= 0 mod
p, then gcd(a, p) = 1. By Bezout, there exist integers m, n such that
am + pn = 1. Thus, am = 1 mod p, and so a−1 = m mod p. Use
EA and Bezout to explicitly find m.
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For example, if p = 5 and a = 2, then −2 · 2 + 1 · 5 = 1 and so in
Z5, 2−1 = −2 = 3.
Lecture 12: Midterm 1
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