Comments on 296 Homework 6 Problem 4 (equivalence classes) [Lecture notes]

2 Pages·2011·0.19 MB·English

by  Ruthi Hortsch

Comments on 296 Homework 6 Problem 4
Course Assistant: Ruthi Hortsch
March 20, 2011
Here are some comments and partial solutions for Homework 6 Problem 4. Please review this carefully
as not only are finite fields important in many areas of mathematics, but these methods/ideas show up
later. In particular, the idea of considering equivalence classes as elements of a set shows up literally
everywhere in math (as quotient spaces/groups/rings/etc). So in particular, make sure you know how to
checkwhethersomething“well-defined”. Someofthisismoreformalandwithmoredetailsthannecessary,
but I wanted to make sure you understand the thinking.
(1), (2), (3) What does “well-defined” mean? For example, how do I show that
A : Z ×Z → Z
n n n
a×b (cid:55)→ a+b
is well-defined?1
Comment: Recall that a = {a+kn | k ∈ Z},2 i.e. elements of Z are subsets of Z. I could just as
n
easily write a(cid:48), where a(cid:48) ∈ Z such that a(cid:48) ∈ a, and then a = a(cid:48) (here a and a(cid:48) are called representatives
of a). Since there is nothing canonical about these choices and the definition of A depends on them, the
definition of A could be ambiguous, i.e. different choices of representatives of a and b could give different
images. So what needs to be checked is that if I choose different representatives, I will get the same image
(i.e. that this really is a function!).
A correct solution: Let a(cid:48) ∈ a and b(cid:48) ∈ b be choices of representatives. That is, a−a(cid:48) = k n and
1
b−b(cid:48) = k nforsomek ,k ∈ Z. ThenIwanttomakesurethatA(a×b) = A(a(cid:48)×b(cid:48)), thatisa+b = a(cid:48)+b(cid:48).
2 1 2
This happens if n divides (a+b)−(a(cid:48)+b(cid:48)). Well
(a+b)−(a(cid:48)+b(cid:48)) = a−a(cid:48)+b−b(cid:48)
= k n+k n
1 2
= (k +k )n
1 2
which implies that a+b = a(cid:48)+b(cid:48). We write a+b for A(a×b).
The situation for multiplication is almost exactly the same, only you need to consider a(cid:48)b(cid:48)−ab instead.
Checking the associative, commutative and distributive properties does not require anything particularly
difficult, but the idea is to simply reduce it to the properties in Z, for example:
a+b = a+b
= b+a
= b+a.
1This notation means A is a map from Z ×Z to Z that takes the element a×b∈Z ×Z (also written (a,b)) to the
n n n n n
element a+b∈Z , that is A(a×b)=a+b.
n
2It is not correct to say a=a+kn for some k∈Z
1
(4) A solution: We will prove the contrapositive: Suppose that n = n n . Then n n = 0 in Z .
1 2 1 2 n
Suppose n has an inverse m. Then n = mn n = m0 = m0 = 0. So by definition, this means that
1 2 1 2
n = kn for some k ∈ Z. But then n = n n = n kn, which implies that n k = 1, so since both n ,k ∈ Z,
2 1 2 1 1 1
this means n = ±1. This shows that if every element of Z has an inverse, then n is prime.
1 n
Note on rings (not required material): If R is a ring,3 the set of elements with multiplicative
inverses in R are called units. This uses that the units of Z are precisely ±1. It also uses the fact that if
a (cid:54)= 0 and ab = ac in Z, then b = c (I’m going to call this property ♣). Some rings have elements such
that ab = 0 but a, b are both nonzero. An element a such that there exists such a b is called a zero divisor.
A ring with no zero divisors is called a division ring and has property ♣ (why?). This solution shows that
if n is not prime, then Z has zero divisors. Another example of a ring with zero divisors is M (k), the
n n×n
matrix ring over a field k. Can you find zero divisors of M (k)? (Note that while M is a ring, it is
n×n n×n
not a commutative ring since multiplication is not commutative.)
(5) Suppose p is prime. Suppose that m ∈ Z such that m (cid:54)= 0. That means that p (cid:54) |m. Since the only
p
positive divisors of p are 1 and p, this means that gcd(m,p) = 1. On a previous homework, you proved
that this means that there are a,b ∈ Z such that am+bp = 1. So, if we take am = am = 1−bp = 1. So
m is the multiplicative inverse of a.
(6) There a few steps that need to be checked in this proof and I leave as an exercise to you.
Let F be a field with p elements. Since it is a field there are elements 0 and 1 ∈ F which act as
F F
additive and multiplicative identities. For k > 0, let k denote the sum of 1 with itself k times (that is
F F
n = 1 +···+1 ). Define a map:
F F F
(cid:124) (cid:123)(cid:122) (cid:125)
k
φ : Z → F
p
k¯ (cid:55)→ k
F
where if k < 0, we can use the division algorithm to chose a nonnegative representative (i.e. there are
q,r ∈ Z such that 0 ≤ r < p and k = qp + r, then define φ(k) = r ). We need to check that this is
F
well-defined. If k = k(cid:48) in Z (k > k(cid:48) ≥ 0), then there is an m ∈ Z such that k−k(cid:48) = mp and m ≥ 0. So
p
φ(k)−φ(k(cid:48)) = k −k(cid:48) = 1 +···+1 −1 +···+1 = 1 +···+1 = 1 +···+1 = (mp) . So to
F F F F F F F F F F F
(cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122) (cid:125)
k k(cid:48) k−k(cid:48) mp
show that φ is well-defined, it suffices to show that in F any multiple of p is zero.
Let f be the characteristic of F. Note it can’t be 0 since F is finite. Then consider the elements
{0 ,1 ,...,(f −1) }. This is a field isomorphic to Z (you should check this!). Since this is a field and
F F F f
F is a field containing it, F is a vector space over Z . So we can fix a basis {x ,...,x }, where d is the
f 1 d
dimension of F as a vector space over f, and note that this implies that p = |F| = fd. But the only
positive integers dividing p are 1 and p. Since 0 (cid:54)= 1 in any field, the characteristic of a field is never 1, so
this means that f = p. As a result of this, φ is well-defined.4
You should check that φ is a field morphism (i.e. preserved multiplication and addition) but this is not
difficult. Additionally, note that this implies that φ is injective, since if φ(k) = φ(k(cid:48)), so k = k(cid:48) . Since
F F
the characteristic of F is p, this implies that p divides k −k(cid:48), and so k = k(cid:48). Since we have an injective
field morphism between two finite fields of the same cardinality, this implies that φ is a field isomorphism.
3If you don’t know what a ring is, you can find a definition on wikipedia under “Ring (mathematics)”.
4This proof in fact shows that any finite field is of the form pd where p is the characteristic of the field. It can be shown
that for every prime p and d > 0, there is a unique field of characteristic p and size pd, which completely classifies all finite
fields. The proof above shows that this is exhaustive and uniqueness of addition, but showing existence and uniqueness of
multiplication uses some algebra that is a little more advanced.
2