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Fields

Now we've had a gentle introduction to abstract mathematics with groups, we're ready to explore a slightly more complex structure: Fields.

Just like groups, fields have a set of elements that can be combined. However, this time they can be combined with not one, but two operations. One operation, \cdot, behaves like multiplication and the other, ++, behaves like addition.

These two operations must obey the following properties:

Closure

(a+b)F(ab)F(a + b) \in F \\[3ex] (a \cdot b) \in F

Very simple one we already saw in groups. Combining any 2 elements from the field under any operation should produce another member of the field.

Associativity

a+(b+c)=(a+b)+ca(bc)=(ab)ca + (b + c) = (a + b) + c \\[3ex] a \cdot (b \cdot c) = (a \cdot b) \cdot c

Another one we've seen already from groups. The brackets must behave the same way with both operations.

Identity

0,a+0=0+a=a1,a1=1a=a\exists 0, a + 0 = 0 + a = a \\[3ex] \exists 1, a \cdot 1 = 1 \cdot a = a

There must exist an additive and multiplicative identity in the field. The elements 00 and 11 are the additive and multiplicative identities, respectively. The decision to label them as 00 and 11 is arbitrary, but it is a common convention.

Inverse

(a),a+(a)=(a)+a=0a1,a0,aa1=a1a=1\exists (-a), a + (-a) = (-a) + a = 0 \\[3ex] \exists a^{-1}, a \neq 0, a \cdot a^{-1} = a^{-1} \cdot a = 1

Every element must have both an additive and multiplicative inverse. The elements a-a and a1a^{-1} are the additive and multiplicative inverses, respectively. Again the decision to label them as such is arbitrary, but a common convention.

Note the special case defined here for multiplication. The additive inverse doesn't need to have a multiplicative inverse. This makes sense becase there's nothing we can multiply by 0 to get 1.

Commutativity

a+b=b+aab=baa + b = b + a \\[3ex] a \cdot b = b \cdot a

We should get the same result from either operation even when we swap the order of the elements.

Remeber that this wasn't a strict requirement for groups, but it will be in fields.

Distributivity

a(b+c)=(ab)+(ac)a \cdot (b + c) = (a \cdot b) + (a \cdot c)

This is a completely new rule that we didn't have with groups. This tells us how the 2 operations interact when used together.


Now we've defined our rules. Let's take a look at an example of a field!

Integers modulo 3 Z3\mathbb{Z}_3

This is like the set of integers, but with a twist. We only care about the remainders when divided by 3. So, we have the following elements:

Z3={0,1,2}\mathbb{Z}_3 = \{0, 1, 2\}

We can think about this as counting on a clock with 3 hours, starting from 0:

A representation of counting mod 3 by a clock with the numbers 0, 1 and 2

Say we want to check what 2+22 + 2 is. Normally,2+2=42 + 2 = 4, but we don't have 4 in our set of elements. So we'll need another way to think about addition.

With normal addition, to calculate 2+22 + 2, we'd point to the number 22 on the number line and then move 2 places further along. Let's do that but using our clock:

A representation of adding 2 and 2 to get 1 when counting mod 3

So in our field, 2+2=12 + 2 = 1! Earlier we said the elements of our field were the integers remainder when divided by 3, so this result makes sense since 4 divided by 3 gives a remainder of 1. This gives us another way to calculate results here. Instead of using the clock, we can just do the calculation on the normal number line and then take the remainder of the result when divided by 3.

And multiplication just behaves as we would expect. Let's draw out the addition and multiplication tables:

++001122
00001122
11112200
22220011
\cdot001122
00000000
11001122
22002211

So does this satisfy our rules? Let's check!

  • Closure
    • We can see every cell of each table is also an element of the field so this is clearly true.
  • Associativity
    • This one's not hard to prove, but it takes a lot of work since we'd need to check every combination. Here's one example: 2(10)=20=(21)02 \cdot (1 \cdot 0) = 2 \cdot 0 = (2 \cdot 1) \cdot 0
  • Identity
    • From the addition table we can see that 00 is the additive identity and from the multiplication table we can see that 11 is the multiplicative identity. This is because they both have the property that when combined with any other element, they don't change the element.
  • Inverse
    • We now know the additive identity is 00, so for each element we need to find another element that when added to it gives 00. We can easily see this in our table 1+2=01 + 2 = 0 which means 1 and 2 are eachother's inverse respectively.
    • For multiplication, we know the identity is 11 and we can see that11=11 \cdot 1 = 1 and 22=12 \cdot 2 = 1 so 1 and 2 are each their own multiplicative inverse.
  • Commutativity
    • There's a very nice visual way we can prove this. If we draw out a line from the top left of the table to the bottom right, you will notice that the table is symmetrical along this line. This means that the order of the elements doesn't matter.
  • Distributivity
    • Like associativity, this one is a bit of a pain to prove. We'd need to check every combination. Here's one example: 2(1+0)=21=(21)+0=(21)+(20)2 \cdot (1 + 0) = 2 \cdot 1 = (2 \cdot 1) + 0 = (2 \cdot 1) + (2 \cdot 0)

So we've proved that Z3\mathbb{Z}_3 is a valid field!

Although we have only a few rules on fields, there are actually many interesting properties that we can prove from these rules. We'll do one example here to give a flavour and then leave some more to the exercises.

One of the things we know from basic maths is that if ab=0a \cdot b = 0 then either a=0a = 0 or b=0b = 0. Here's how we can prove this using the rules above:

We'll do a proof by contradiction. We'll assume the opposite is true and then show that it implies a contradiction. So we'll assume a0a \neq 0 and b0b \neq 0. Then we have:

ab=0(ab)+a=0+a(ab)+(a1)=aa(b+1)=aa \cdot b = 0 \\[3ex] (a \cdot b) + a = 0 + a \\[3ex] (a \cdot b) + (a \cdot 1) = a \\[3ex] a \cdot (b + 1) = a

Since a0a \neq 0 we know it must have a multiplicative inverse so we can multiply both sides by it to get:

a1(a(b+1))=a1a(a1a)(b+1)=11(b+1)=1b+1=1(b+1)+(1)=1+(1)b+(1+(1))=0b+0=0b=0a^{-1} \cdot (a \cdot (b + 1)) = a^{-1} \cdot a \\[3ex] (a^{-1} \cdot a) \cdot (b + 1) = 1 \\[3ex] 1 \cdot (b + 1) = 1 \\[3ex] b + 1 = 1 \\[3ex] (b + 1) + (-1) = 1 + (-1) \\[3ex] b + (1 + (-1)) = 0 \\[3ex] b + 0 = 0 \\[3ex] b = 0

This violates our assumption! An equivalent argument could be made for aa. So we have proved our statement!

Move on to the exercises to see if you can prove some more interesting facts!

Exercises

Exercise 1

Prove that 0a=00 \cdot a = 0

Seems fairly obvious in normal math, but requires careful proof when we're dealing with more abstract objects.

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Exercise 2

Prove that (1)a=a(-1) \cdot a = -a.

Note here the usage of the term 1-1. Fields are abstract so don't have to represent numbers. What we're actually describing here is the additive inverse of the multiplicative identity. The notation 1-1 is just for convenience.

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Exercise 3

Prove that, for our Z3\mathbb{Z}_3 field, that following holds: (a+b)3=a3+b3(a+b)^3 = a^3 + b^3 where we define x3=xxxx^3 = x \cdot x \cdot x. Note here I'm leaving out the brackets for simplicity.

There are 2 ways to prove this. You can check every possible combination, or you can use the rules we defined. See if you can find both proofs.

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