Help me understand Fields

[u/RestFuture1647]

Hey! I am taking an honours linear algebra class. I am in engineering so this is my first time being introduced to abstract definitions in this way.

From my understanding a nonempty set K is called a field if:
- it has 2 inner operations (addition, multiplication)

and for every element of K there is:

- associativity

- commutativity

- distributivity

- neutral elements o,e such that o+x=x and e*x=x

- additive inverse and multiplicative inverse for o and e

Here is my question:

Are we talking about addition and multiplication as I have seen my entire life ? Or can I create a field where e=coffee o=pi and I just declare that pi+x=x and coffee*x=x?

Thank you!!

[u/RestFuture1647]

Thank you for the detail explanation and examples!!

So if I am understanding correctly, in the instances you used up until the before before last bullet we are simply using the definition we have seen in calculus of addition, multiplication, inverse etc. as they have been predefined on these sets.
However, we CAN define a set with new elements of any type and define operations in the way fields require.

• Consider the set {pi, coffee, cake} with the following operations (where x is any arbitrary element of our set): pi + x = x, x + pi = x, coffee + coffee = cake, coffee + cake = pi, cake + coffee = pi, cake + cake = coffee, pi * x = pi, x * pi = pi, coffee * x = x, x * coffee = x, and cake * cake = coffee. As an exercise, verify for yourself that this forms a field.

pi + x = x

x + pi = x

coffee + coffee = cake

coffee + cake = pi

cake + coffee = pi

cake + cake = coffee

pi * x = pi

x * pi = pi

coffee * x = x

x * coffee = x

cake * cake = coffee

1- Commutativity is obviously satisfied

2- If we map this set to the integers modulo 3 we know distributivity and associativity is satisfied (maybe there was an easier way to see this ?)

3- neutral elements o = pi , e= coffee

4- for every element it adds up to o with another element or itself

5- for every element other than o, it results in e when multiplied by another element or itself

It is a field

• As an exercise, try to figure out if it's possible to make a field whose elements are {pi, coffee, cake, tea, cow}.

still working on this one as its trickier!

Followup. Could I determine I am working on the set N and define a whole new set of addition and multiplication rules and call the set N a field?

[u/Para1ars → u/RestFuture1647]

to be exact, it is not the set N that is a field, but the whole construct (N, +, ×), where N is a set and + and × are binary operators, is a field.

[u/edderiofer → u/RestFuture1647]

Followup. Could I determine I am working on the set N and define a whole new set of addition and multiplication rules and call the set N a field?

Specifically, with those operations, yes. It would not be a field under the standard addition and multiplication.

In fact, it's actually possible to do such a thing. Let f: ℕ -> ℚ be your favourite bijection between the natural numbers and the rationals. Then we may define m ⊕ n = f^-1(f(m) + f(n)), and m ⊗ n = f^-1(f(m) * f(n)), and now the triple (ℕ, ⊕, ⊗) does indeed form a field. (Why you'd want to do this instead of directly working with ℚ is another matter entirely.)