# Proving an axiom

1. Oct 2, 2009

### sara_87

1. The problem statement, all variables and given/known data

I was asked to prove the axioms of a field.
so, if we look at the first one:
commutativity: a+b=b+a and a*b=b*a where a and b belong in the set of the field

2. Relevant equations

3. The attempt at a solution

it's tempting to just substitute values but i know this is the wrong approach since this wouldnt prove the axiom for all values in the set. so im stuck.

2. Oct 2, 2009

### aPhilosopher

It would help a lot if we knew what the set was.

3. Oct 2, 2009

### sara_87

the question just says: prove the axioms of a field.
but the chapter we are studying is complex numbers so maybe i have to prove them for complex numbers. if so, would this proof be good enough:
for: a+b=b+a: where a=u+vi and b=x+yi
u+vi+x+yi=x+yi+u+vi
??

4. Oct 2, 2009

### aPhilosopher

mmmmm. I suppose you can't just say that C is isomorphic to R[x]/(x2 + 1) and that because (x2 + 1) is maximal, it must be a field?

How you write the proof depends on how you've defined complex numbers and how you've defined addition and multiplication. Is the definition something like a + bi = (a, b)?

5. Oct 2, 2009

### sara_87

what do you mean '(x^2+1) i maximal' here?
yes that's the definition we use for complex numbers.

6. Oct 2, 2009

### aPhilosopher

Ok, then what you want to do for commutativity of addition is (a, b) + (c, d) = (a + c, b + d) = (c + a, d + b) = (c, d) + (a, b).

Do you notice how we "got a free ride" once we got a and c in the same tuple? That just reduces to commutativity of the reals, which I assume you can take for granted.

As far as the maximal thing, it just means that the ideal generated by x1 + 1 in the polynomial ring over R isn't contained in any others. You'll probably learn about it if you're taking an algebra class. It was mainly just a joke.

7. Oct 2, 2009

### sara_87

:) ok
I understand the proof now.
for a*b=b*a, would this be convincing:

8. Oct 2, 2009

### aPhilosopher

You don't need to stick the term that I struck out in there. Remember, i is just (0, 1). Other than that, it looks great though.

9. Oct 2, 2009

### sara_87

thank you. now i can prove the other axioms.