# Proving C (complex) is a field.

1. May 28, 2010

### jspectral

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

Prove that C is a field.

2. Relevant equations

Field Axioms.

3. The attempt at a solution

I know there are 10 field axioms, but is it enough to just say something like:
Let a,b,c be in C
(a+b) + c = a + (b + c)
So satisfies addition associativity, and then go ahead and do all the others?
Just doesn't seem like proof...How else could I go about proving it?

2. May 29, 2010

### talolard

I think thats how you do it, but ina little more detail.
let x,y,z belong to C.
then $$x= a+ia', y=b+ib', z=c+ic'$$
then $$(x+y)+z = (a+ia'+b+ib')+c+ic'=a+ia'+b+ib'+c+ic'=a+ia'+(b+ib'+c+ic')$$

3. May 29, 2010

### jambaugh

I think in part the answer to "How does one go about proving it?" depends on the form of the construction of the complex numbers that you're using.

If you are constructing the complex numbers as the set of ordered pairs of real numbers with a specific + and x operation then you need to parse each Field axiom with these C operations.

Thus to show associativity of addition:
-------------
Let: $$c_1 = (a_1,b_1), c_2 = (a_2,b_2), c_3 = (a_3,b_3)$$ be 3 complex numbers.
Observe
$$[ c_1 + c_2] + c_3 = [ (a_1,b_1) + (a_2,b_2)] + (a_3,b_3)$$
$$= [(a_1+a_2, b_1+b_2)] + (a_3,b_3)$$ (by def of + in C)
$$= ([a_1+a_2]+a_3,[b_1+b_2]+b_3)$$ (by def of + in C)
$$= (a_1 + [a_2+a_3], b_1 + [b_2 +b_3])$$ (by assoc. in R)
...
$$= (a_1,b_1) + [(a_2,b_2)+(a_3,b_3)]$$ (by def of + in C twice applied)
$$= c_1 + [c_2+c_3]$$
hence C is associative under +.
-------------
Rather tedious but those are the steps if you define C this way.

Or you may construct complex numbers as 2-dim vectors with a specific multiplication operation, as vectors you already have the addition axioms satisfied. One would then have to parse the definition of the product to show it satisfies the field product axioms.

You may construct complex numbers as "the ring of polynomials in 1 variable ( i ) modulo the identity: ( i^2 + 1 = 0)". One then has all the ring axioms automatically satisfied and you only need to prove that this ring is in fact a field.

I think the critical point in most cases is showing each non-zero element has a multiplicative inverse, i.e. that the ring C is in fact a field.

4. May 29, 2010

### jambaugh

It may be helpful to use a C or R subscript to show the distinct operations in C vs in R, at least in your scatch work. This way you'll understand where you are working with the operation whose property you wish to prove and when you are working with an operation you may already assume has a given property.

Thus for my first two lines below you'd write:
$$[ c_1 +_C c_2] +_C c_3 = [ (a_1,b_1) +_C (a_2,b_2)] +_C (a_3,b_3)$$
$$= [(a_1+_R a_2, b_1+_R b_2)] +_C (a_3,b_3)$$

Or you can just forget the R subscripts, and also you could just use a circled + for the complex addition operation. That I think can help you understand when you've parsed down to the point where you can automatically use the axioms of the real numbers or where you've not yet expanded the new operation in terms of its definition.

5. Jan 26, 2011

### aya2002

guide me to a book that explain the complex numbers over GF(q).

Regards

6. Jan 26, 2011

### mathstew

If you have already been doing group theory then you should be aware that

C forms an abelian group under addition and the nonzero elements of C forms an abelian group under multiplication.

So, obviously C is a commutative ring with 1. Now every nonzero element is in C's multiplicative group implies every nonzero element is a unit or has an inverse in C which implies that C is a field.

Building C from the ring R (the quadratic ring R adjoin i) would be the easiest way if you have to prove everything from scratch.

Otherwise you could prove this equivlent statement. Since an ideal M of a commutative ring R is maximal iff R/M is a field.

You would only need to prove that the principal ideal generated by (x) is maximal in the polynomial ring in C[x]. This is because C[x]/(x) is isomorphic to C.