# Math Help - Commutativity of Addition

Is commutativity of addition necessary as an axiom in the Field Axioms or can it be proved by the others?

2. Originally Posted by rualin
Is commutativity of addition necessary as an axiom in the Field Axioms or can it be proved by the others?
Yes it is necessay*.
It cannot be proved by others.

In fact, I can prove that it cannot be proved
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Define $\mathbb{H} = \mathbb{R}\times \mathbb{R}\times \mathbb{R} \times \mathbb{R} = \{(a,b,c,d)|a,b,c,d\in \mathbb{R}\}$.

So any element in $\mathbb{H}$ is an ordered 4-tuple $(a,b,c,d)$ but it is more convient to write $a+bi+cj+dk$. We define the following products:
$i^2 = j^2 = k^2 = -1$
And,
$ij=k \ \ jk = i \ \ ki = j \ \ ji = -k \ \ kj=-i \ \ ik = - j$

So for example if we need to multiply $(2+i)(1+j)$ we would write $2+i+j+ij$ and get $2+i+2j+k$.

But if $(1+j)(2+i)=2+2j+i+ji=2+i+2j-k$.
So $(1+j)(2+i)\not = (2+i)(1+j)$.

I leave it to you to verify that $\mathbb{H}$ has the following properties:
$h_1+h_2=h_2+h_1$
$h_1+0=h_1$
$\forall h_1 \exists h_2 \ h_1+h_2=0$
$h_1+(h_2+h_3)=(h_1+h_2)+h_3$
$h_1(h_2h_3)=(h_1h_2)h_3$ --- > Boring and Long.
$h_1(h_2+h_3)=h_1h_2+h_1h_3$ ---> Boring and Long.
$(h_2+h_3)h_1=h_2h_1+h_3h_1$ --->Boring and Long.
$1h_1=h_11=h_1$
$\forall h_1\not = 0 \exists h_2 \ h_1h_2=h_2h_1=1$
This show it has everything EXCEPT $h_1h_2=h_2h_1$

*)If the field is finite it is not necesarry. This supprising result is not easy to prove.

3. According to a professor's notes I found online, Leonard Dickson discovered it can be proved from the other axioms. Here is how the proof in the notes goes:

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Let $F$ be any field. Then $+$ is a commutative operation on $F$.

Proof: Let $x,y\in F$. Then,

$(1+x)(1+y)=(1+y)(1+x)$ by commutativity of multiplication
$(1+x)\cdot 1 + (1+x)\cdot y = (1+y)\cdot 1 + (1+y)\cdot x$ by distributive law
$(1+x) + (1+x)\cdot y = (1+y) + (1+y)\cdot x$ by definition of multiplicative identity
$1 + (x + (1+x)\cdot y) = 1 + (y + (1+y)\cdot x)$ by associativity of addition
** $x + (1+x)\cdot y = y + (1+y)\cdot x$ by cancellation law for addition
$x+(1\cdot y) + (x\cdot y) = y + (1\cdot x) + (y\cdot x)$ by distributive law
$x+(y+(x\cdot y)) = y+(x+(y\cdot x)$ by definition of multiplicative identity
$(x+y)+(x\cdot y) = (y+x)+(y\cdot x)$ by associativity of addition
$(x+y)+(x\cdot y) = (y+x)+(x\cdot y)$ by commutativity of multiplication
** $x+y = y+x$ by cancellation law of addition

Hence, $+$ is commutative.

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** What doesn't completely convince me is the cancellation step but what do I know.

What do you think?

4. Originally Posted by rualin
What do you think?
I looked it over quickly and it looks right.

The only problem is the way the theorem is stated it should say let F be a field "expect for possibly commutativity of addition" then it must be commutative under addition.

But the problem is that is not the "right" definition.

In fact a more appropriate definition is that a field is a commutative division ring.

See Here for the definition of a field.