18
$\begingroup$

I seem to have found a proof that the commutativity of $+$ follows from the other field axioms. It is as follows:

Let $(k,+,\cdot)$ be a structure satisfying all field axioms except commutativity of addition, with $a,b\in k$. Then

1) $(a + b)\in k$ and $(b+a)\in k$ (closure of addition)

2) $-(a+b)\in k$ and $-(b+a)\in k$ (invertible addition)

3) $(a+b) + [-(b+a)] =(a+b) + (-b) + (-a)$ (distributivity of $\cdot$ over $+$)

4) $(a+b) + (-b) + (-a) = a + (b + (-b)) + (-a)$ (associativity of $+$)

5) $a + (b + (-b)) + (-a) = a + 0 + (-a) = a + (-a)$ (invertibility and identity of $+$)

6) $(a + b) + [-(b+a)] = 0$ (identity of $+$)

7) $(a+b) = (b+a)$ (invertibility of $+$) $\,\square$

But commutativity is a field axiom, so it must be necessary. Given that, what is wrong with this proof? Can the final conclusion (7) not be drawn without commutativity?

$\endgroup$
  • 3
    $\begingroup$ I'd use more care around line 3. Are you assuming that $-x = (-1) \cdot x$? $\endgroup$ – aschepler Oct 27 '14 at 21:11
  • $\begingroup$ Yes, I was. When I wrote this down initially I wasn't really paying attention and overused subtraction. $\endgroup$ – theage Oct 27 '14 at 21:14
  • $\begingroup$ We can still prove $-x = (-1)\cdot x$ for every $x$, though. $\endgroup$ – aschepler Oct 27 '14 at 21:20
  • $\begingroup$ @aschepler Curious about how you'd show that... $\endgroup$ – AlexR Oct 27 '14 at 21:23
  • $\begingroup$ @aschepler I'm pretty sure that needs commutativity, right? $\endgroup$ – theage Oct 27 '14 at 21:23
16
$\begingroup$

Yes, commutativity of addition is a logical consequence of the other field axioms. Nothing wrong with being a little bit redundant.

The missing lemma for the proof:

Let $x \in k$. Then

$$0 = 0 \cdot x = (1 + (-1))\cdot x = 1 \cdot x + (-1) \cdot x = x + (-1)\cdot x$$

Similarly,

$$0 = 0 \cdot x = (-1 + 1)\cdot x = (-1) \cdot x + 1 \cdot x = (-1)\cdot x + x$$

So $(-1)\cdot x$ is the additive inverse of $x$,

$$-x = (-1) \cdot x$$

$\endgroup$
20
$\begingroup$

Since we are talking about Fields here and the field axioms are an extension of the ring axioms (a field can be defined as a commutative ring with $1$), you are correct to see redundancy.
However, since the field axioms are constructed from ring axioms and in rings the property is not redundant, it's justified to keep this redundancy. The essential part of the field axioms that generates the redundancy is that $1\in F$ (the field has a multiplicative identity) and it is a ring.

Without this property $-(b+a)$ (the additive inverse of $b+a$) is not guaranteed to be $-1 \cdot (b+a)$ since $-1$ may not even exist. Instead, $-(b+a) = (-a)+(-b)$ because $b+a+(-a)+(-b)=0$.

We can prove that for a Ring to have commutative addition it suffices that it has a $1$:

$$\begin{align*} a+a+b+b & = (1+1)\cdot a+(1+1)\cdot b & \text{left-distributivity}; 1\cdot a = a \\ & = (1+1)\cdot(a+b) & \text{right-distributivity}\\ &= 1\cdot(a+b) + 1\cdot(a+b) & \text{left-distributivity}\\ & = a+b+a+b &\text{right-distributivity} \\ \Rightarrow a+b&=b+a & \text{cancellation} \end{align*}$$ Proof taken from here.
Now although the property is redundant for fields, it isn't in the context of a commutative ring so it's a good idea to still keep all the (non-redundant) ring axioms for consistency since every field is a ring.

$\endgroup$
  • $\begingroup$ So commutativity is redundant then, in the sense that it is a logical consequence of the complete set of field axioms modulo commutativity? $\endgroup$ – theage Oct 27 '14 at 21:40
  • $\begingroup$ @theage Yes, it is indeed. $\endgroup$ – Daniel Fischer Oct 27 '14 at 21:41
  • $\begingroup$ @theage Yes, it is in the setting of fields. Your proof is valid when given together with (or extended by) the proof of $-x = (-1) \cdot x$ $\endgroup$ – AlexR Oct 27 '14 at 21:41
  • $\begingroup$ Okay, I see... that's actually somewhat interesting! I'm changing the accepted answer, but thanks for the help. $\endgroup$ – theage Oct 27 '14 at 21:42
  • $\begingroup$ @theage Always happy to help. The (now) accepted answer is more direct so I agree on changing that. Sorry for any confusion that arose from me overlooking the 'field' in your post. I hope it became clear now. $\endgroup$ – AlexR Oct 27 '14 at 21:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.