# If commutativity of vector space is omitted, can we still use other axioms to prove the commutativity?

Here I am thinking of using $-(x+y)$ and show that it equals $-(y+x)$.

$-(x+y)=-x-y$ by distributivity

=$-x-y+0=...$ Here I don't know how to continue, could someone suggest?

• If that would be possible, then why not drop that axiom?
– mvw
Commented Feb 8, 2016 at 21:33
• @mvw The axiom that the sum in a unital ring is commutative is redundant, but most people don't drop it when defining a unital ring.
– Pedro
Commented Feb 8, 2016 at 21:39
• @PedroTamaroff Of course, because it has a nice ring to it. Commented Feb 8, 2016 at 21:42
• @PedroTamaroff math.stackexchange.com/a/1284795/86776 claims the same. But really, why then not drop this "axiom"? Or at least demote it, like Pluto, into a theorem?
– mvw
Commented Feb 8, 2016 at 22:00

Yes this is possible even for a (left) module $M$ over a ring $R$ with a multiplicative identity $1$: $$(1+1)(x+y)=x+y+x+y$$ by the left distributive law. But also $$(1+1)(x+y)=2(x+y)=2x+2y=x+x+y+y.$$Cancelling $x$ and $y$ at both sides yields $x+y=y+x$, that is, $M$ must be an abelian group.
• Is it also true if $R$ has characteristic $2$? Commented Feb 8, 2016 at 21:47
• In char 2 the proof is even easier. In char 2, $x=-x$. So $-(x+y)=-y+(-x)$ (always true for the inverse of a product) and so $x+y=-(x+y)=-y+(-x)=y+x$. :) Commented Feb 8, 2016 at 21:52
• Yes exactly, in characteristic $2$, $(x+y)+(x+y)=0=(x+x)+(y+y)$, so same reasoning. Commented Feb 8, 2016 at 21:53