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?
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2$\begingroup$ If that would be possible, then why not drop that axiom? $\endgroup$– mvwFeb 8, 2016 at 21:33
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7$\begingroup$ @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. $\endgroup$– Pedro ♦Feb 8, 2016 at 21:39
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1$\begingroup$ @PedroTamaroff Of course, because it has a nice ring to it. $\endgroup$– Clement C.Feb 8, 2016 at 21:42
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$\begingroup$ @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? $\endgroup$– mvwFeb 8, 2016 at 22:00
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1 Answer
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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.
answered Feb 8, 2016 at 21:45
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$\begingroup$ Is it also true if $R$ has characteristic $2$? $\endgroup$ Feb 8, 2016 at 21:47
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$\begingroup$ The proof above doesn't use anything about characteristic 2, so yes. (+1) $\endgroup$ Feb 8, 2016 at 21:50
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3$\begingroup$ 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 $x+y=-(x+y)=-y+(-x)=y+x$. :) $\endgroup$ Feb 8, 2016 at 21:52
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2$\begingroup$ Yes exactly, in characteristic $2$, $(x+y)+(x+y)=0=(x+x)+(y+y)$, so same reasoning. $\endgroup$ Feb 8, 2016 at 21:53
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$\begingroup$ (Of course the very same proof applies to prove a ring with unit must have commutative addition) $\endgroup$– Pedro ♦Feb 8, 2016 at 22:04