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Let $R$ be a ring except that addition and multiplication in $R$ are not assumed to be commutative. Show that addition must be commutative.

(Hint : for $a,b,c,d \in R$ look at $(a+b)(c+d)$)

The question is not very clear to me, especially the first statement and haven't been able to make much progress, looking for some hints to finish the proof.

spaceman_spiff
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  • The question is answered to death by Andrew Ursitti in the linked question, even considering rings without identity. – rschwieb Apr 13 '17 at 20:27

2 Answers2

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Let $a,b \in R$. Look at $$x := (1+1) (a+b).$$ On the one hand, this is equal to $x=(1+1)a + (1+1)b$. On the other hand, this is equal to $x=1(a+b) + 1(a+b)$.

If you develop these two expressions, you will get $a+b=b+a$ as desired.

Watson
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As Watson points out, considering $(1+1)(a+b)$ and simplifying proves the ring to be abelian.

Remark that, in a non-unital ring, the addition need not be abelian. Take your favorite non-abelian group $G$, denote its group law by $\cdot$, and denote the "ring multiplication" $\circ : G \times G \to G$ by just $(x, y) \mapsto e_G$. This is easily verified to be a rng, with non-abelian addition (although it does have commutative multiplication)

MT_
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  • I think the terminology here is incorrectly used. A ring in which the underlying $+$ group is not abelian is called a nearring not a rng. Rng just means no identity is assumed (but the underlying group is still abelian.) Btw, an abelian ring commonly means that the ring is either commutative or else its idempotents are all central. – rschwieb Apr 13 '17 at 20:24
  • @rschwieb I guess it was unclear. I meant "if you take OP's premise, replacing occurrences of the word "Ring" with "non-unital Ring", then the result does not necessarily hold." In particular, the assumption of the existence of 1 in the Watson's proof is necessary. – MT_ Apr 14 '17 at 22:00