# Multiplication of rings is an abelian group homomorphism

Let $R$ be a ring without identity.

Suppose that the multiplication $\cdot : R \times R \rightarrow R$ is an abelian group homomorphism.

For $a, b \in R$ what can we conclude about the product of $a \cdot b$ ?

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If you call the multiplication abelian group homomorphism it then must be that under that multiplication $\;R\;$ is a group , so how come it has no (multiplicative) unit, what you probably call "identity"? – DonAntonio Jun 9 '13 at 10:14
@DonAntonio $R$ is an abelian group under addition! – Zhen Lin Jun 9 '13 at 10:23
@DonAntonio I don't think that's right, I think that here multiplication is a homomorphism between the two groups $(R\times R,+)$ and $(R,+)$. – Tom Oldfield Jun 9 '13 at 10:30
I know that, @Zhen ! But if the problem states that the product is a group homom. I think it follows that is under that multiplication...or else let the OP address and clear out this, of course. – DonAntonio Jun 9 '13 at 10:56
Perhaps so, @TomOldfield...yet I think there's room for quite some confussion here – DonAntonio Jun 9 '13 at 10:58

Let $m\colon R\times R\to R$ be the multiplication of $R$ and suppose $m$ is an abelian group homomorphism on the addition groups of $R\times R$ and $R$. Then, for any $a,b\in R$: $$ab = m(a,b) = m((a,0)+(0,b)) = m(a,0)+m(0,b) = 0+0 = 0.$$