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Suppose we have a ring $R$, $r \in R$ and we make the additive and multiplicative identities coincide $1 = 0$. How does then the following proof of it being the zero ring hold: $$r = 1r = 0r = 0$$

My confusion arises from the fact that in the definition(axioms) of the ring there is no property that says that the binary operation of multiplication with a $0$ will produce a $0$.

I might have this all wrong, can anyone clear this up?

Michael Munta
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    One can prove that the multiplication with $0$ in the ring gives you $0$ as answer. You could look it up at e.g Wikipedia – Luke Mar 10 '20 at 10:41
  • Can you provide a link? – Michael Munta Mar 10 '20 at 12:43
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    I can even provide you a stackexchange link, here you go: https://math.stackexchange.com/questions/1483716/in-a-ring-how-do-we-prove-that-a-0-0 – Luke Mar 10 '20 at 12:48

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You can show this is the case since: $$ 0r = (0+0)r = 0r + 0r \\ \implies 0r - 0r = 0r + 0r - 0r \\ \implies 0 = 0r $$

Alternatively: $$ r = 1r = (1+0)r = 1r + 0r = r + 0r \\ r = 1r = (0 + 1)r = 0r + 1r = 0r + r $$ So $0r$ is indeed the additive identity of the ring, thus it's equal to $0$ (since there can only be one identity of a group).

Niki Di Giano
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  • I dont understand this. Why does this show that $0r=0$? – Michael Munta Mar 10 '20 at 10:46
  • You can see from this proof that the element $0r$ is the identity in the operation $+:R\times R \to R$, thus it's the zero of the ring. I'll add another proof that shows this a little bit more clearly. – Niki Di Giano Mar 10 '20 at 10:50