Given a left module $M$ over a ring $R$ where $0_M$ is the additive identity in $M$ and $0_R$ is the additive identity in $R$, I am trying to work out why $r0_M= 0_M$ for all $r \in R$. So far I have \begin{align*} & r0_M = r(0_M+0_M)\\ \Rightarrow & r0_M = r0_M + r0_M\\ \Rightarrow & r0_M + (-r)0_M = r0_M + r0_M + (-r)0_M\\ \Rightarrow & (r - r)0_M = r0_M + (r - r)0_M\\ \Rightarrow & 0_R0_M = r0_M + 0_R0_M\\ \Rightarrow & 0_R0_M = (r + 0_R)0_M\\ \Rightarrow & 0_R0_M = r0_M\\ \end{align*} I am having trouble seeing how $0_R0_M = 0_M$
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Can you prove that a function $f:G\to G$ from an abelian group to itself such that $f(g)f(h)=f(gh)$ for all g and all h In G maps the identity of G to itself? – Mariano Suárez-Álvarez Jul 04 '16 at 16:28
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Am I overlooking something? Why in the second line can't you just add directly the additive inverse of $r0_M$ to both sides? – Aloizio Macedo Jul 04 '16 at 17:12
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It's just as easy to prove the more general result that for any $m \in M$, $0_R m = 0_M$.
Proof: $$\begin{align} 0_R m &= (0_R + 0_R) m \\ &= 0_R m+ 0_R m \end{align}$$
Now add $-(0_R m)$ to both sides, and get $$0_R m + -(0_R m) = 0_R m + 0_R m + -(0_Rm)$$ $$0_M = 0_R m + 0_M$$ $$0_M = 0_R m$$
However, as noted in Aloizio Macedoo's comment on the original question, you really don't need to know that $0_R 0_M =0_M$ order to solve your original question, because from the second line of your proof you can just add $-(r0_M)$ to both sides. The argument is virtually identical to the one above.
mweiss
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