# Equality in rng with no zero divisors.

I'm working on this problem, but I'm missing some manipulation.

Suppose $R$ is a rng without zero divisors and has elements $a$ and $b\neq 0$ such that $ab+kb=0$ for some $k\in\mathbb{N}$ (that is, $ab+\underbrace{b+b+\cdots+b}_{k\ \textrm{times}}=0$.) I'm trying to show that $ca+kc=0=ac+kc$ for all $c\in R$.

I observe that $$(ca+kc)b=cab+kcb=c(ab+kb)=c\cdot 0=0$$ so $ca+kc=0$ since $b\neq 0$ and $R$ has no zero divisors. I can't get the other equation. I note it's equivalent to showing $ac=ca$ for all $c\in R$, so I was trying to multiply $ac-ca$ by something nonzero to get $0$ to conclude but with no luck. Does anyone see what to do to get the other equality? Thanks.

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Why do you write $0ac$ rather than $0$? Did you intend to write "=" between $0$ and $ac$? Also is your rng assumed to be commutatitive? If not you cannot extract a left factor $c$ from $kcb$. – Marc van Leeuwen Jul 31 '12 at 20:56
Marc, $k \in \mathbb{N}$ here, so $kcb=cb+cb+...+cb.$ Threw me as well. – Kevin Carlson Jul 31 '12 at 21:17
Sorry, I must have missed the $=$ between $0$ and $ac$. My rng is not assumed to be commutative, but I figure I can extract a left factor of $c$ from $kcb$ since it is a sum of $k$ many $cb$ terms. – Chelsea Dirks Jul 31 '12 at 23:37

Since you've already shown $ca+kc=0,$ let's ignore the information about $b$ now and work directly with $c \neq 0$ (noting the result is trivial for zero.)
Try this: $c(ac+kc)=cac+kc^2=(ca+kc)c=0$ by the previous result, and since the rng's integral, $ac+kc=0.$
Thanks Kevin! ${}$ – Chelsea Dirks Jul 31 '12 at 23:36