Let $R$ be a ring and $S$ be a subring of $R$. Suppose that $R$ does not have unity, but $S$ does. Let $1_S$ be the unity of S. Show that $1_S$ is a zero divisor of $R$.

I've been stuck on this for a bit, and I'm not sure how to approach it. I know how to show that if $R$ has unity, then every element of S is a zero divisor, but I can't seem to nail this case. Any help or guidance would be helpful.


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  • $\begingroup$ I don't follow your claim that is $S$ is a subring of $R$ and $R$ is unital then every element of $S$ is a zero divisor. What about $\Bbb Z\subseteq \Bbb Q$? Are you assuming something else about $S$? $\endgroup$ – Pedro Tamaroff Jan 29 '15 at 21:05
  • $\begingroup$ $S$ also has unity. $\endgroup$ – Calculus08 Jan 29 '15 at 21:10
  • $\begingroup$ Still, $\Bbb Z$ is unital! $\endgroup$ – Pedro Tamaroff Jan 29 '15 at 21:17

write $1_S=e$

if $e$ is not a unity in $R$ then $\exists r \in R$ with $er \ne r$. if $er=0$ then we are done, otherwise set $a=er-r$

then $a \ne 0$ and $$ ea = e(er-r) = e^2r-er =er-er =0 $$

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  • $\begingroup$ This makes sense. Thank you. $\endgroup$ – Calculus08 Jan 29 '15 at 21:15

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