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So, my underlying subset is all the elements in R that contain a multiplicative inverse. I need to show closure under subtraction and multiplication.

I dont think this is a subring, but I'm having trouble showing this. Suppose $r_1, r_2 \in S$. Then $r_1^{-1}, r_2^{-1}\in S$. This would mean that for element $(r_1-r_2)$, $(r_1-r_2)^{-1}$ needs to be in $S$.

But since $r_1^{-1}-r_2^{-1}\neq (r_1-r_2)^{-1}$, this doesnt hold. Is this how the proof is finished? I dont feel like this is right, but I feel this is the subring condition that is violated. I feel as though i need to show more to prove the subtraction closure is not upheld. It would be closed under multiplication as

$$(r_1r_2)^-1=r_2^{-1}r_1^{-1}$$

How do I procede?

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    $\begingroup$ just take $r_1=r_2=s \in S$, then $r_1-r_2=0$ and hence not invertible. $\endgroup$ – Anurag A Jan 29 '17 at 14:26
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Your idea is correct but it'd be better if you gave an explicit counterexample. For example, if you take $R=\Bbb{Z}$, and $r_1=r_2=1$, then $r_1-r_2=0$, which is not invertible in $R$.

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  • $\begingroup$ you know, I was thinking that somehow the 0 element might be involved since 0 is noninvertible. Thank you!! $\endgroup$ – AveryJessup Jan 29 '17 at 14:25
  • $\begingroup$ @AveryJessup no problem. Note you can actually turn my "proof" into the fact that for every single ring $R$, this subset is not a subring. This is because you can always take $1\neq0$, and consider $1-1$. $\endgroup$ – Alex Mathers Jan 29 '17 at 14:27
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    $\begingroup$ thats why i liked it...the subring must contain both of those elements, so letting it equal 1 makes perfect sense as you said. I knew I was on the right track. Thank you again. $\endgroup$ – AveryJessup Jan 29 '17 at 14:28

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