the question is as follows

(TRUE or FALSE.) If R is a commutative ring with unity, then the set of units in R forms a subring. (If true, give a short proof. If false, give a specic counter-example.)

I don't think this forms a sub-ring but i don't entirely understand why i think that. i feel like im kinda looking at $Z_{12}$ and $U_{12}$ to me one these groups is all happy under addition --> $Z_{12}$ and one of these groups is happy under multiplication -->$U_{12}$ but to me $U_{12}$ doesn't make any sense under addition as u can land on things that aren't in $U_{12}$ and $Z_{12}$ has 0 divisors and well those break math.

i think what i want to say is true as long as R is also a subring of the subring of units. and i have no idea why. As for a counter example to show that this is not true i have no idea.

i really would like an answer with this is true only when x,y,z are satisfied and it is false because ( insert counter example here)

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    $\begingroup$ Well, $\,1\,$ is a unit, but $\,1-1\,$ is...not? $\endgroup$ – DonAntonio Mar 17 '13 at 6:09
  • $\begingroup$ uh can you define what you mean by 1-1? and quantify why its not a unit? i understand that 0 is not a unit but its never been a unit. perhaps your implying that a ring cannot exist because of only units it must have 0 in it? but that seems obvious, odd and confusing all at once i am having trouble with the 2 operations at once thing you mean that things won't have inverse's under addition in $U_{12}$ because it removed 0? $\endgroup$ – Faust Mar 17 '13 at 6:23
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    $\begingroup$ @Faust7, Don gave you enough answer, If set of units forms a subring, then that should be closed under addition, 1 is a unit, and also -1 so, 1 + (-1) should be unit if it is a subring, but 0 is not a unit, hence set of units don't for a subring. I suggest you to be more polite in the forum. $\endgroup$ – Ram Mar 17 '13 at 7:32

Counterexample: Let $R=\mathbb{Z}$ be the ring of integers. Then $R^*=\{1,-1\}$. Subrings must be closed under multiplication and addition. $R^*$ is closed under multiplication (in fact it is a group under multiplication) but it is not closed under addition. For example, $1+(-1)=0\notin R^*$.

In fact, the result of the above counterexample is much more general. Suppose $R$ is a nonzero ring. $R^*$ is always a group under multiplication, but never a subring, and this follows because subrings must be subgroups under addition, but $0\notin R^*$ for $R$ nonzero..

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    $\begingroup$ I think you should improve the wording of your last paragraph. First you claim that $R^*$ is never a subring and then, in the very next sentence, you say that the group of units in the zero ring is a subring. $\endgroup$ – kahen Mar 17 '13 at 8:00
  • $\begingroup$ Hi Kahen, you're absolutely right. I only stated I was assuming a nonzero ring after I had made my claim (which was false without the assumption). A small oversight. I've made the appropriate edit. If you'd like, in the future please be willing to make this small edit for me if I forget. Thanks! $\endgroup$ – Jared Mar 17 '13 at 21:30

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