Does every ID with subring which has a unity have an unity?












2












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For an arbitrary ID (integral domain) $R$ with subring $S$, assume that $S$ has an unity. Then does $R$ have a unity too? If not, please provide a counter-example.






ring-theory integral-domain







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    2












    $begingroup$


    For an arbitrary ID (integral domain) $R$ with subring $S$, assume that $S$ has an unity. Then does $R$ have a unity too? If not, please provide a counter-example.






    ring-theory integral-domain







    share|cite|improve this question











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      2












      2








      2


      0



      $begingroup$


      For an arbitrary ID (integral domain) $R$ with subring $S$, assume that $S$ has an unity. Then does $R$ have a unity too? If not, please provide a counter-example.






      ring-theory integral-domain







      share|cite|improve this question











      $endgroup$




      For an arbitrary ID (integral domain) $R$ with subring $S$, assume that $S$ has an unity. Then does $R$ have a unity too? If not, please provide a counter-example.






      ring-theory integral-domain




      ring-theory integral-domain






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      edited Aug 6 '15 at 12:43









      man_in_green_shirt

      8431028




      8431028










      asked Aug 6 '15 at 12:09









      jawlangjawlang

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      30715






















          1 Answer
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          $begingroup$

          If $S$ is nontrivial (that is, has a nonzero identity $e$) then yes.



          A nonzero idempotent in a domain must act as the identity for the domain.



          To see this, just examine what $e(ex-x)$ and $(xe-x)e$ must be, where $x$ is an arbitrary element of the domain.



          This came up under slightly different circumstances here.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
            $endgroup$
            – man_in_green_shirt
            Aug 6 '15 at 12:55






          • 1




            $begingroup$
            @man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
            $endgroup$
            – WillO
            Aug 6 '15 at 12:57










          • $begingroup$
            Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
            $endgroup$
            – man_in_green_shirt
            Aug 6 '15 at 13:07






          • 1




            $begingroup$
            @man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
            $endgroup$
            – WillO
            Aug 6 '15 at 13:29










          • $begingroup$
            @man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
            $endgroup$
            – rschwieb
            Aug 6 '15 at 14:01













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          1 Answer
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          1 Answer
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          active

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          $begingroup$

          If $S$ is nontrivial (that is, has a nonzero identity $e$) then yes.



          A nonzero idempotent in a domain must act as the identity for the domain.



          To see this, just examine what $e(ex-x)$ and $(xe-x)e$ must be, where $x$ is an arbitrary element of the domain.



          This came up under slightly different circumstances here.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
            $endgroup$
            – man_in_green_shirt
            Aug 6 '15 at 12:55






          • 1




            $begingroup$
            @man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
            $endgroup$
            – WillO
            Aug 6 '15 at 12:57










          • $begingroup$
            Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
            $endgroup$
            – man_in_green_shirt
            Aug 6 '15 at 13:07






          • 1




            $begingroup$
            @man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
            $endgroup$
            – WillO
            Aug 6 '15 at 13:29










          • $begingroup$
            @man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
            $endgroup$
            – rschwieb
            Aug 6 '15 at 14:01


















          1












          $begingroup$

          If $S$ is nontrivial (that is, has a nonzero identity $e$) then yes.



          A nonzero idempotent in a domain must act as the identity for the domain.



          To see this, just examine what $e(ex-x)$ and $(xe-x)e$ must be, where $x$ is an arbitrary element of the domain.



          This came up under slightly different circumstances here.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
            $endgroup$
            – man_in_green_shirt
            Aug 6 '15 at 12:55






          • 1




            $begingroup$
            @man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
            $endgroup$
            – WillO
            Aug 6 '15 at 12:57










          • $begingroup$
            Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
            $endgroup$
            – man_in_green_shirt
            Aug 6 '15 at 13:07






          • 1




            $begingroup$
            @man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
            $endgroup$
            – WillO
            Aug 6 '15 at 13:29










          • $begingroup$
            @man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
            $endgroup$
            – rschwieb
            Aug 6 '15 at 14:01
















          1












          1








          1





          $begingroup$

          If $S$ is nontrivial (that is, has a nonzero identity $e$) then yes.



          A nonzero idempotent in a domain must act as the identity for the domain.



          To see this, just examine what $e(ex-x)$ and $(xe-x)e$ must be, where $x$ is an arbitrary element of the domain.



          This came up under slightly different circumstances here.






          share|cite|improve this answer











          $endgroup$



          If $S$ is nontrivial (that is, has a nonzero identity $e$) then yes.



          A nonzero idempotent in a domain must act as the identity for the domain.



          To see this, just examine what $e(ex-x)$ and $(xe-x)e$ must be, where $x$ is an arbitrary element of the domain.



          This came up under slightly different circumstances here.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 13 '17 at 12:20









          Community♦

          1




          1










          answered Aug 6 '15 at 12:47









          rschwiebrschwieb

          107k12102251




          107k12102251












          • $begingroup$
            If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
            $endgroup$
            – man_in_green_shirt
            Aug 6 '15 at 12:55






          • 1




            $begingroup$
            @man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
            $endgroup$
            – WillO
            Aug 6 '15 at 12:57










          • $begingroup$
            Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
            $endgroup$
            – man_in_green_shirt
            Aug 6 '15 at 13:07






          • 1




            $begingroup$
            @man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
            $endgroup$
            – WillO
            Aug 6 '15 at 13:29










          • $begingroup$
            @man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
            $endgroup$
            – rschwieb
            Aug 6 '15 at 14:01




















          • $begingroup$
            If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
            $endgroup$
            – man_in_green_shirt
            Aug 6 '15 at 12:55






          • 1




            $begingroup$
            @man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
            $endgroup$
            – WillO
            Aug 6 '15 at 12:57










          • $begingroup$
            Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
            $endgroup$
            – man_in_green_shirt
            Aug 6 '15 at 13:07






          • 1




            $begingroup$
            @man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
            $endgroup$
            – WillO
            Aug 6 '15 at 13:29










          • $begingroup$
            @man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
            $endgroup$
            – rschwieb
            Aug 6 '15 at 14:01


















          $begingroup$
          If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
          $endgroup$
          – man_in_green_shirt
          Aug 6 '15 at 12:55




          $begingroup$
          If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
          $endgroup$
          – man_in_green_shirt
          Aug 6 '15 at 12:55




          1




          1




          $begingroup$
          @man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
          $endgroup$
          – WillO
          Aug 6 '15 at 12:57




          $begingroup$
          @man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
          $endgroup$
          – WillO
          Aug 6 '15 at 12:57












          $begingroup$
          Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
          $endgroup$
          – man_in_green_shirt
          Aug 6 '15 at 13:07




          $begingroup$
          Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
          $endgroup$
          – man_in_green_shirt
          Aug 6 '15 at 13:07




          1




          1




          $begingroup$
          @man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
          $endgroup$
          – WillO
          Aug 6 '15 at 13:29




          $begingroup$
          @man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
          $endgroup$
          – WillO
          Aug 6 '15 at 13:29












          $begingroup$
          @man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
          $endgroup$
          – rschwieb
          Aug 6 '15 at 14:01






          $begingroup$
          @man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
          $endgroup$
          – rschwieb
          Aug 6 '15 at 14:01




















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