A subset $R'$ of a ring $R$ which is a ring but does not contain $1 in R$.












1












$begingroup$


Please give me an example of a subset $R'$ of a ring $R$ with the following properties:





  • $R'$ is a ring.


  • $R'$ does not contain $1 in R$.






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is multiplicative identity part of the ring structure?
    $endgroup$
    – Berci
    Dec 31 '18 at 10:09










  • $begingroup$
    In the book I am reading now, a ring $R$ contains multiplicative identity.
    $endgroup$
    – tchappy ha
    Dec 31 '18 at 10:17










  • $begingroup$
    and $R$ is commutative.
    $endgroup$
    – tchappy ha
    Dec 31 '18 at 10:18








  • 1




    $begingroup$
    Just a remark: $R'$ is input as R' and not R^{'}.
    $endgroup$
    – egreg
    Dec 31 '18 at 10:44






  • 1




    $begingroup$
    Another place where the four element ring $F_2times F_2$ works.
    $endgroup$
    – rschwieb
    Dec 31 '18 at 11:20
















1












$begingroup$


Please give me an example of a subset $R'$ of a ring $R$ with the following properties:





  • $R'$ is a ring.


  • $R'$ does not contain $1 in R$.






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is multiplicative identity part of the ring structure?
    $endgroup$
    – Berci
    Dec 31 '18 at 10:09










  • $begingroup$
    In the book I am reading now, a ring $R$ contains multiplicative identity.
    $endgroup$
    – tchappy ha
    Dec 31 '18 at 10:17










  • $begingroup$
    and $R$ is commutative.
    $endgroup$
    – tchappy ha
    Dec 31 '18 at 10:18








  • 1




    $begingroup$
    Just a remark: $R'$ is input as R' and not R^{'}.
    $endgroup$
    – egreg
    Dec 31 '18 at 10:44






  • 1




    $begingroup$
    Another place where the four element ring $F_2times F_2$ works.
    $endgroup$
    – rschwieb
    Dec 31 '18 at 11:20














1












1








1





$begingroup$


Please give me an example of a subset $R'$ of a ring $R$ with the following properties:





  • $R'$ is a ring.


  • $R'$ does not contain $1 in R$.






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$




Please give me an example of a subset $R'$ of a ring $R$ with the following properties:





  • $R'$ is a ring.


  • $R'$ does not contain $1 in R$.






abstract-algebra ring-theory




abstract-algebra ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 10:43









egreg

186k1486209




186k1486209










asked Dec 31 '18 at 10:05









tchappy hatchappy ha

788412




788412












  • $begingroup$
    Is multiplicative identity part of the ring structure?
    $endgroup$
    – Berci
    Dec 31 '18 at 10:09










  • $begingroup$
    In the book I am reading now, a ring $R$ contains multiplicative identity.
    $endgroup$
    – tchappy ha
    Dec 31 '18 at 10:17










  • $begingroup$
    and $R$ is commutative.
    $endgroup$
    – tchappy ha
    Dec 31 '18 at 10:18








  • 1




    $begingroup$
    Just a remark: $R'$ is input as R' and not R^{'}.
    $endgroup$
    – egreg
    Dec 31 '18 at 10:44






  • 1




    $begingroup$
    Another place where the four element ring $F_2times F_2$ works.
    $endgroup$
    – rschwieb
    Dec 31 '18 at 11:20


















  • $begingroup$
    Is multiplicative identity part of the ring structure?
    $endgroup$
    – Berci
    Dec 31 '18 at 10:09










  • $begingroup$
    In the book I am reading now, a ring $R$ contains multiplicative identity.
    $endgroup$
    – tchappy ha
    Dec 31 '18 at 10:17










  • $begingroup$
    and $R$ is commutative.
    $endgroup$
    – tchappy ha
    Dec 31 '18 at 10:18








  • 1




    $begingroup$
    Just a remark: $R'$ is input as R' and not R^{'}.
    $endgroup$
    – egreg
    Dec 31 '18 at 10:44






  • 1




    $begingroup$
    Another place where the four element ring $F_2times F_2$ works.
    $endgroup$
    – rschwieb
    Dec 31 '18 at 11:20
















$begingroup$
Is multiplicative identity part of the ring structure?
$endgroup$
– Berci
Dec 31 '18 at 10:09




$begingroup$
Is multiplicative identity part of the ring structure?
$endgroup$
– Berci
Dec 31 '18 at 10:09












$begingroup$
In the book I am reading now, a ring $R$ contains multiplicative identity.
$endgroup$
– tchappy ha
Dec 31 '18 at 10:17




$begingroup$
In the book I am reading now, a ring $R$ contains multiplicative identity.
$endgroup$
– tchappy ha
Dec 31 '18 at 10:17












$begingroup$
and $R$ is commutative.
$endgroup$
– tchappy ha
Dec 31 '18 at 10:18






$begingroup$
and $R$ is commutative.
$endgroup$
– tchappy ha
Dec 31 '18 at 10:18






1




1




$begingroup$
Just a remark: $R'$ is input as R' and not R^{'}.
$endgroup$
– egreg
Dec 31 '18 at 10:44




$begingroup$
Just a remark: $R'$ is input as R' and not R^{'}.
$endgroup$
– egreg
Dec 31 '18 at 10:44




1




1




$begingroup$
Another place where the four element ring $F_2times F_2$ works.
$endgroup$
– rschwieb
Dec 31 '18 at 11:20




$begingroup$
Another place where the four element ring $F_2times F_2$ works.
$endgroup$
– rschwieb
Dec 31 '18 at 11:20










3 Answers
3






active

oldest

votes


















3












$begingroup$

$$ Bbb Z times {0} subseteq Bbb Z times Bbb Z $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much for a very simple nice example. Kenny Lau.
    $endgroup$
    – tchappy ha
    Dec 31 '18 at 10:13






  • 3




    $begingroup$
    More generally, whenever $e$ is an idempotent of a (not necessarily commutative) ring $R$, then $eRe={ere:rin R}$ satisfies the requested conditions.
    $endgroup$
    – egreg
    Dec 31 '18 at 10:45



















4












$begingroup$

An obvious example is $R'={0}$, but there are more interesting examples.



Suppose you found it; then call $e$ the identity of $R'$. By the property of the identity, $e^2=e$, so $e$ is an idempotent element of $R$.



Conversely, if $ein R$ is idempotent, then $eR$ satisfies your requirements (if $R$ is commutative).



If $R$ is not commutative, choose $eRe$, sometimes called a Peirce corner, see https://math.berkeley.edu/~lam/html/corner1.pdf



Not all rings have nontrivial idempotents, that is, different from $0$ and $1$. If $e$ is idempotent, then also $1-e$ is. In the case $R$ is commutative, there is a ring isomorphism
$$
Rcong eRtimes (1-e)R
$$
given by $rmapsto(er,(1-e)r)$. Any product of rings gives rise to idempotents: if $R=S_1times S_2$, then $(1,0)$ and $(0,1)$ are idempotents.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for perfect answer, egreg. But unfortunately it is too difficult for me.
    $endgroup$
    – tchappy ha
    Jan 10 at 12:19





















0












$begingroup$

Another easy example: $2 Bbb Z subseteq Bbb Z.$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you very much, Alex Sanger.
    $endgroup$
    – tchappy ha
    Jan 10 at 12:20












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

$$ Bbb Z times {0} subseteq Bbb Z times Bbb Z $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much for a very simple nice example. Kenny Lau.
    $endgroup$
    – tchappy ha
    Dec 31 '18 at 10:13






  • 3




    $begingroup$
    More generally, whenever $e$ is an idempotent of a (not necessarily commutative) ring $R$, then $eRe={ere:rin R}$ satisfies the requested conditions.
    $endgroup$
    – egreg
    Dec 31 '18 at 10:45
















3












$begingroup$

$$ Bbb Z times {0} subseteq Bbb Z times Bbb Z $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you very much for a very simple nice example. Kenny Lau.
    $endgroup$
    – tchappy ha
    Dec 31 '18 at 10:13






  • 3




    $begingroup$
    More generally, whenever $e$ is an idempotent of a (not necessarily commutative) ring $R$, then $eRe={ere:rin R}$ satisfies the requested conditions.
    $endgroup$
    – egreg
    Dec 31 '18 at 10:45














3












3








3





$begingroup$

$$ Bbb Z times {0} subseteq Bbb Z times Bbb Z $$






share|cite|improve this answer











$endgroup$



$$ Bbb Z times {0} subseteq Bbb Z times Bbb Z $$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 31 '18 at 10:10

























answered Dec 31 '18 at 10:06









Kenny LauKenny Lau

20k2260




20k2260












  • $begingroup$
    Thank you very much for a very simple nice example. Kenny Lau.
    $endgroup$
    – tchappy ha
    Dec 31 '18 at 10:13






  • 3




    $begingroup$
    More generally, whenever $e$ is an idempotent of a (not necessarily commutative) ring $R$, then $eRe={ere:rin R}$ satisfies the requested conditions.
    $endgroup$
    – egreg
    Dec 31 '18 at 10:45


















  • $begingroup$
    Thank you very much for a very simple nice example. Kenny Lau.
    $endgroup$
    – tchappy ha
    Dec 31 '18 at 10:13






  • 3




    $begingroup$
    More generally, whenever $e$ is an idempotent of a (not necessarily commutative) ring $R$, then $eRe={ere:rin R}$ satisfies the requested conditions.
    $endgroup$
    – egreg
    Dec 31 '18 at 10:45
















$begingroup$
Thank you very much for a very simple nice example. Kenny Lau.
$endgroup$
– tchappy ha
Dec 31 '18 at 10:13




$begingroup$
Thank you very much for a very simple nice example. Kenny Lau.
$endgroup$
– tchappy ha
Dec 31 '18 at 10:13




3




3




$begingroup$
More generally, whenever $e$ is an idempotent of a (not necessarily commutative) ring $R$, then $eRe={ere:rin R}$ satisfies the requested conditions.
$endgroup$
– egreg
Dec 31 '18 at 10:45




$begingroup$
More generally, whenever $e$ is an idempotent of a (not necessarily commutative) ring $R$, then $eRe={ere:rin R}$ satisfies the requested conditions.
$endgroup$
– egreg
Dec 31 '18 at 10:45











4












$begingroup$

An obvious example is $R'={0}$, but there are more interesting examples.



Suppose you found it; then call $e$ the identity of $R'$. By the property of the identity, $e^2=e$, so $e$ is an idempotent element of $R$.



Conversely, if $ein R$ is idempotent, then $eR$ satisfies your requirements (if $R$ is commutative).



If $R$ is not commutative, choose $eRe$, sometimes called a Peirce corner, see https://math.berkeley.edu/~lam/html/corner1.pdf



Not all rings have nontrivial idempotents, that is, different from $0$ and $1$. If $e$ is idempotent, then also $1-e$ is. In the case $R$ is commutative, there is a ring isomorphism
$$
Rcong eRtimes (1-e)R
$$
given by $rmapsto(er,(1-e)r)$. Any product of rings gives rise to idempotents: if $R=S_1times S_2$, then $(1,0)$ and $(0,1)$ are idempotents.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for perfect answer, egreg. But unfortunately it is too difficult for me.
    $endgroup$
    – tchappy ha
    Jan 10 at 12:19


















4












$begingroup$

An obvious example is $R'={0}$, but there are more interesting examples.



Suppose you found it; then call $e$ the identity of $R'$. By the property of the identity, $e^2=e$, so $e$ is an idempotent element of $R$.



Conversely, if $ein R$ is idempotent, then $eR$ satisfies your requirements (if $R$ is commutative).



If $R$ is not commutative, choose $eRe$, sometimes called a Peirce corner, see https://math.berkeley.edu/~lam/html/corner1.pdf



Not all rings have nontrivial idempotents, that is, different from $0$ and $1$. If $e$ is idempotent, then also $1-e$ is. In the case $R$ is commutative, there is a ring isomorphism
$$
Rcong eRtimes (1-e)R
$$
given by $rmapsto(er,(1-e)r)$. Any product of rings gives rise to idempotents: if $R=S_1times S_2$, then $(1,0)$ and $(0,1)$ are idempotents.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for perfect answer, egreg. But unfortunately it is too difficult for me.
    $endgroup$
    – tchappy ha
    Jan 10 at 12:19
















4












4








4





$begingroup$

An obvious example is $R'={0}$, but there are more interesting examples.



Suppose you found it; then call $e$ the identity of $R'$. By the property of the identity, $e^2=e$, so $e$ is an idempotent element of $R$.



Conversely, if $ein R$ is idempotent, then $eR$ satisfies your requirements (if $R$ is commutative).



If $R$ is not commutative, choose $eRe$, sometimes called a Peirce corner, see https://math.berkeley.edu/~lam/html/corner1.pdf



Not all rings have nontrivial idempotents, that is, different from $0$ and $1$. If $e$ is idempotent, then also $1-e$ is. In the case $R$ is commutative, there is a ring isomorphism
$$
Rcong eRtimes (1-e)R
$$
given by $rmapsto(er,(1-e)r)$. Any product of rings gives rise to idempotents: if $R=S_1times S_2$, then $(1,0)$ and $(0,1)$ are idempotents.






share|cite|improve this answer









$endgroup$



An obvious example is $R'={0}$, but there are more interesting examples.



Suppose you found it; then call $e$ the identity of $R'$. By the property of the identity, $e^2=e$, so $e$ is an idempotent element of $R$.



Conversely, if $ein R$ is idempotent, then $eR$ satisfies your requirements (if $R$ is commutative).



If $R$ is not commutative, choose $eRe$, sometimes called a Peirce corner, see https://math.berkeley.edu/~lam/html/corner1.pdf



Not all rings have nontrivial idempotents, that is, different from $0$ and $1$. If $e$ is idempotent, then also $1-e$ is. In the case $R$ is commutative, there is a ring isomorphism
$$
Rcong eRtimes (1-e)R
$$
given by $rmapsto(er,(1-e)r)$. Any product of rings gives rise to idempotents: if $R=S_1times S_2$, then $(1,0)$ and $(0,1)$ are idempotents.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 31 '18 at 10:54









egregegreg

186k1486209




186k1486209












  • $begingroup$
    Thank you very much for perfect answer, egreg. But unfortunately it is too difficult for me.
    $endgroup$
    – tchappy ha
    Jan 10 at 12:19




















  • $begingroup$
    Thank you very much for perfect answer, egreg. But unfortunately it is too difficult for me.
    $endgroup$
    – tchappy ha
    Jan 10 at 12:19


















$begingroup$
Thank you very much for perfect answer, egreg. But unfortunately it is too difficult for me.
$endgroup$
– tchappy ha
Jan 10 at 12:19






$begingroup$
Thank you very much for perfect answer, egreg. But unfortunately it is too difficult for me.
$endgroup$
– tchappy ha
Jan 10 at 12:19













0












$begingroup$

Another easy example: $2 Bbb Z subseteq Bbb Z.$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you very much, Alex Sanger.
    $endgroup$
    – tchappy ha
    Jan 10 at 12:20
















0












$begingroup$

Another easy example: $2 Bbb Z subseteq Bbb Z.$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you very much, Alex Sanger.
    $endgroup$
    – tchappy ha
    Jan 10 at 12:20














0












0








0





$begingroup$

Another easy example: $2 Bbb Z subseteq Bbb Z.$






share|cite|improve this answer









$endgroup$



Another easy example: $2 Bbb Z subseteq Bbb Z.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 7 at 17:50









Alex SangerAlex Sanger

10329




10329








  • 1




    $begingroup$
    Thank you very much, Alex Sanger.
    $endgroup$
    – tchappy ha
    Jan 10 at 12:20














  • 1




    $begingroup$
    Thank you very much, Alex Sanger.
    $endgroup$
    – tchappy ha
    Jan 10 at 12:20








1




1




$begingroup$
Thank you very much, Alex Sanger.
$endgroup$
– tchappy ha
Jan 10 at 12:20




$begingroup$
Thank you very much, Alex Sanger.
$endgroup$
– tchappy ha
Jan 10 at 12:20


















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