Prove that the center of a ring is a subring.












1












$begingroup$


The center of a ring $R$ is ${cin R : cr=rc $ for every $ r in R}$. Prove that the center of a ring is a subring. What is the center of a commutative ring?



Is my solution right?



solution



You just need to prove that the centre is a ring within itself.




  1. Associativity of both addition and multiplication is inherited from R, and distributivity of multiplication over addition is inherited from R.


  2. Show that it is a group under addition:
    Take a and b in the centre and r in R.
    Then
    $(a + b)r = ar + br = ra + rb = r(a + b)$



hence it is closed under addition.
Show that $0$ (additive identity) is in the centre.



$0 = 0.r = r.0$
so $0$ is in the centre.



For a in the centre, there exists $-a$ in R such that $a + (-a) = 1 = (-a) + a$



Show that $-a$ is in the centre.
Let r in R
$0 = 0.r = (a + (-a))r = ar + (-a)r $
and
$0 = r.0 = r(a + (-a)) = ra + r(-a) $



as $ ra = ar$, it follows that $(-a)r = r(-a) $
Hence inverses exist in the centre.
So the centre is a group under +.




  1. Show that the centre is closed under multiplication:
    For a,b in the centre and r in R,#
    $(ab)r = a(br) = a(rb) = (ar)b = (ra)b = r(ab) $


thus multiplication is closed.
And show that 1 (the multiplicative identity) is in the centre.
$1.r = r = r.1$

hence 1 is in the centre.



So the center is a subring of R.



The center of a commutative ring is the ring itself. (By definition the centre is the ring of all commutative elements.)






abstract-algebra proof-verification ring-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is correct but showing the identity is superflous, as rings do not necciserily have an identity.
    $endgroup$
    – Zelos Malum
    Nov 27 '16 at 17:47






  • 1




    $begingroup$
    @ZelosMalum some authors impose that rings must have an identity.
    $endgroup$
    – Xam
    Nov 27 '16 at 18:31










  • $begingroup$
    Possible duplicate of On the center of a ring
    $endgroup$
    – user2371916
    Jul 26 '17 at 19:42
















1












$begingroup$


The center of a ring $R$ is ${cin R : cr=rc $ for every $ r in R}$. Prove that the center of a ring is a subring. What is the center of a commutative ring?



Is my solution right?



solution



You just need to prove that the centre is a ring within itself.




  1. Associativity of both addition and multiplication is inherited from R, and distributivity of multiplication over addition is inherited from R.


  2. Show that it is a group under addition:
    Take a and b in the centre and r in R.
    Then
    $(a + b)r = ar + br = ra + rb = r(a + b)$



hence it is closed under addition.
Show that $0$ (additive identity) is in the centre.



$0 = 0.r = r.0$
so $0$ is in the centre.



For a in the centre, there exists $-a$ in R such that $a + (-a) = 1 = (-a) + a$



Show that $-a$ is in the centre.
Let r in R
$0 = 0.r = (a + (-a))r = ar + (-a)r $
and
$0 = r.0 = r(a + (-a)) = ra + r(-a) $



as $ ra = ar$, it follows that $(-a)r = r(-a) $
Hence inverses exist in the centre.
So the centre is a group under +.




  1. Show that the centre is closed under multiplication:
    For a,b in the centre and r in R,#
    $(ab)r = a(br) = a(rb) = (ar)b = (ra)b = r(ab) $


thus multiplication is closed.
And show that 1 (the multiplicative identity) is in the centre.
$1.r = r = r.1$

hence 1 is in the centre.



So the center is a subring of R.



The center of a commutative ring is the ring itself. (By definition the centre is the ring of all commutative elements.)






abstract-algebra proof-verification ring-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is correct but showing the identity is superflous, as rings do not necciserily have an identity.
    $endgroup$
    – Zelos Malum
    Nov 27 '16 at 17:47






  • 1




    $begingroup$
    @ZelosMalum some authors impose that rings must have an identity.
    $endgroup$
    – Xam
    Nov 27 '16 at 18:31










  • $begingroup$
    Possible duplicate of On the center of a ring
    $endgroup$
    – user2371916
    Jul 26 '17 at 19:42














1












1








1





$begingroup$


The center of a ring $R$ is ${cin R : cr=rc $ for every $ r in R}$. Prove that the center of a ring is a subring. What is the center of a commutative ring?



Is my solution right?



solution



You just need to prove that the centre is a ring within itself.




  1. Associativity of both addition and multiplication is inherited from R, and distributivity of multiplication over addition is inherited from R.


  2. Show that it is a group under addition:
    Take a and b in the centre and r in R.
    Then
    $(a + b)r = ar + br = ra + rb = r(a + b)$



hence it is closed under addition.
Show that $0$ (additive identity) is in the centre.



$0 = 0.r = r.0$
so $0$ is in the centre.



For a in the centre, there exists $-a$ in R such that $a + (-a) = 1 = (-a) + a$



Show that $-a$ is in the centre.
Let r in R
$0 = 0.r = (a + (-a))r = ar + (-a)r $
and
$0 = r.0 = r(a + (-a)) = ra + r(-a) $



as $ ra = ar$, it follows that $(-a)r = r(-a) $
Hence inverses exist in the centre.
So the centre is a group under +.




  1. Show that the centre is closed under multiplication:
    For a,b in the centre and r in R,#
    $(ab)r = a(br) = a(rb) = (ar)b = (ra)b = r(ab) $


thus multiplication is closed.
And show that 1 (the multiplicative identity) is in the centre.
$1.r = r = r.1$

hence 1 is in the centre.



So the center is a subring of R.



The center of a commutative ring is the ring itself. (By definition the centre is the ring of all commutative elements.)






abstract-algebra proof-verification ring-theory







share|cite|improve this question











$endgroup$




The center of a ring $R$ is ${cin R : cr=rc $ for every $ r in R}$. Prove that the center of a ring is a subring. What is the center of a commutative ring?



Is my solution right?



solution



You just need to prove that the centre is a ring within itself.




  1. Associativity of both addition and multiplication is inherited from R, and distributivity of multiplication over addition is inherited from R.


  2. Show that it is a group under addition:
    Take a and b in the centre and r in R.
    Then
    $(a + b)r = ar + br = ra + rb = r(a + b)$



hence it is closed under addition.
Show that $0$ (additive identity) is in the centre.



$0 = 0.r = r.0$
so $0$ is in the centre.



For a in the centre, there exists $-a$ in R such that $a + (-a) = 1 = (-a) + a$



Show that $-a$ is in the centre.
Let r in R
$0 = 0.r = (a + (-a))r = ar + (-a)r $
and
$0 = r.0 = r(a + (-a)) = ra + r(-a) $



as $ ra = ar$, it follows that $(-a)r = r(-a) $
Hence inverses exist in the centre.
So the centre is a group under +.




  1. Show that the centre is closed under multiplication:
    For a,b in the centre and r in R,#
    $(ab)r = a(br) = a(rb) = (ar)b = (ra)b = r(ab) $


thus multiplication is closed.
And show that 1 (the multiplicative identity) is in the centre.
$1.r = r = r.1$

hence 1 is in the centre.



So the center is a subring of R.



The center of a commutative ring is the ring itself. (By definition the centre is the ring of all commutative elements.)






abstract-algebra proof-verification ring-theory




abstract-algebra proof-verification ring-theory






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edited Nov 27 '16 at 18:25









user26857

39.3k124183




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asked Nov 27 '16 at 17:43









sh.alzoubish.alzoubi

4217




4217












  • $begingroup$
    It is correct but showing the identity is superflous, as rings do not necciserily have an identity.
    $endgroup$
    – Zelos Malum
    Nov 27 '16 at 17:47






  • 1




    $begingroup$
    @ZelosMalum some authors impose that rings must have an identity.
    $endgroup$
    – Xam
    Nov 27 '16 at 18:31










  • $begingroup$
    Possible duplicate of On the center of a ring
    $endgroup$
    – user2371916
    Jul 26 '17 at 19:42


















  • $begingroup$
    It is correct but showing the identity is superflous, as rings do not necciserily have an identity.
    $endgroup$
    – Zelos Malum
    Nov 27 '16 at 17:47






  • 1




    $begingroup$
    @ZelosMalum some authors impose that rings must have an identity.
    $endgroup$
    – Xam
    Nov 27 '16 at 18:31










  • $begingroup$
    Possible duplicate of On the center of a ring
    $endgroup$
    – user2371916
    Jul 26 '17 at 19:42
















$begingroup$
It is correct but showing the identity is superflous, as rings do not necciserily have an identity.
$endgroup$
– Zelos Malum
Nov 27 '16 at 17:47




$begingroup$
It is correct but showing the identity is superflous, as rings do not necciserily have an identity.
$endgroup$
– Zelos Malum
Nov 27 '16 at 17:47




1




1




$begingroup$
@ZelosMalum some authors impose that rings must have an identity.
$endgroup$
– Xam
Nov 27 '16 at 18:31




$begingroup$
@ZelosMalum some authors impose that rings must have an identity.
$endgroup$
– Xam
Nov 27 '16 at 18:31












$begingroup$
Possible duplicate of On the center of a ring
$endgroup$
– user2371916
Jul 26 '17 at 19:42




$begingroup$
Possible duplicate of On the center of a ring
$endgroup$
– user2371916
Jul 26 '17 at 19:42










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$begingroup$

Your proof is right, but unnecessarily long. When you want to prove that some nonempty set is a subring you have to use the subring test. Denote the center of your ring by $Z(R)$, you only have to prove that $1in Z(R)$ and if $x,yin Z(R)$, then $x-y, xcdot yin Z(R)$. Since you have proved all that, then $Z(R)$ is a subring of $R$.



The answer to the other question is right too, the center of a commutative ring it's the ring itself.






share|cite|improve this answer











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    $begingroup$

    Your proof is right, but unnecessarily long. When you want to prove that some nonempty set is a subring you have to use the subring test. Denote the center of your ring by $Z(R)$, you only have to prove that $1in Z(R)$ and if $x,yin Z(R)$, then $x-y, xcdot yin Z(R)$. Since you have proved all that, then $Z(R)$ is a subring of $R$.



    The answer to the other question is right too, the center of a commutative ring it's the ring itself.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Your proof is right, but unnecessarily long. When you want to prove that some nonempty set is a subring you have to use the subring test. Denote the center of your ring by $Z(R)$, you only have to prove that $1in Z(R)$ and if $x,yin Z(R)$, then $x-y, xcdot yin Z(R)$. Since you have proved all that, then $Z(R)$ is a subring of $R$.



      The answer to the other question is right too, the center of a commutative ring it's the ring itself.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Your proof is right, but unnecessarily long. When you want to prove that some nonempty set is a subring you have to use the subring test. Denote the center of your ring by $Z(R)$, you only have to prove that $1in Z(R)$ and if $x,yin Z(R)$, then $x-y, xcdot yin Z(R)$. Since you have proved all that, then $Z(R)$ is a subring of $R$.



        The answer to the other question is right too, the center of a commutative ring it's the ring itself.






        share|cite|improve this answer











        $endgroup$



        Your proof is right, but unnecessarily long. When you want to prove that some nonempty set is a subring you have to use the subring test. Denote the center of your ring by $Z(R)$, you only have to prove that $1in Z(R)$ and if $x,yin Z(R)$, then $x-y, xcdot yin Z(R)$. Since you have proved all that, then $Z(R)$ is a subring of $R$.



        The answer to the other question is right too, the center of a commutative ring it's the ring itself.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 13 '17 at 12:20









        Community♦

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        answered Nov 27 '16 at 18:43









        XamXam

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