Characteristic of $3mathbb{Z}$, $mathbb{Z} times 5mathbb{Z}$, $mathbb{Z}_5 times mathbb{Z}_3$












2












$begingroup$


I want to find the characteristic of $3mathbb{Z}$, $mathbb{Z} times 5mathbb{Z}$, $mathbb{Z}_5 times mathbb{Z}_3$. Need a confirm for my solutions:



$bullet$ $text{char}(3mathbb{Z}) = 0$



$bullet$ $text{char}(mathbb{Z} times 5mathbb{Z}) = 0$ because both $mathbb{Z}$ and $5mathbb{Z}$ have characteristic $0$.



$bullet$ $text{char}(mathbb{Z}_5 times mathbb{Z}_3)=15$ because they are finite rings with order $5$ and $3$ and $(5,3) = 1$.



Thanks






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Arthur Rings, right? I've edited the main post thx
    $endgroup$
    – Alessar
    Dec 12 '18 at 7:39






  • 1




    $begingroup$
    $3mathbb{Z}$ does not have a multiplicative identity, so characteristic does not make sense for it.
    $endgroup$
    – Aniruddh Agarwal
    Dec 12 '18 at 7:44












  • $begingroup$
    @AniruddhAgarwal you're right, because the "unity" of $mathbb{Z}$ became $3$ in this new ring. Thanks for let me see this!
    $endgroup$
    – Alessar
    Dec 12 '18 at 7:46






  • 1




    $begingroup$
    The ring $mathbb{Z} times 5mathbb{Z}$ also does not have a multiplicative identity. Your solution for $mathbb{Z}_5 times mathbb{Z}_3$ is indeed correct (by the Chinese Remainder Theorem.)
    $endgroup$
    – Aniruddh Agarwal
    Dec 12 '18 at 7:49


















2












$begingroup$


I want to find the characteristic of $3mathbb{Z}$, $mathbb{Z} times 5mathbb{Z}$, $mathbb{Z}_5 times mathbb{Z}_3$. Need a confirm for my solutions:



$bullet$ $text{char}(3mathbb{Z}) = 0$



$bullet$ $text{char}(mathbb{Z} times 5mathbb{Z}) = 0$ because both $mathbb{Z}$ and $5mathbb{Z}$ have characteristic $0$.



$bullet$ $text{char}(mathbb{Z}_5 times mathbb{Z}_3)=15$ because they are finite rings with order $5$ and $3$ and $(5,3) = 1$.



Thanks






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Arthur Rings, right? I've edited the main post thx
    $endgroup$
    – Alessar
    Dec 12 '18 at 7:39






  • 1




    $begingroup$
    $3mathbb{Z}$ does not have a multiplicative identity, so characteristic does not make sense for it.
    $endgroup$
    – Aniruddh Agarwal
    Dec 12 '18 at 7:44












  • $begingroup$
    @AniruddhAgarwal you're right, because the "unity" of $mathbb{Z}$ became $3$ in this new ring. Thanks for let me see this!
    $endgroup$
    – Alessar
    Dec 12 '18 at 7:46






  • 1




    $begingroup$
    The ring $mathbb{Z} times 5mathbb{Z}$ also does not have a multiplicative identity. Your solution for $mathbb{Z}_5 times mathbb{Z}_3$ is indeed correct (by the Chinese Remainder Theorem.)
    $endgroup$
    – Aniruddh Agarwal
    Dec 12 '18 at 7:49
















2












2








2





$begingroup$


I want to find the characteristic of $3mathbb{Z}$, $mathbb{Z} times 5mathbb{Z}$, $mathbb{Z}_5 times mathbb{Z}_3$. Need a confirm for my solutions:



$bullet$ $text{char}(3mathbb{Z}) = 0$



$bullet$ $text{char}(mathbb{Z} times 5mathbb{Z}) = 0$ because both $mathbb{Z}$ and $5mathbb{Z}$ have characteristic $0$.



$bullet$ $text{char}(mathbb{Z}_5 times mathbb{Z}_3)=15$ because they are finite rings with order $5$ and $3$ and $(5,3) = 1$.



Thanks






abstract-algebra







share|cite|improve this question











$endgroup$




I want to find the characteristic of $3mathbb{Z}$, $mathbb{Z} times 5mathbb{Z}$, $mathbb{Z}_5 times mathbb{Z}_3$. Need a confirm for my solutions:



$bullet$ $text{char}(3mathbb{Z}) = 0$



$bullet$ $text{char}(mathbb{Z} times 5mathbb{Z}) = 0$ because both $mathbb{Z}$ and $5mathbb{Z}$ have characteristic $0$.



$bullet$ $text{char}(mathbb{Z}_5 times mathbb{Z}_3)=15$ because they are finite rings with order $5$ and $3$ and $(5,3) = 1$.



Thanks






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 7:47









Aniruddh Agarwal

1218




1218










asked Dec 12 '18 at 7:08









AlessarAlessar

298115




298115












  • $begingroup$
    @Arthur Rings, right? I've edited the main post thx
    $endgroup$
    – Alessar
    Dec 12 '18 at 7:39






  • 1




    $begingroup$
    $3mathbb{Z}$ does not have a multiplicative identity, so characteristic does not make sense for it.
    $endgroup$
    – Aniruddh Agarwal
    Dec 12 '18 at 7:44












  • $begingroup$
    @AniruddhAgarwal you're right, because the "unity" of $mathbb{Z}$ became $3$ in this new ring. Thanks for let me see this!
    $endgroup$
    – Alessar
    Dec 12 '18 at 7:46






  • 1




    $begingroup$
    The ring $mathbb{Z} times 5mathbb{Z}$ also does not have a multiplicative identity. Your solution for $mathbb{Z}_5 times mathbb{Z}_3$ is indeed correct (by the Chinese Remainder Theorem.)
    $endgroup$
    – Aniruddh Agarwal
    Dec 12 '18 at 7:49




















  • $begingroup$
    @Arthur Rings, right? I've edited the main post thx
    $endgroup$
    – Alessar
    Dec 12 '18 at 7:39






  • 1




    $begingroup$
    $3mathbb{Z}$ does not have a multiplicative identity, so characteristic does not make sense for it.
    $endgroup$
    – Aniruddh Agarwal
    Dec 12 '18 at 7:44












  • $begingroup$
    @AniruddhAgarwal you're right, because the "unity" of $mathbb{Z}$ became $3$ in this new ring. Thanks for let me see this!
    $endgroup$
    – Alessar
    Dec 12 '18 at 7:46






  • 1




    $begingroup$
    The ring $mathbb{Z} times 5mathbb{Z}$ also does not have a multiplicative identity. Your solution for $mathbb{Z}_5 times mathbb{Z}_3$ is indeed correct (by the Chinese Remainder Theorem.)
    $endgroup$
    – Aniruddh Agarwal
    Dec 12 '18 at 7:49


















$begingroup$
@Arthur Rings, right? I've edited the main post thx
$endgroup$
– Alessar
Dec 12 '18 at 7:39




$begingroup$
@Arthur Rings, right? I've edited the main post thx
$endgroup$
– Alessar
Dec 12 '18 at 7:39




1




1




$begingroup$
$3mathbb{Z}$ does not have a multiplicative identity, so characteristic does not make sense for it.
$endgroup$
– Aniruddh Agarwal
Dec 12 '18 at 7:44






$begingroup$
$3mathbb{Z}$ does not have a multiplicative identity, so characteristic does not make sense for it.
$endgroup$
– Aniruddh Agarwal
Dec 12 '18 at 7:44














$begingroup$
@AniruddhAgarwal you're right, because the "unity" of $mathbb{Z}$ became $3$ in this new ring. Thanks for let me see this!
$endgroup$
– Alessar
Dec 12 '18 at 7:46




$begingroup$
@AniruddhAgarwal you're right, because the "unity" of $mathbb{Z}$ became $3$ in this new ring. Thanks for let me see this!
$endgroup$
– Alessar
Dec 12 '18 at 7:46




1




1




$begingroup$
The ring $mathbb{Z} times 5mathbb{Z}$ also does not have a multiplicative identity. Your solution for $mathbb{Z}_5 times mathbb{Z}_3$ is indeed correct (by the Chinese Remainder Theorem.)
$endgroup$
– Aniruddh Agarwal
Dec 12 '18 at 7:49






$begingroup$
The ring $mathbb{Z} times 5mathbb{Z}$ also does not have a multiplicative identity. Your solution for $mathbb{Z}_5 times mathbb{Z}_3$ is indeed correct (by the Chinese Remainder Theorem.)
$endgroup$
– Aniruddh Agarwal
Dec 12 '18 at 7:49












2 Answers
2






active

oldest

votes


















1












$begingroup$

You intuitions are right. But according to definition,




  • there is no $n$ so that $n cdot 3=0$

  • there is no $n$ so that $n cdot(1,5)=(0,0)$


  • For all $(a,b) in Bbb Z_5 times Bbb Z_3$, $$15.(a,b)=(0,0)$$



where $n cdot a=underbrace{a+a+cdots+a}_{n;text{summands}}$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    $textbf{More explanation for third:}$



    Let $text{char}(mathbb{Z}_5 times mathbb{Z}_3)=n$. So,
    for all $(a,b) in Bbb Z_5 times Bbb Z_3$ $$n(a,b)=(na,nb)=(0,0). $$
    begin{align}
    n(a,b)=(na,nb)=(0,0) &iff 5mid na quad mbox{and} quad 3mid nb\
    &iff 5mid n quad mbox{and} 3mid n (mbox{$5$ prime and $5nmid a$ so $5mid n$.
    Similar situation is for $3mid nb$}) \
    &iff nmid 15
    &iff n=15
    end{align}






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      You intuitions are right. But according to definition,




      • there is no $n$ so that $n cdot 3=0$

      • there is no $n$ so that $n cdot(1,5)=(0,0)$


      • For all $(a,b) in Bbb Z_5 times Bbb Z_3$, $$15.(a,b)=(0,0)$$



      where $n cdot a=underbrace{a+a+cdots+a}_{n;text{summands}}$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        You intuitions are right. But according to definition,




        • there is no $n$ so that $n cdot 3=0$

        • there is no $n$ so that $n cdot(1,5)=(0,0)$


        • For all $(a,b) in Bbb Z_5 times Bbb Z_3$, $$15.(a,b)=(0,0)$$



        where $n cdot a=underbrace{a+a+cdots+a}_{n;text{summands}}$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          You intuitions are right. But according to definition,




          • there is no $n$ so that $n cdot 3=0$

          • there is no $n$ so that $n cdot(1,5)=(0,0)$


          • For all $(a,b) in Bbb Z_5 times Bbb Z_3$, $$15.(a,b)=(0,0)$$



          where $n cdot a=underbrace{a+a+cdots+a}_{n;text{summands}}$






          share|cite|improve this answer











          $endgroup$



          You intuitions are right. But according to definition,




          • there is no $n$ so that $n cdot 3=0$

          • there is no $n$ so that $n cdot(1,5)=(0,0)$


          • For all $(a,b) in Bbb Z_5 times Bbb Z_3$, $$15.(a,b)=(0,0)$$



          where $n cdot a=underbrace{a+a+cdots+a}_{n;text{summands}}$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 7:53

























          answered Dec 12 '18 at 7:48









          Chinnapparaj RChinnapparaj R

          5,4331928




          5,4331928























              1












              $begingroup$

              $textbf{More explanation for third:}$



              Let $text{char}(mathbb{Z}_5 times mathbb{Z}_3)=n$. So,
              for all $(a,b) in Bbb Z_5 times Bbb Z_3$ $$n(a,b)=(na,nb)=(0,0). $$
              begin{align}
              n(a,b)=(na,nb)=(0,0) &iff 5mid na quad mbox{and} quad 3mid nb\
              &iff 5mid n quad mbox{and} 3mid n (mbox{$5$ prime and $5nmid a$ so $5mid n$.
              Similar situation is for $3mid nb$}) \
              &iff nmid 15
              &iff n=15
              end{align}






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $textbf{More explanation for third:}$



                Let $text{char}(mathbb{Z}_5 times mathbb{Z}_3)=n$. So,
                for all $(a,b) in Bbb Z_5 times Bbb Z_3$ $$n(a,b)=(na,nb)=(0,0). $$
                begin{align}
                n(a,b)=(na,nb)=(0,0) &iff 5mid na quad mbox{and} quad 3mid nb\
                &iff 5mid n quad mbox{and} 3mid n (mbox{$5$ prime and $5nmid a$ so $5mid n$.
                Similar situation is for $3mid nb$}) \
                &iff nmid 15
                &iff n=15
                end{align}






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $textbf{More explanation for third:}$



                  Let $text{char}(mathbb{Z}_5 times mathbb{Z}_3)=n$. So,
                  for all $(a,b) in Bbb Z_5 times Bbb Z_3$ $$n(a,b)=(na,nb)=(0,0). $$
                  begin{align}
                  n(a,b)=(na,nb)=(0,0) &iff 5mid na quad mbox{and} quad 3mid nb\
                  &iff 5mid n quad mbox{and} 3mid n (mbox{$5$ prime and $5nmid a$ so $5mid n$.
                  Similar situation is for $3mid nb$}) \
                  &iff nmid 15
                  &iff n=15
                  end{align}






                  share|cite|improve this answer









                  $endgroup$



                  $textbf{More explanation for third:}$



                  Let $text{char}(mathbb{Z}_5 times mathbb{Z}_3)=n$. So,
                  for all $(a,b) in Bbb Z_5 times Bbb Z_3$ $$n(a,b)=(na,nb)=(0,0). $$
                  begin{align}
                  n(a,b)=(na,nb)=(0,0) &iff 5mid na quad mbox{and} quad 3mid nb\
                  &iff 5mid n quad mbox{and} 3mid n (mbox{$5$ prime and $5nmid a$ so $5mid n$.
                  Similar situation is for $3mid nb$}) \
                  &iff nmid 15
                  &iff n=15
                  end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 9:23









                  1Spectre11Spectre1

                  999




                  999






























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