Dummit and Foote 10.4.10












0












$begingroup$


It is a problem 10.4.10 of Dummit and Foote.





  1. Suppose $R$ is commutative and $Ncong R^n$ is a free $R-$module of rank $n$ with $R-$module basis $e_1,e_2,dots, e_n$.


(a) For any nonzero $R$-module $M$ show that every element of $Motimes N$ can be written uniquely in the form $sum_{i=1}^{n}m_iotimes e_i$ where $m_iin M$. Deduce that if $sum_{i=1}^{n}m_iotimes e_i=0$ in $Motimes N$ then $m_i=0$ for $i=1,dots,n$.




Actually I already proved the first part.



Since any element in $Motimes N$ is a finite sum of simple tensors, for any ${m^j}_{j=1}^{p}subset M$ and ${n^j}_{j=1}^psubset N$ observe that
begin{align*}
sum_{j=1}^{p}m^jotimes n^j=sum_{j=1}^{p} sum_{i=1}^{n}m^jotimes r_i^je_i=sum_{i=1}^{n}sum_{j=1}^{p}(m^jr^j_i)otimes e_i=sum_{i=1}^{n}left( sum_{j=1}^{p}m^jr^j_iright)otimes e_i
end{align*}



where $$n^j=sum_{i=1}^{n}r^j_ie_i.$$



In other words, $$m_i=sum_{j=1}^{p}m^jr^j_i.$$



But I cannot see why $sum_{i=1}^{n}m_iotimes e_i=0$ implies $m_i=0$ for all $i$. Since the problem says 'Deduce', I think I need to use the result... But how?



I thank to any help in advance.






abstract-algebra tensor-products







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you address the word "uniquely" in your answer to the first part?
    $endgroup$
    – Lord Shark the Unknown
    Dec 20 '18 at 7:57










  • $begingroup$
    @LordSharktheUnknown Oh! I did not. Does that mean I need to change the way to prove that?
    $endgroup$
    – Lev Ban
    Dec 20 '18 at 7:59










  • $begingroup$
    Ensure that the title contains some actual words, and not only LaTeX -- the search functionality of MSE does not interact well with typeset mathematics, making questions comprising only LaTeX (much) harder to find. Furthermore, the MathJax context menu (which pops up when you right-click on some typeset expression) overrides the browser's link context menu, making e.g. opening the question in a new tab difficult.
    $endgroup$
    – user10354138
    Dec 20 '18 at 8:04
















0












$begingroup$


It is a problem 10.4.10 of Dummit and Foote.





  1. Suppose $R$ is commutative and $Ncong R^n$ is a free $R-$module of rank $n$ with $R-$module basis $e_1,e_2,dots, e_n$.


(a) For any nonzero $R$-module $M$ show that every element of $Motimes N$ can be written uniquely in the form $sum_{i=1}^{n}m_iotimes e_i$ where $m_iin M$. Deduce that if $sum_{i=1}^{n}m_iotimes e_i=0$ in $Motimes N$ then $m_i=0$ for $i=1,dots,n$.




Actually I already proved the first part.



Since any element in $Motimes N$ is a finite sum of simple tensors, for any ${m^j}_{j=1}^{p}subset M$ and ${n^j}_{j=1}^psubset N$ observe that
begin{align*}
sum_{j=1}^{p}m^jotimes n^j=sum_{j=1}^{p} sum_{i=1}^{n}m^jotimes r_i^je_i=sum_{i=1}^{n}sum_{j=1}^{p}(m^jr^j_i)otimes e_i=sum_{i=1}^{n}left( sum_{j=1}^{p}m^jr^j_iright)otimes e_i
end{align*}



where $$n^j=sum_{i=1}^{n}r^j_ie_i.$$



In other words, $$m_i=sum_{j=1}^{p}m^jr^j_i.$$



But I cannot see why $sum_{i=1}^{n}m_iotimes e_i=0$ implies $m_i=0$ for all $i$. Since the problem says 'Deduce', I think I need to use the result... But how?



I thank to any help in advance.






abstract-algebra tensor-products







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you address the word "uniquely" in your answer to the first part?
    $endgroup$
    – Lord Shark the Unknown
    Dec 20 '18 at 7:57










  • $begingroup$
    @LordSharktheUnknown Oh! I did not. Does that mean I need to change the way to prove that?
    $endgroup$
    – Lev Ban
    Dec 20 '18 at 7:59










  • $begingroup$
    Ensure that the title contains some actual words, and not only LaTeX -- the search functionality of MSE does not interact well with typeset mathematics, making questions comprising only LaTeX (much) harder to find. Furthermore, the MathJax context menu (which pops up when you right-click on some typeset expression) overrides the browser's link context menu, making e.g. opening the question in a new tab difficult.
    $endgroup$
    – user10354138
    Dec 20 '18 at 8:04














0












0








0





$begingroup$


It is a problem 10.4.10 of Dummit and Foote.





  1. Suppose $R$ is commutative and $Ncong R^n$ is a free $R-$module of rank $n$ with $R-$module basis $e_1,e_2,dots, e_n$.


(a) For any nonzero $R$-module $M$ show that every element of $Motimes N$ can be written uniquely in the form $sum_{i=1}^{n}m_iotimes e_i$ where $m_iin M$. Deduce that if $sum_{i=1}^{n}m_iotimes e_i=0$ in $Motimes N$ then $m_i=0$ for $i=1,dots,n$.




Actually I already proved the first part.



Since any element in $Motimes N$ is a finite sum of simple tensors, for any ${m^j}_{j=1}^{p}subset M$ and ${n^j}_{j=1}^psubset N$ observe that
begin{align*}
sum_{j=1}^{p}m^jotimes n^j=sum_{j=1}^{p} sum_{i=1}^{n}m^jotimes r_i^je_i=sum_{i=1}^{n}sum_{j=1}^{p}(m^jr^j_i)otimes e_i=sum_{i=1}^{n}left( sum_{j=1}^{p}m^jr^j_iright)otimes e_i
end{align*}



where $$n^j=sum_{i=1}^{n}r^j_ie_i.$$



In other words, $$m_i=sum_{j=1}^{p}m^jr^j_i.$$



But I cannot see why $sum_{i=1}^{n}m_iotimes e_i=0$ implies $m_i=0$ for all $i$. Since the problem says 'Deduce', I think I need to use the result... But how?



I thank to any help in advance.






abstract-algebra tensor-products







share|cite|improve this question











$endgroup$




It is a problem 10.4.10 of Dummit and Foote.





  1. Suppose $R$ is commutative and $Ncong R^n$ is a free $R-$module of rank $n$ with $R-$module basis $e_1,e_2,dots, e_n$.


(a) For any nonzero $R$-module $M$ show that every element of $Motimes N$ can be written uniquely in the form $sum_{i=1}^{n}m_iotimes e_i$ where $m_iin M$. Deduce that if $sum_{i=1}^{n}m_iotimes e_i=0$ in $Motimes N$ then $m_i=0$ for $i=1,dots,n$.




Actually I already proved the first part.



Since any element in $Motimes N$ is a finite sum of simple tensors, for any ${m^j}_{j=1}^{p}subset M$ and ${n^j}_{j=1}^psubset N$ observe that
begin{align*}
sum_{j=1}^{p}m^jotimes n^j=sum_{j=1}^{p} sum_{i=1}^{n}m^jotimes r_i^je_i=sum_{i=1}^{n}sum_{j=1}^{p}(m^jr^j_i)otimes e_i=sum_{i=1}^{n}left( sum_{j=1}^{p}m^jr^j_iright)otimes e_i
end{align*}



where $$n^j=sum_{i=1}^{n}r^j_ie_i.$$



In other words, $$m_i=sum_{j=1}^{p}m^jr^j_i.$$



But I cannot see why $sum_{i=1}^{n}m_iotimes e_i=0$ implies $m_i=0$ for all $i$. Since the problem says 'Deduce', I think I need to use the result... But how?



I thank to any help in advance.






abstract-algebra tensor-products




abstract-algebra tensor-products






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 8:09







Lev Ban

















asked Dec 20 '18 at 7:56









Lev BanLev Ban

1,0671317




1,0671317












  • $begingroup$
    Did you address the word "uniquely" in your answer to the first part?
    $endgroup$
    – Lord Shark the Unknown
    Dec 20 '18 at 7:57










  • $begingroup$
    @LordSharktheUnknown Oh! I did not. Does that mean I need to change the way to prove that?
    $endgroup$
    – Lev Ban
    Dec 20 '18 at 7:59










  • $begingroup$
    Ensure that the title contains some actual words, and not only LaTeX -- the search functionality of MSE does not interact well with typeset mathematics, making questions comprising only LaTeX (much) harder to find. Furthermore, the MathJax context menu (which pops up when you right-click on some typeset expression) overrides the browser's link context menu, making e.g. opening the question in a new tab difficult.
    $endgroup$
    – user10354138
    Dec 20 '18 at 8:04


















  • $begingroup$
    Did you address the word "uniquely" in your answer to the first part?
    $endgroup$
    – Lord Shark the Unknown
    Dec 20 '18 at 7:57










  • $begingroup$
    @LordSharktheUnknown Oh! I did not. Does that mean I need to change the way to prove that?
    $endgroup$
    – Lev Ban
    Dec 20 '18 at 7:59










  • $begingroup$
    Ensure that the title contains some actual words, and not only LaTeX -- the search functionality of MSE does not interact well with typeset mathematics, making questions comprising only LaTeX (much) harder to find. Furthermore, the MathJax context menu (which pops up when you right-click on some typeset expression) overrides the browser's link context menu, making e.g. opening the question in a new tab difficult.
    $endgroup$
    – user10354138
    Dec 20 '18 at 8:04
















$begingroup$
Did you address the word "uniquely" in your answer to the first part?
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 7:57




$begingroup$
Did you address the word "uniquely" in your answer to the first part?
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 7:57












$begingroup$
@LordSharktheUnknown Oh! I did not. Does that mean I need to change the way to prove that?
$endgroup$
– Lev Ban
Dec 20 '18 at 7:59




$begingroup$
@LordSharktheUnknown Oh! I did not. Does that mean I need to change the way to prove that?
$endgroup$
– Lev Ban
Dec 20 '18 at 7:59












$begingroup$
Ensure that the title contains some actual words, and not only LaTeX -- the search functionality of MSE does not interact well with typeset mathematics, making questions comprising only LaTeX (much) harder to find. Furthermore, the MathJax context menu (which pops up when you right-click on some typeset expression) overrides the browser's link context menu, making e.g. opening the question in a new tab difficult.
$endgroup$
– user10354138
Dec 20 '18 at 8:04




$begingroup$
Ensure that the title contains some actual words, and not only LaTeX -- the search functionality of MSE does not interact well with typeset mathematics, making questions comprising only LaTeX (much) harder to find. Furthermore, the MathJax context menu (which pops up when you right-click on some typeset expression) overrides the browser's link context menu, making e.g. opening the question in a new tab difficult.
$endgroup$
– user10354138
Dec 20 '18 at 8:04










1 Answer
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There is a bilinear map $b:Mtimes Nto M$ defined by
$$bleft(m,sum_{i=1}^n r_ie_iright)=r_1m.$$
This induces a linear map $beta:Motimes Nto M$ with
$$betaleft(motimessum_{i=1}^n r_ie_iright)=r_1m.$$
Then $beta(m_1otimes e_1)=m_1$ and $beta(m_iotimes e_i)=0$
for $ine 1$. Then
$$betaleft(sum_{i=1}^n m_iotimes e_iright)=m_1$$
so $sum_{i=1}^n m_iotimes e_i$ determines $m_1$ etc.






share|cite|improve this answer











$endgroup$













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    active

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    1












    $begingroup$

    There is a bilinear map $b:Mtimes Nto M$ defined by
    $$bleft(m,sum_{i=1}^n r_ie_iright)=r_1m.$$
    This induces a linear map $beta:Motimes Nto M$ with
    $$betaleft(motimessum_{i=1}^n r_ie_iright)=r_1m.$$
    Then $beta(m_1otimes e_1)=m_1$ and $beta(m_iotimes e_i)=0$
    for $ine 1$. Then
    $$betaleft(sum_{i=1}^n m_iotimes e_iright)=m_1$$
    so $sum_{i=1}^n m_iotimes e_i$ determines $m_1$ etc.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      There is a bilinear map $b:Mtimes Nto M$ defined by
      $$bleft(m,sum_{i=1}^n r_ie_iright)=r_1m.$$
      This induces a linear map $beta:Motimes Nto M$ with
      $$betaleft(motimessum_{i=1}^n r_ie_iright)=r_1m.$$
      Then $beta(m_1otimes e_1)=m_1$ and $beta(m_iotimes e_i)=0$
      for $ine 1$. Then
      $$betaleft(sum_{i=1}^n m_iotimes e_iright)=m_1$$
      so $sum_{i=1}^n m_iotimes e_i$ determines $m_1$ etc.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        There is a bilinear map $b:Mtimes Nto M$ defined by
        $$bleft(m,sum_{i=1}^n r_ie_iright)=r_1m.$$
        This induces a linear map $beta:Motimes Nto M$ with
        $$betaleft(motimessum_{i=1}^n r_ie_iright)=r_1m.$$
        Then $beta(m_1otimes e_1)=m_1$ and $beta(m_iotimes e_i)=0$
        for $ine 1$. Then
        $$betaleft(sum_{i=1}^n m_iotimes e_iright)=m_1$$
        so $sum_{i=1}^n m_iotimes e_i$ determines $m_1$ etc.






        share|cite|improve this answer











        $endgroup$



        There is a bilinear map $b:Mtimes Nto M$ defined by
        $$bleft(m,sum_{i=1}^n r_ie_iright)=r_1m.$$
        This induces a linear map $beta:Motimes Nto M$ with
        $$betaleft(motimessum_{i=1}^n r_ie_iright)=r_1m.$$
        Then $beta(m_1otimes e_1)=m_1$ and $beta(m_iotimes e_i)=0$
        for $ine 1$. Then
        $$betaleft(sum_{i=1}^n m_iotimes e_iright)=m_1$$
        so $sum_{i=1}^n m_iotimes e_i$ determines $m_1$ etc.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 20 '18 at 17:55

























        answered Dec 20 '18 at 8:03









        Lord Shark the UnknownLord Shark the Unknown

        105k1160133




        105k1160133






























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