equality of inverse limits in $R=k[x_1,x_2,…]$












1












$begingroup$


Let $k$ be a field, $R=k[x_1,x_2,x_3...]=k[x_i]_{mathbb{N}_0}$ and $mathfrak{m}=(x_1,x_2,x_3,...)$.



I want to check whether the following equality is true or not:




$varprojlim{mathfrak{m}/mathfrak{m}^{i+1}}= mathfrak{m}(varprojlim{R/mathfrak{m}^i})$




I claim that this equality is true. And I claim




$varprojlim{mathfrak{m}/mathfrak{m}^{i+1}}=L= mathfrak{m}(varprojlim{R/mathfrak{m}^i})$, where $L={fin k[[x_i]]_{i in mathbb{N}_0} | f(0)=0}$




My try:



I define $phi_i: L longrightarrow mathfrak{m}/mathfrak{m}^{i+1}: p mapsto p + mathfrak{m}^{i+1}$.



Let $F:mathbb{N}^{op} longrightarrow mathcal{C}$ such that $F(i)=mathfrak{m}/mathfrak{m}^{i+1}$ and $F(igeq j)= mathfrak{m}/mathfrak{m}^{i+1} longrightarrow mathfrak{m}/mathfrak{m}^{j+1}$.



We want to check that $F(igeq j) circ phi_j=phi_i$:



$F(igeq j): mathfrak{m}/mathfrak{m}^{i+1} longrightarrow mathfrak{m}/mathfrak{m}^{j+1}: p + mathfrak{m}^{i+1} mapsto p + mathfrak{m}^{j+1}$



$F(igeq j) circ phi_i: L longrightarrow p/mathfrak{m}^{j+1}: p mapsto p + mathfrak{m}^{j+1}$



$phi_j: L longrightarrow mathfrak{m}/mathfrak{m}^{j+1}$



And this is $F(igeq j)circ phi_i =phi_j$.



Now we have to see that if there is a $(L',alpha)$ satisfying the same condition, then there exists a unique morphism $u:Llongrightarrow L'$ such that $phi circ u = alpha$. But I do not know how to check that.



I know I still need to prove that the right inverse limit is equal to $L$.



But how could I prove the fact mentioned above?



Thank you.






abstract-algebra commutative-algebra limits-colimits







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $k$ be a field, $R=k[x_1,x_2,x_3...]=k[x_i]_{mathbb{N}_0}$ and $mathfrak{m}=(x_1,x_2,x_3,...)$.



    I want to check whether the following equality is true or not:




    $varprojlim{mathfrak{m}/mathfrak{m}^{i+1}}= mathfrak{m}(varprojlim{R/mathfrak{m}^i})$




    I claim that this equality is true. And I claim




    $varprojlim{mathfrak{m}/mathfrak{m}^{i+1}}=L= mathfrak{m}(varprojlim{R/mathfrak{m}^i})$, where $L={fin k[[x_i]]_{i in mathbb{N}_0} | f(0)=0}$




    My try:



    I define $phi_i: L longrightarrow mathfrak{m}/mathfrak{m}^{i+1}: p mapsto p + mathfrak{m}^{i+1}$.



    Let $F:mathbb{N}^{op} longrightarrow mathcal{C}$ such that $F(i)=mathfrak{m}/mathfrak{m}^{i+1}$ and $F(igeq j)= mathfrak{m}/mathfrak{m}^{i+1} longrightarrow mathfrak{m}/mathfrak{m}^{j+1}$.



    We want to check that $F(igeq j) circ phi_j=phi_i$:



    $F(igeq j): mathfrak{m}/mathfrak{m}^{i+1} longrightarrow mathfrak{m}/mathfrak{m}^{j+1}: p + mathfrak{m}^{i+1} mapsto p + mathfrak{m}^{j+1}$



    $F(igeq j) circ phi_i: L longrightarrow p/mathfrak{m}^{j+1}: p mapsto p + mathfrak{m}^{j+1}$



    $phi_j: L longrightarrow mathfrak{m}/mathfrak{m}^{j+1}$



    And this is $F(igeq j)circ phi_i =phi_j$.



    Now we have to see that if there is a $(L',alpha)$ satisfying the same condition, then there exists a unique morphism $u:Llongrightarrow L'$ such that $phi circ u = alpha$. But I do not know how to check that.



    I know I still need to prove that the right inverse limit is equal to $L$.



    But how could I prove the fact mentioned above?



    Thank you.






    abstract-algebra commutative-algebra limits-colimits







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $k$ be a field, $R=k[x_1,x_2,x_3...]=k[x_i]_{mathbb{N}_0}$ and $mathfrak{m}=(x_1,x_2,x_3,...)$.



      I want to check whether the following equality is true or not:




      $varprojlim{mathfrak{m}/mathfrak{m}^{i+1}}= mathfrak{m}(varprojlim{R/mathfrak{m}^i})$




      I claim that this equality is true. And I claim




      $varprojlim{mathfrak{m}/mathfrak{m}^{i+1}}=L= mathfrak{m}(varprojlim{R/mathfrak{m}^i})$, where $L={fin k[[x_i]]_{i in mathbb{N}_0} | f(0)=0}$




      My try:



      I define $phi_i: L longrightarrow mathfrak{m}/mathfrak{m}^{i+1}: p mapsto p + mathfrak{m}^{i+1}$.



      Let $F:mathbb{N}^{op} longrightarrow mathcal{C}$ such that $F(i)=mathfrak{m}/mathfrak{m}^{i+1}$ and $F(igeq j)= mathfrak{m}/mathfrak{m}^{i+1} longrightarrow mathfrak{m}/mathfrak{m}^{j+1}$.



      We want to check that $F(igeq j) circ phi_j=phi_i$:



      $F(igeq j): mathfrak{m}/mathfrak{m}^{i+1} longrightarrow mathfrak{m}/mathfrak{m}^{j+1}: p + mathfrak{m}^{i+1} mapsto p + mathfrak{m}^{j+1}$



      $F(igeq j) circ phi_i: L longrightarrow p/mathfrak{m}^{j+1}: p mapsto p + mathfrak{m}^{j+1}$



      $phi_j: L longrightarrow mathfrak{m}/mathfrak{m}^{j+1}$



      And this is $F(igeq j)circ phi_i =phi_j$.



      Now we have to see that if there is a $(L',alpha)$ satisfying the same condition, then there exists a unique morphism $u:Llongrightarrow L'$ such that $phi circ u = alpha$. But I do not know how to check that.



      I know I still need to prove that the right inverse limit is equal to $L$.



      But how could I prove the fact mentioned above?



      Thank you.






      abstract-algebra commutative-algebra limits-colimits







      share|cite|improve this question











      $endgroup$




      Let $k$ be a field, $R=k[x_1,x_2,x_3...]=k[x_i]_{mathbb{N}_0}$ and $mathfrak{m}=(x_1,x_2,x_3,...)$.



      I want to check whether the following equality is true or not:




      $varprojlim{mathfrak{m}/mathfrak{m}^{i+1}}= mathfrak{m}(varprojlim{R/mathfrak{m}^i})$




      I claim that this equality is true. And I claim




      $varprojlim{mathfrak{m}/mathfrak{m}^{i+1}}=L= mathfrak{m}(varprojlim{R/mathfrak{m}^i})$, where $L={fin k[[x_i]]_{i in mathbb{N}_0} | f(0)=0}$




      My try:



      I define $phi_i: L longrightarrow mathfrak{m}/mathfrak{m}^{i+1}: p mapsto p + mathfrak{m}^{i+1}$.



      Let $F:mathbb{N}^{op} longrightarrow mathcal{C}$ such that $F(i)=mathfrak{m}/mathfrak{m}^{i+1}$ and $F(igeq j)= mathfrak{m}/mathfrak{m}^{i+1} longrightarrow mathfrak{m}/mathfrak{m}^{j+1}$.



      We want to check that $F(igeq j) circ phi_j=phi_i$:



      $F(igeq j): mathfrak{m}/mathfrak{m}^{i+1} longrightarrow mathfrak{m}/mathfrak{m}^{j+1}: p + mathfrak{m}^{i+1} mapsto p + mathfrak{m}^{j+1}$



      $F(igeq j) circ phi_i: L longrightarrow p/mathfrak{m}^{j+1}: p mapsto p + mathfrak{m}^{j+1}$



      $phi_j: L longrightarrow mathfrak{m}/mathfrak{m}^{j+1}$



      And this is $F(igeq j)circ phi_i =phi_j$.



      Now we have to see that if there is a $(L',alpha)$ satisfying the same condition, then there exists a unique morphism $u:Llongrightarrow L'$ such that $phi circ u = alpha$. But I do not know how to check that.



      I know I still need to prove that the right inverse limit is equal to $L$.



      But how could I prove the fact mentioned above?



      Thank you.






      abstract-algebra commutative-algebra limits-colimits




      abstract-algebra commutative-algebra limits-colimits






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      share|cite|improve this question













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      edited Jan 16 at 10:00







      idriskameni

















      asked Jan 2 at 10:28









      idriskameniidriskameni

      585318




      585318






















          1 Answer
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          $begingroup$

          In case someone come back to this post one day, I will answer what I found.



          The equality does not hold.



          $mathfrak{m}(varprojlim{R/mathfrak{m}^i})subset varprojlim{mathfrak{m}/mathfrak{m}^{i+1}} $



          To see it, we can check that $$g(x_1,x_2,x_3,...)=sum_{n=1}^{infty} x_n^n$$ is not in $mathfrak{m}(varprojlim{R/mathfrak{m}^i})$. But it is in $varprojlim{mathfrak{m}/mathfrak{m}^{i+1}}$.






          share|cite|improve this answer









          $endgroup$













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            1 Answer
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            active

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            1












            $begingroup$

            In case someone come back to this post one day, I will answer what I found.



            The equality does not hold.



            $mathfrak{m}(varprojlim{R/mathfrak{m}^i})subset varprojlim{mathfrak{m}/mathfrak{m}^{i+1}} $



            To see it, we can check that $$g(x_1,x_2,x_3,...)=sum_{n=1}^{infty} x_n^n$$ is not in $mathfrak{m}(varprojlim{R/mathfrak{m}^i})$. But it is in $varprojlim{mathfrak{m}/mathfrak{m}^{i+1}}$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              In case someone come back to this post one day, I will answer what I found.



              The equality does not hold.



              $mathfrak{m}(varprojlim{R/mathfrak{m}^i})subset varprojlim{mathfrak{m}/mathfrak{m}^{i+1}} $



              To see it, we can check that $$g(x_1,x_2,x_3,...)=sum_{n=1}^{infty} x_n^n$$ is not in $mathfrak{m}(varprojlim{R/mathfrak{m}^i})$. But it is in $varprojlim{mathfrak{m}/mathfrak{m}^{i+1}}$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                In case someone come back to this post one day, I will answer what I found.



                The equality does not hold.



                $mathfrak{m}(varprojlim{R/mathfrak{m}^i})subset varprojlim{mathfrak{m}/mathfrak{m}^{i+1}} $



                To see it, we can check that $$g(x_1,x_2,x_3,...)=sum_{n=1}^{infty} x_n^n$$ is not in $mathfrak{m}(varprojlim{R/mathfrak{m}^i})$. But it is in $varprojlim{mathfrak{m}/mathfrak{m}^{i+1}}$.






                share|cite|improve this answer









                $endgroup$



                In case someone come back to this post one day, I will answer what I found.



                The equality does not hold.



                $mathfrak{m}(varprojlim{R/mathfrak{m}^i})subset varprojlim{mathfrak{m}/mathfrak{m}^{i+1}} $



                To see it, we can check that $$g(x_1,x_2,x_3,...)=sum_{n=1}^{infty} x_n^n$$ is not in $mathfrak{m}(varprojlim{R/mathfrak{m}^i})$. But it is in $varprojlim{mathfrak{m}/mathfrak{m}^{i+1}}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 16 at 10:04









                idriskameniidriskameni

                585318




                585318






























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