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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 16 at 10:00







      idriskameni

















      asked Jan 2 at 10:28









      idriskameniidriskameni

      585318




      585318






















          1 Answer
          1






          active

          oldest

          votes


















          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









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059315%2fequality-of-inverse-limits-in-r-kx-1-x-2%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059315%2fequality-of-inverse-limits-in-r-kx-1-x-2%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Mario Kart Wii

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
                    Read more

                    The Binding of Isaac: Rebirth/Afterbirth

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
                    Read more

                    What does “Dominus providebit” mean?

                    Image
                    5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
                    Read more