Unique prime ideal factorization in domains?












5












$begingroup$


This is a follow-up to this question.




Let $A$ be a domain; let $mathfrak p_1,dots,mathfrak p_k$ be distinct prime ideals of $A$ such that $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $1le ile k$, $jge1$; and let $m$ and $n$ be elements of $mathbb N^k$ such that $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$ Can we conclude that $m$ and $n$ are equal?




user26857 proved that the answer is Yes if $A$ is noetherian (see this answer). (Of course in this case the condition $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $jge1$ is equivalent to $mathfrak p_ine(0)$.)






commutative-algebra maximal-and-prime-ideals integral-domain







share|cite|improve this question









$endgroup$

















    5












    $begingroup$


    This is a follow-up to this question.




    Let $A$ be a domain; let $mathfrak p_1,dots,mathfrak p_k$ be distinct prime ideals of $A$ such that $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $1le ile k$, $jge1$; and let $m$ and $n$ be elements of $mathbb N^k$ such that $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$ Can we conclude that $m$ and $n$ are equal?




    user26857 proved that the answer is Yes if $A$ is noetherian (see this answer). (Of course in this case the condition $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $jge1$ is equivalent to $mathfrak p_ine(0)$.)






    commutative-algebra maximal-and-prime-ideals integral-domain







    share|cite|improve this question









    $endgroup$















      5












      5








      5


      2



      $begingroup$


      This is a follow-up to this question.




      Let $A$ be a domain; let $mathfrak p_1,dots,mathfrak p_k$ be distinct prime ideals of $A$ such that $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $1le ile k$, $jge1$; and let $m$ and $n$ be elements of $mathbb N^k$ such that $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$ Can we conclude that $m$ and $n$ are equal?




      user26857 proved that the answer is Yes if $A$ is noetherian (see this answer). (Of course in this case the condition $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $jge1$ is equivalent to $mathfrak p_ine(0)$.)






      commutative-algebra maximal-and-prime-ideals integral-domain







      share|cite|improve this question









      $endgroup$




      This is a follow-up to this question.




      Let $A$ be a domain; let $mathfrak p_1,dots,mathfrak p_k$ be distinct prime ideals of $A$ such that $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $1le ile k$, $jge1$; and let $m$ and $n$ be elements of $mathbb N^k$ such that $$mathfrak p_1^{m_1}cdotsmathfrak p_k^{m_k}=mathfrak p_1^{n_1}cdotsmathfrak p_k^{n_k}$$ Can we conclude that $m$ and $n$ are equal?




      user26857 proved that the answer is Yes if $A$ is noetherian (see this answer). (Of course in this case the condition $mathfrak p_i^{j+1}nemathfrak p_i^j$ for all $jge1$ is equivalent to $mathfrak p_ine(0)$.)






      commutative-algebra maximal-and-prime-ideals integral-domain




      commutative-algebra maximal-and-prime-ideals integral-domain






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 31 '18 at 13:21









      Pierre-Yves GaillardPierre-Yves Gaillard

      13.2k23181




      13.2k23181






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          The answer is no.



          Let $G$ be the abelian group $mathbb{Z}times mathbb{Z}$ with the lexicographic order, and let $Msubset G$ be the sub-monoid of elements that are greater than or equal to $(0,0)$. Define monoid ideals $I_1:={(m,n):mgeq 1text{ or }ngeq 1}subset M$ and $I_2:={(m,n):mgeq 1}subset M$. For $jinmathbb{Z}_{geq 0}$ and $Isubset M$ an ideal, write $jI:={i_1+ldots+i_j:i_1,ldots,i_jin I}subset M$. Check that for all $j$, we have $jI_1neq (j+1)I_1$ and $jI_2neq (j+1)I_2$, but $I_1+I_2=I_2$.



          We get a counterexample to your question by taking $A=k[M]$ to be the monoid ring of $M$ (where $k$ is any field), $mathfrak{p}_1=k[I_1]$, and $mathfrak{p}_2=k[I_2]$. Then $mathfrak{p}_1^jneq mathfrak{p}_1^{j+1}$ and $mathfrak{p}_2^jneq mathfrak{p}_2^{j+1}$, but $mathfrak{p}_1mathfrak{p}_2=mathfrak{p}_2$.



          An alternate way to describe this example is $A=k[x,y,yx^{-1},yx^{-2},ldots]$, $mathfrak{p}_1=(x)$, $mathfrak{p}_2=(y,yx^{-1},yx^{-2},ldots)$.






          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%2f3057707%2funique-prime-ideal-factorization-in-domains%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









            4












            $begingroup$

            The answer is no.



            Let $G$ be the abelian group $mathbb{Z}times mathbb{Z}$ with the lexicographic order, and let $Msubset G$ be the sub-monoid of elements that are greater than or equal to $(0,0)$. Define monoid ideals $I_1:={(m,n):mgeq 1text{ or }ngeq 1}subset M$ and $I_2:={(m,n):mgeq 1}subset M$. For $jinmathbb{Z}_{geq 0}$ and $Isubset M$ an ideal, write $jI:={i_1+ldots+i_j:i_1,ldots,i_jin I}subset M$. Check that for all $j$, we have $jI_1neq (j+1)I_1$ and $jI_2neq (j+1)I_2$, but $I_1+I_2=I_2$.



            We get a counterexample to your question by taking $A=k[M]$ to be the monoid ring of $M$ (where $k$ is any field), $mathfrak{p}_1=k[I_1]$, and $mathfrak{p}_2=k[I_2]$. Then $mathfrak{p}_1^jneq mathfrak{p}_1^{j+1}$ and $mathfrak{p}_2^jneq mathfrak{p}_2^{j+1}$, but $mathfrak{p}_1mathfrak{p}_2=mathfrak{p}_2$.



            An alternate way to describe this example is $A=k[x,y,yx^{-1},yx^{-2},ldots]$, $mathfrak{p}_1=(x)$, $mathfrak{p}_2=(y,yx^{-1},yx^{-2},ldots)$.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              The answer is no.



              Let $G$ be the abelian group $mathbb{Z}times mathbb{Z}$ with the lexicographic order, and let $Msubset G$ be the sub-monoid of elements that are greater than or equal to $(0,0)$. Define monoid ideals $I_1:={(m,n):mgeq 1text{ or }ngeq 1}subset M$ and $I_2:={(m,n):mgeq 1}subset M$. For $jinmathbb{Z}_{geq 0}$ and $Isubset M$ an ideal, write $jI:={i_1+ldots+i_j:i_1,ldots,i_jin I}subset M$. Check that for all $j$, we have $jI_1neq (j+1)I_1$ and $jI_2neq (j+1)I_2$, but $I_1+I_2=I_2$.



              We get a counterexample to your question by taking $A=k[M]$ to be the monoid ring of $M$ (where $k$ is any field), $mathfrak{p}_1=k[I_1]$, and $mathfrak{p}_2=k[I_2]$. Then $mathfrak{p}_1^jneq mathfrak{p}_1^{j+1}$ and $mathfrak{p}_2^jneq mathfrak{p}_2^{j+1}$, but $mathfrak{p}_1mathfrak{p}_2=mathfrak{p}_2$.



              An alternate way to describe this example is $A=k[x,y,yx^{-1},yx^{-2},ldots]$, $mathfrak{p}_1=(x)$, $mathfrak{p}_2=(y,yx^{-1},yx^{-2},ldots)$.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                The answer is no.



                Let $G$ be the abelian group $mathbb{Z}times mathbb{Z}$ with the lexicographic order, and let $Msubset G$ be the sub-monoid of elements that are greater than or equal to $(0,0)$. Define monoid ideals $I_1:={(m,n):mgeq 1text{ or }ngeq 1}subset M$ and $I_2:={(m,n):mgeq 1}subset M$. For $jinmathbb{Z}_{geq 0}$ and $Isubset M$ an ideal, write $jI:={i_1+ldots+i_j:i_1,ldots,i_jin I}subset M$. Check that for all $j$, we have $jI_1neq (j+1)I_1$ and $jI_2neq (j+1)I_2$, but $I_1+I_2=I_2$.



                We get a counterexample to your question by taking $A=k[M]$ to be the monoid ring of $M$ (where $k$ is any field), $mathfrak{p}_1=k[I_1]$, and $mathfrak{p}_2=k[I_2]$. Then $mathfrak{p}_1^jneq mathfrak{p}_1^{j+1}$ and $mathfrak{p}_2^jneq mathfrak{p}_2^{j+1}$, but $mathfrak{p}_1mathfrak{p}_2=mathfrak{p}_2$.



                An alternate way to describe this example is $A=k[x,y,yx^{-1},yx^{-2},ldots]$, $mathfrak{p}_1=(x)$, $mathfrak{p}_2=(y,yx^{-1},yx^{-2},ldots)$.






                share|cite|improve this answer









                $endgroup$



                The answer is no.



                Let $G$ be the abelian group $mathbb{Z}times mathbb{Z}$ with the lexicographic order, and let $Msubset G$ be the sub-monoid of elements that are greater than or equal to $(0,0)$. Define monoid ideals $I_1:={(m,n):mgeq 1text{ or }ngeq 1}subset M$ and $I_2:={(m,n):mgeq 1}subset M$. For $jinmathbb{Z}_{geq 0}$ and $Isubset M$ an ideal, write $jI:={i_1+ldots+i_j:i_1,ldots,i_jin I}subset M$. Check that for all $j$, we have $jI_1neq (j+1)I_1$ and $jI_2neq (j+1)I_2$, but $I_1+I_2=I_2$.



                We get a counterexample to your question by taking $A=k[M]$ to be the monoid ring of $M$ (where $k$ is any field), $mathfrak{p}_1=k[I_1]$, and $mathfrak{p}_2=k[I_2]$. Then $mathfrak{p}_1^jneq mathfrak{p}_1^{j+1}$ and $mathfrak{p}_2^jneq mathfrak{p}_2^{j+1}$, but $mathfrak{p}_1mathfrak{p}_2=mathfrak{p}_2$.



                An alternate way to describe this example is $A=k[x,y,yx^{-1},yx^{-2},ldots]$, $mathfrak{p}_1=(x)$, $mathfrak{p}_2=(y,yx^{-1},yx^{-2},ldots)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 3 at 0:46









                Julian RosenJulian Rosen

                11.8k12348




                11.8k12348






























                    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%2f3057707%2funique-prime-ideal-factorization-in-domains%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

                    Human spaceflight

                    Image
                    Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
                    Read more

                    Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                    Image
                    0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
                    Read more

                    File:DeusFollowingSea.jpg

                    Image
                    File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
                    Read more