Show that $mathbb{Z}[x] to mathbb{Z}[x]/(2) times mathbb{Z}[x]/(x)$ is not surjective.












3












$begingroup$


In Aluffi's book it says that $mathbb{Z}[x] to mathbb{Z}[x]/(2) times mathbb{Z}[x]/(x)$ is not surjective. To prove that I need to know how that means at all! Why it is not surjective if for any polynomial $q(x)$ we can write $(q(x)+2n_1, q(x)+n_2x)$?






abstract-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    How is the map defined?
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:47






  • 1




    $begingroup$
    @Servaes, natural projection. It's an example to a theorem for $phi$ a natural projection.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:48


















3












$begingroup$


In Aluffi's book it says that $mathbb{Z}[x] to mathbb{Z}[x]/(2) times mathbb{Z}[x]/(x)$ is not surjective. To prove that I need to know how that means at all! Why it is not surjective if for any polynomial $q(x)$ we can write $(q(x)+2n_1, q(x)+n_2x)$?






abstract-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    How is the map defined?
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:47






  • 1




    $begingroup$
    @Servaes, natural projection. It's an example to a theorem for $phi$ a natural projection.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:48
















3












3








3





$begingroup$


In Aluffi's book it says that $mathbb{Z}[x] to mathbb{Z}[x]/(2) times mathbb{Z}[x]/(x)$ is not surjective. To prove that I need to know how that means at all! Why it is not surjective if for any polynomial $q(x)$ we can write $(q(x)+2n_1, q(x)+n_2x)$?






abstract-algebra







share|cite|improve this question









$endgroup$




In Aluffi's book it says that $mathbb{Z}[x] to mathbb{Z}[x]/(2) times mathbb{Z}[x]/(x)$ is not surjective. To prove that I need to know how that means at all! Why it is not surjective if for any polynomial $q(x)$ we can write $(q(x)+2n_1, q(x)+n_2x)$?






abstract-algebra




abstract-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 7:44









72D72D

512116




512116












  • $begingroup$
    How is the map defined?
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:47






  • 1




    $begingroup$
    @Servaes, natural projection. It's an example to a theorem for $phi$ a natural projection.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:48




















  • $begingroup$
    How is the map defined?
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:47






  • 1




    $begingroup$
    @Servaes, natural projection. It's an example to a theorem for $phi$ a natural projection.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:48


















$begingroup$
How is the map defined?
$endgroup$
– Servaes
Dec 3 '18 at 7:47




$begingroup$
How is the map defined?
$endgroup$
– Servaes
Dec 3 '18 at 7:47




1




1




$begingroup$
@Servaes, natural projection. It's an example to a theorem for $phi$ a natural projection.
$endgroup$
– 72D
Dec 3 '18 at 7:48






$begingroup$
@Servaes, natural projection. It's an example to a theorem for $phi$ a natural projection.
$endgroup$
– 72D
Dec 3 '18 at 7:48












2 Answers
2






active

oldest

votes


















3












$begingroup$

The product on the right hand side is generated by $(1,0)$ and $(0,1)$. So the map is surjective if and only if these are in the image of the map, i.e. if there exist polynomials $f,ginBbb{Z}[x]$ such that
$$f longmapsto (1,0)qquadtext{ and }qquad g longmapsto (0,1).$$
Show that this is impossible.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
    $endgroup$
    – 72D
    Dec 3 '18 at 7:50






  • 1




    $begingroup$
    They are polynomials with integer coefficients mod $2$, and simply integers.
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:51












  • $begingroup$
    I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:59










  • $begingroup$
    I've expanded my hint into a sketch of an answer.
    $endgroup$
    – Servaes
    Dec 3 '18 at 8:05






  • 1




    $begingroup$
    "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
    $endgroup$
    – 72D
    Dec 3 '18 at 8:05



















0












$begingroup$

Hint $ R to R/I times R/J $ surjective $Rightarrow I + J = 1$



since $,rto (1,0),Rightarrow, 1 = (1!-!r) + rin I + J$






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%2f3023749%2fshow-that-mathbbzx-to-mathbbzx-2-times-mathbbzx-x-is-not%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    The product on the right hand side is generated by $(1,0)$ and $(0,1)$. So the map is surjective if and only if these are in the image of the map, i.e. if there exist polynomials $f,ginBbb{Z}[x]$ such that
    $$f longmapsto (1,0)qquadtext{ and }qquad g longmapsto (0,1).$$
    Show that this is impossible.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
      $endgroup$
      – 72D
      Dec 3 '18 at 7:50






    • 1




      $begingroup$
      They are polynomials with integer coefficients mod $2$, and simply integers.
      $endgroup$
      – Servaes
      Dec 3 '18 at 7:51












    • $begingroup$
      I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
      $endgroup$
      – 72D
      Dec 3 '18 at 7:59










    • $begingroup$
      I've expanded my hint into a sketch of an answer.
      $endgroup$
      – Servaes
      Dec 3 '18 at 8:05






    • 1




      $begingroup$
      "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
      $endgroup$
      – 72D
      Dec 3 '18 at 8:05
















    3












    $begingroup$

    The product on the right hand side is generated by $(1,0)$ and $(0,1)$. So the map is surjective if and only if these are in the image of the map, i.e. if there exist polynomials $f,ginBbb{Z}[x]$ such that
    $$f longmapsto (1,0)qquadtext{ and }qquad g longmapsto (0,1).$$
    Show that this is impossible.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
      $endgroup$
      – 72D
      Dec 3 '18 at 7:50






    • 1




      $begingroup$
      They are polynomials with integer coefficients mod $2$, and simply integers.
      $endgroup$
      – Servaes
      Dec 3 '18 at 7:51












    • $begingroup$
      I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
      $endgroup$
      – 72D
      Dec 3 '18 at 7:59










    • $begingroup$
      I've expanded my hint into a sketch of an answer.
      $endgroup$
      – Servaes
      Dec 3 '18 at 8:05






    • 1




      $begingroup$
      "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
      $endgroup$
      – 72D
      Dec 3 '18 at 8:05














    3












    3








    3





    $begingroup$

    The product on the right hand side is generated by $(1,0)$ and $(0,1)$. So the map is surjective if and only if these are in the image of the map, i.e. if there exist polynomials $f,ginBbb{Z}[x]$ such that
    $$f longmapsto (1,0)qquadtext{ and }qquad g longmapsto (0,1).$$
    Show that this is impossible.






    share|cite|improve this answer











    $endgroup$



    The product on the right hand side is generated by $(1,0)$ and $(0,1)$. So the map is surjective if and only if these are in the image of the map, i.e. if there exist polynomials $f,ginBbb{Z}[x]$ such that
    $$f longmapsto (1,0)qquadtext{ and }qquad g longmapsto (0,1).$$
    Show that this is impossible.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 3 '18 at 8:05

























    answered Dec 3 '18 at 7:48









    ServaesServaes

    23.4k33893




    23.4k33893












    • $begingroup$
      How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
      $endgroup$
      – 72D
      Dec 3 '18 at 7:50






    • 1




      $begingroup$
      They are polynomials with integer coefficients mod $2$, and simply integers.
      $endgroup$
      – Servaes
      Dec 3 '18 at 7:51












    • $begingroup$
      I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
      $endgroup$
      – 72D
      Dec 3 '18 at 7:59










    • $begingroup$
      I've expanded my hint into a sketch of an answer.
      $endgroup$
      – Servaes
      Dec 3 '18 at 8:05






    • 1




      $begingroup$
      "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
      $endgroup$
      – 72D
      Dec 3 '18 at 8:05


















    • $begingroup$
      How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
      $endgroup$
      – 72D
      Dec 3 '18 at 7:50






    • 1




      $begingroup$
      They are polynomials with integer coefficients mod $2$, and simply integers.
      $endgroup$
      – Servaes
      Dec 3 '18 at 7:51












    • $begingroup$
      I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
      $endgroup$
      – 72D
      Dec 3 '18 at 7:59










    • $begingroup$
      I've expanded my hint into a sketch of an answer.
      $endgroup$
      – Servaes
      Dec 3 '18 at 8:05






    • 1




      $begingroup$
      "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
      $endgroup$
      – 72D
      Dec 3 '18 at 8:05
















    $begingroup$
    How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
    $endgroup$
    – 72D
    Dec 3 '18 at 7:50




    $begingroup$
    How are the typical/general elements of $mathbb{Z}[x]/(2)$ and $mathbb{Z}[x]/(x)$?
    $endgroup$
    – 72D
    Dec 3 '18 at 7:50




    1




    1




    $begingroup$
    They are polynomials with integer coefficients mod $2$, and simply integers.
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:51






    $begingroup$
    They are polynomials with integer coefficients mod $2$, and simply integers.
    $endgroup$
    – Servaes
    Dec 3 '18 at 7:51














    $begingroup$
    I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:59




    $begingroup$
    I could understand now their meanings but still don't understand the meaning for surjection esp because the rhs is 'double'-element. To try to show that (first) it is surjective (then fail), so we try to find a polynomial such that simultaneously (??) is an integer and a polynomial with integer coefficients mod 2? But it is not intersection it is product.
    $endgroup$
    – 72D
    Dec 3 '18 at 7:59












    $begingroup$
    I've expanded my hint into a sketch of an answer.
    $endgroup$
    – Servaes
    Dec 3 '18 at 8:05




    $begingroup$
    I've expanded my hint into a sketch of an answer.
    $endgroup$
    – Servaes
    Dec 3 '18 at 8:05




    1




    1




    $begingroup$
    "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
    $endgroup$
    – 72D
    Dec 3 '18 at 8:05




    $begingroup$
    "Which elements of Z[x] map to (1,0)" : so is this a counterexample? : for any polynomial q(x) in Z[x], 2q(x)+1 is the only element of Z[x] such that it is 1 in Z[x]/(2) but 2q(x)+1 is never 0 in Z[x]/(x); am I right?
    $endgroup$
    – 72D
    Dec 3 '18 at 8:05











    0












    $begingroup$

    Hint $ R to R/I times R/J $ surjective $Rightarrow I + J = 1$



    since $,rto (1,0),Rightarrow, 1 = (1!-!r) + rin I + J$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint $ R to R/I times R/J $ surjective $Rightarrow I + J = 1$



      since $,rto (1,0),Rightarrow, 1 = (1!-!r) + rin I + J$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint $ R to R/I times R/J $ surjective $Rightarrow I + J = 1$



        since $,rto (1,0),Rightarrow, 1 = (1!-!r) + rin I + J$






        share|cite|improve this answer









        $endgroup$



        Hint $ R to R/I times R/J $ surjective $Rightarrow I + J = 1$



        since $,rto (1,0),Rightarrow, 1 = (1!-!r) + rin I + J$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 23:03









        Bill DubuqueBill Dubuque

        210k29191639




        210k29191639






























            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%2f3023749%2fshow-that-mathbbzx-to-mathbbzx-2-times-mathbbzx-x-is-not%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more