$mathbb{Z_9}$ is not a subring of $mathbb{Z_{12}}$ because Choose the correct option












0














$mathbb{Z_9}$ is not a subring of $mathbb{Z_{12}}$ because



Choose the correct option



$a)$ $mathbb{Z_9}$ is not a subset of $mathbb{Z_{12}}$



$b)$GCD$(9,12) = 3neq 1$



$c)$ $9$ doesnot divide $12$



I thinks option $c)$ will be correct by Lagrange Theorem






abstract-algebra







share|cite|improve this question




















  • 2




    Can you explain why you think that?
    – fleablood
    Nov 19 at 23:34






  • 1




    So Legrange Theorem says the order of every subgroup of a finite group divides the order of a group. And as rings are groups (under their primary operation (in this case addition)) that would apply to rings as well. And the order of $mathbb Z_{12}$ is $12$ and the order of $mathbb Z_9$ is $9$ and $9not mid 12$. Yep.... seems like you've figured it out.
    – fleablood
    Nov 19 at 23:39








  • 1




    .... and a) is obviously false as $Z_9 subset Z_{12}$ and b) is not a requirement of subrings. In fact by legrange the gcd of (non trivial) subrings must never be $1$.
    – fleablood
    Nov 19 at 23:43






  • 1




    I don't really have any more to add to i707107's hint. Need to show it is closed undo addition and multiplication, and that the identitys of C[x] are in C[x^5] and that every element has an additive inverse.
    – fleablood
    Nov 19 at 23:56






  • 1




    @fleablood (a) isn't obviously false as you have suggested. It depends on what definition of $Bbb{Z}_{12}$ and $Bbb{Z}_{9}$ is one using. If you consider $Bbb{Z}_{9}={[0]_9, [1]_9, ldots, [8]_9}$ and $Bbb{Z}_{12}={[0]_{12}, [1]_{12}, ldots, [11]_{12}}$ as quotients of $Bbb{Z}$ by the relevant congruence relation, then $Bbb{Z}_{9}$ isn't a subset of $Bbb{Z}_{12}$.
    – Anurag A
    Nov 20 at 0:00


















0














$mathbb{Z_9}$ is not a subring of $mathbb{Z_{12}}$ because



Choose the correct option



$a)$ $mathbb{Z_9}$ is not a subset of $mathbb{Z_{12}}$



$b)$GCD$(9,12) = 3neq 1$



$c)$ $9$ doesnot divide $12$



I thinks option $c)$ will be correct by Lagrange Theorem






abstract-algebra







share|cite|improve this question




















  • 2




    Can you explain why you think that?
    – fleablood
    Nov 19 at 23:34






  • 1




    So Legrange Theorem says the order of every subgroup of a finite group divides the order of a group. And as rings are groups (under their primary operation (in this case addition)) that would apply to rings as well. And the order of $mathbb Z_{12}$ is $12$ and the order of $mathbb Z_9$ is $9$ and $9not mid 12$. Yep.... seems like you've figured it out.
    – fleablood
    Nov 19 at 23:39








  • 1




    .... and a) is obviously false as $Z_9 subset Z_{12}$ and b) is not a requirement of subrings. In fact by legrange the gcd of (non trivial) subrings must never be $1$.
    – fleablood
    Nov 19 at 23:43






  • 1




    I don't really have any more to add to i707107's hint. Need to show it is closed undo addition and multiplication, and that the identitys of C[x] are in C[x^5] and that every element has an additive inverse.
    – fleablood
    Nov 19 at 23:56






  • 1




    @fleablood (a) isn't obviously false as you have suggested. It depends on what definition of $Bbb{Z}_{12}$ and $Bbb{Z}_{9}$ is one using. If you consider $Bbb{Z}_{9}={[0]_9, [1]_9, ldots, [8]_9}$ and $Bbb{Z}_{12}={[0]_{12}, [1]_{12}, ldots, [11]_{12}}$ as quotients of $Bbb{Z}$ by the relevant congruence relation, then $Bbb{Z}_{9}$ isn't a subset of $Bbb{Z}_{12}$.
    – Anurag A
    Nov 20 at 0:00
















0












0








0







$mathbb{Z_9}$ is not a subring of $mathbb{Z_{12}}$ because



Choose the correct option



$a)$ $mathbb{Z_9}$ is not a subset of $mathbb{Z_{12}}$



$b)$GCD$(9,12) = 3neq 1$



$c)$ $9$ doesnot divide $12$



I thinks option $c)$ will be correct by Lagrange Theorem






abstract-algebra







share|cite|improve this question















$mathbb{Z_9}$ is not a subring of $mathbb{Z_{12}}$ because



Choose the correct option



$a)$ $mathbb{Z_9}$ is not a subset of $mathbb{Z_{12}}$



$b)$GCD$(9,12) = 3neq 1$



$c)$ $9$ doesnot divide $12$



I thinks option $c)$ will be correct by Lagrange Theorem






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 at 23:35

























asked Nov 19 at 23:29









jasmine

1,522416




1,522416








  • 2




    Can you explain why you think that?
    – fleablood
    Nov 19 at 23:34






  • 1




    So Legrange Theorem says the order of every subgroup of a finite group divides the order of a group. And as rings are groups (under their primary operation (in this case addition)) that would apply to rings as well. And the order of $mathbb Z_{12}$ is $12$ and the order of $mathbb Z_9$ is $9$ and $9not mid 12$. Yep.... seems like you've figured it out.
    – fleablood
    Nov 19 at 23:39








  • 1




    .... and a) is obviously false as $Z_9 subset Z_{12}$ and b) is not a requirement of subrings. In fact by legrange the gcd of (non trivial) subrings must never be $1$.
    – fleablood
    Nov 19 at 23:43






  • 1




    I don't really have any more to add to i707107's hint. Need to show it is closed undo addition and multiplication, and that the identitys of C[x] are in C[x^5] and that every element has an additive inverse.
    – fleablood
    Nov 19 at 23:56






  • 1




    @fleablood (a) isn't obviously false as you have suggested. It depends on what definition of $Bbb{Z}_{12}$ and $Bbb{Z}_{9}$ is one using. If you consider $Bbb{Z}_{9}={[0]_9, [1]_9, ldots, [8]_9}$ and $Bbb{Z}_{12}={[0]_{12}, [1]_{12}, ldots, [11]_{12}}$ as quotients of $Bbb{Z}$ by the relevant congruence relation, then $Bbb{Z}_{9}$ isn't a subset of $Bbb{Z}_{12}$.
    – Anurag A
    Nov 20 at 0:00
















  • 2




    Can you explain why you think that?
    – fleablood
    Nov 19 at 23:34






  • 1




    So Legrange Theorem says the order of every subgroup of a finite group divides the order of a group. And as rings are groups (under their primary operation (in this case addition)) that would apply to rings as well. And the order of $mathbb Z_{12}$ is $12$ and the order of $mathbb Z_9$ is $9$ and $9not mid 12$. Yep.... seems like you've figured it out.
    – fleablood
    Nov 19 at 23:39








  • 1




    .... and a) is obviously false as $Z_9 subset Z_{12}$ and b) is not a requirement of subrings. In fact by legrange the gcd of (non trivial) subrings must never be $1$.
    – fleablood
    Nov 19 at 23:43






  • 1




    I don't really have any more to add to i707107's hint. Need to show it is closed undo addition and multiplication, and that the identitys of C[x] are in C[x^5] and that every element has an additive inverse.
    – fleablood
    Nov 19 at 23:56






  • 1




    @fleablood (a) isn't obviously false as you have suggested. It depends on what definition of $Bbb{Z}_{12}$ and $Bbb{Z}_{9}$ is one using. If you consider $Bbb{Z}_{9}={[0]_9, [1]_9, ldots, [8]_9}$ and $Bbb{Z}_{12}={[0]_{12}, [1]_{12}, ldots, [11]_{12}}$ as quotients of $Bbb{Z}$ by the relevant congruence relation, then $Bbb{Z}_{9}$ isn't a subset of $Bbb{Z}_{12}$.
    – Anurag A
    Nov 20 at 0:00










2




2




Can you explain why you think that?
– fleablood
Nov 19 at 23:34




Can you explain why you think that?
– fleablood
Nov 19 at 23:34




1




1




So Legrange Theorem says the order of every subgroup of a finite group divides the order of a group. And as rings are groups (under their primary operation (in this case addition)) that would apply to rings as well. And the order of $mathbb Z_{12}$ is $12$ and the order of $mathbb Z_9$ is $9$ and $9not mid 12$. Yep.... seems like you've figured it out.
– fleablood
Nov 19 at 23:39






So Legrange Theorem says the order of every subgroup of a finite group divides the order of a group. And as rings are groups (under their primary operation (in this case addition)) that would apply to rings as well. And the order of $mathbb Z_{12}$ is $12$ and the order of $mathbb Z_9$ is $9$ and $9not mid 12$. Yep.... seems like you've figured it out.
– fleablood
Nov 19 at 23:39






1




1




.... and a) is obviously false as $Z_9 subset Z_{12}$ and b) is not a requirement of subrings. In fact by legrange the gcd of (non trivial) subrings must never be $1$.
– fleablood
Nov 19 at 23:43




.... and a) is obviously false as $Z_9 subset Z_{12}$ and b) is not a requirement of subrings. In fact by legrange the gcd of (non trivial) subrings must never be $1$.
– fleablood
Nov 19 at 23:43




1




1




I don't really have any more to add to i707107's hint. Need to show it is closed undo addition and multiplication, and that the identitys of C[x] are in C[x^5] and that every element has an additive inverse.
– fleablood
Nov 19 at 23:56




I don't really have any more to add to i707107's hint. Need to show it is closed undo addition and multiplication, and that the identitys of C[x] are in C[x^5] and that every element has an additive inverse.
– fleablood
Nov 19 at 23:56




1




1




@fleablood (a) isn't obviously false as you have suggested. It depends on what definition of $Bbb{Z}_{12}$ and $Bbb{Z}_{9}$ is one using. If you consider $Bbb{Z}_{9}={[0]_9, [1]_9, ldots, [8]_9}$ and $Bbb{Z}_{12}={[0]_{12}, [1]_{12}, ldots, [11]_{12}}$ as quotients of $Bbb{Z}$ by the relevant congruence relation, then $Bbb{Z}_{9}$ isn't a subset of $Bbb{Z}_{12}$.
– Anurag A
Nov 20 at 0:00






@fleablood (a) isn't obviously false as you have suggested. It depends on what definition of $Bbb{Z}_{12}$ and $Bbb{Z}_{9}$ is one using. If you consider $Bbb{Z}_{9}={[0]_9, [1]_9, ldots, [8]_9}$ and $Bbb{Z}_{12}={[0]_{12}, [1]_{12}, ldots, [11]_{12}}$ as quotients of $Bbb{Z}$ by the relevant congruence relation, then $Bbb{Z}_{9}$ isn't a subset of $Bbb{Z}_{12}$.
– Anurag A
Nov 20 at 0:00












1 Answer
1






active

oldest

votes


















1














Your choice and justification are fine. In particular: in order for $Bbb Z_9$ to be a subring of $Bbb Z_{12}$, the group $(Bbb Z_9,+)$ would have to be a subgroup of $(Bbb Z_{12},+)$. However, Lagrange's theorem says that since $9$ does not divide $12$, this cannot be the case.






share|cite|improve this answer





















    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%2f3005684%2fmathbbz-9-is-not-a-subring-of-mathbbz-12-because-choose-the-correct%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














    Your choice and justification are fine. In particular: in order for $Bbb Z_9$ to be a subring of $Bbb Z_{12}$, the group $(Bbb Z_9,+)$ would have to be a subgroup of $(Bbb Z_{12},+)$. However, Lagrange's theorem says that since $9$ does not divide $12$, this cannot be the case.






    share|cite|improve this answer


























      1














      Your choice and justification are fine. In particular: in order for $Bbb Z_9$ to be a subring of $Bbb Z_{12}$, the group $(Bbb Z_9,+)$ would have to be a subgroup of $(Bbb Z_{12},+)$. However, Lagrange's theorem says that since $9$ does not divide $12$, this cannot be the case.






      share|cite|improve this answer
























        1












        1








        1






        Your choice and justification are fine. In particular: in order for $Bbb Z_9$ to be a subring of $Bbb Z_{12}$, the group $(Bbb Z_9,+)$ would have to be a subgroup of $(Bbb Z_{12},+)$. However, Lagrange's theorem says that since $9$ does not divide $12$, this cannot be the case.






        share|cite|improve this answer












        Your choice and justification are fine. In particular: in order for $Bbb Z_9$ to be a subring of $Bbb Z_{12}$, the group $(Bbb Z_9,+)$ would have to be a subgroup of $(Bbb Z_{12},+)$. However, Lagrange's theorem says that since $9$ does not divide $12$, this cannot be the case.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 at 23:37









        Omnomnomnom

        126k788176




        126k788176






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3005684%2fmathbbz-9-is-not-a-subring-of-mathbbz-12-because-choose-the-correct%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

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
            Read more