Is $F(x)/$ isomorphic to a subfield of the splitting field of $f(x)$ over $F$?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












$F$ is a field. $F[x]$ denotes the ring which consists of all the polynomials with coefficients in $F$. Suppose $f(x)in F[x]$ and $f(x)$ is irreducible over $F$, $a$ is a zero of $f(x)$ in some extension field $E$ of $F$. $F(a)$ denote the smallest subfield of $E$ that contains both $F$ and $a$. It is well known that all such $F(a)$ is isomorphic to
$F(x)/<f(x)>$, and all splitting fields of $f(x)$ over $F$ are isomorphic. I am quite curious that is it the case that $F(x)/<f(x)>$ is isomorphic to a subfield of the splitting fields? Some enlightening would be very helpful!










share|cite|improve this question























  • You’re forgetting the assumption that $f$ is irreducible.
    – egreg
    Sep 1 at 7:13










  • Thank you, editing now.
    – qiang heng
    Sep 1 at 7:14














up vote
1
down vote

favorite












$F$ is a field. $F[x]$ denotes the ring which consists of all the polynomials with coefficients in $F$. Suppose $f(x)in F[x]$ and $f(x)$ is irreducible over $F$, $a$ is a zero of $f(x)$ in some extension field $E$ of $F$. $F(a)$ denote the smallest subfield of $E$ that contains both $F$ and $a$. It is well known that all such $F(a)$ is isomorphic to
$F(x)/<f(x)>$, and all splitting fields of $f(x)$ over $F$ are isomorphic. I am quite curious that is it the case that $F(x)/<f(x)>$ is isomorphic to a subfield of the splitting fields? Some enlightening would be very helpful!










share|cite|improve this question























  • You’re forgetting the assumption that $f$ is irreducible.
    – egreg
    Sep 1 at 7:13










  • Thank you, editing now.
    – qiang heng
    Sep 1 at 7:14












up vote
1
down vote

favorite









up vote
1
down vote

favorite











$F$ is a field. $F[x]$ denotes the ring which consists of all the polynomials with coefficients in $F$. Suppose $f(x)in F[x]$ and $f(x)$ is irreducible over $F$, $a$ is a zero of $f(x)$ in some extension field $E$ of $F$. $F(a)$ denote the smallest subfield of $E$ that contains both $F$ and $a$. It is well known that all such $F(a)$ is isomorphic to
$F(x)/<f(x)>$, and all splitting fields of $f(x)$ over $F$ are isomorphic. I am quite curious that is it the case that $F(x)/<f(x)>$ is isomorphic to a subfield of the splitting fields? Some enlightening would be very helpful!










share|cite|improve this question















$F$ is a field. $F[x]$ denotes the ring which consists of all the polynomials with coefficients in $F$. Suppose $f(x)in F[x]$ and $f(x)$ is irreducible over $F$, $a$ is a zero of $f(x)$ in some extension field $E$ of $F$. $F(a)$ denote the smallest subfield of $E$ that contains both $F$ and $a$. It is well known that all such $F(a)$ is isomorphic to
$F(x)/<f(x)>$, and all splitting fields of $f(x)$ over $F$ are isomorphic. I am quite curious that is it the case that $F(x)/<f(x)>$ is isomorphic to a subfield of the splitting fields? Some enlightening would be very helpful!







abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 1 at 7:25

























asked Sep 1 at 7:12









qiang heng

17715




17715











  • You’re forgetting the assumption that $f$ is irreducible.
    – egreg
    Sep 1 at 7:13










  • Thank you, editing now.
    – qiang heng
    Sep 1 at 7:14
















  • You’re forgetting the assumption that $f$ is irreducible.
    – egreg
    Sep 1 at 7:13










  • Thank you, editing now.
    – qiang heng
    Sep 1 at 7:14















You’re forgetting the assumption that $f$ is irreducible.
– egreg
Sep 1 at 7:13




You’re forgetting the assumption that $f$ is irreducible.
– egreg
Sep 1 at 7:13












Thank you, editing now.
– qiang heng
Sep 1 at 7:14




Thank you, editing now.
– qiang heng
Sep 1 at 7:14










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










The splitting field $K$ of $f$ by definition contains a root $b$ of $f$ (actually all roots). Since $F(b)cong F[x]/langle f(x)rangle$, you’re done.






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',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    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%2f2901453%2fis-fx-fx-isomorphic-to-a-subfield-of-the-splitting-field-of-fx-over%23new-answer', 'question_page');

    );

    Post as a guest






























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    The splitting field $K$ of $f$ by definition contains a root $b$ of $f$ (actually all roots). Since $F(b)cong F[x]/langle f(x)rangle$, you’re done.






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      The splitting field $K$ of $f$ by definition contains a root $b$ of $f$ (actually all roots). Since $F(b)cong F[x]/langle f(x)rangle$, you’re done.






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        The splitting field $K$ of $f$ by definition contains a root $b$ of $f$ (actually all roots). Since $F(b)cong F[x]/langle f(x)rangle$, you’re done.






        share|cite|improve this answer












        The splitting field $K$ of $f$ by definition contains a root $b$ of $f$ (actually all roots). Since $F(b)cong F[x]/langle f(x)rangle$, you’re done.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 1 at 7:28









        egreg

        166k1180189




        166k1180189



























             

            draft saved


            draft discarded















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2901453%2fis-fx-fx-isomorphic-to-a-subfield-of-the-splitting-field-of-fx-over%23new-answer', 'question_page');

            );

            Post as a guest













































































            這個網誌中的熱門文章

            How to combine Bézier curves to a surface?

            Mutual Information Always Non-negative

            Why am i infinitely getting the same tweet with the Twitter Search API?