On the irreducibility of a polynomial and Gauss lemma












0












$begingroup$


Let $P (X) = X^5 − 6X + 3$.



Prove that it is irreducible over $mathbb{Q}$.



In the solution I have for this exercice, I have litterally:



P is irreducible over $mathbb{F}_5$, therefore, by Gauss Lemma, it is irreducible over $mathbb{Q}$.



I am lacking the necessary knowledge about irreducibility by reducing modulo p. All I know is that there is a bijection between roots of P in $overline{mathbb{Q}}$ and the roots of $overline{P}$ in $overline{mathbb{F}_p}$.



I also know this generalization of Gauss Lemma: in a factorial ring A, with its fraction field K, a primitive $P$ is irreducible in $A[X] Leftrightarrow $ $P$ is irreducible in K[X].



How does this apply to P above?



Thank you for any directions or help.






abstract-algebra







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Because $P$ is irreducible over $mathbb{Z}_5 Rightarrow P$ is irreducible over $mathbb{Z}$. And then $P$ is irreducible over $mathbb{Q}$ by Gauss's Lemma.
    $endgroup$
    – mouthetics
    Dec 23 '18 at 3:50










  • $begingroup$
    Thank you! is this a general fact? could please give some link to the litterature or a proof that P is irreducible over $mathbb{F}_pRightarrow$ P is irreducible over $mathbb{Z}$?
    $endgroup$
    – PerelMan
    Dec 23 '18 at 4:17








  • 2




    $begingroup$
    @PerelMan A factorisation over $Bbb Z$ induces one over $Bbb F_p$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 23 '18 at 5:19






  • 2




    $begingroup$
    PerelMan, I try and describe the use of modular reduction in irreducibility proofs here. Also observe that Eisenstein's criterion with $p=3$ works here as well (and is the go-to technique for many).
    $endgroup$
    – Jyrki Lahtonen
    Dec 23 '18 at 7:57
















0












$begingroup$


Let $P (X) = X^5 − 6X + 3$.



Prove that it is irreducible over $mathbb{Q}$.



In the solution I have for this exercice, I have litterally:



P is irreducible over $mathbb{F}_5$, therefore, by Gauss Lemma, it is irreducible over $mathbb{Q}$.



I am lacking the necessary knowledge about irreducibility by reducing modulo p. All I know is that there is a bijection between roots of P in $overline{mathbb{Q}}$ and the roots of $overline{P}$ in $overline{mathbb{F}_p}$.



I also know this generalization of Gauss Lemma: in a factorial ring A, with its fraction field K, a primitive $P$ is irreducible in $A[X] Leftrightarrow $ $P$ is irreducible in K[X].



How does this apply to P above?



Thank you for any directions or help.






abstract-algebra







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Because $P$ is irreducible over $mathbb{Z}_5 Rightarrow P$ is irreducible over $mathbb{Z}$. And then $P$ is irreducible over $mathbb{Q}$ by Gauss's Lemma.
    $endgroup$
    – mouthetics
    Dec 23 '18 at 3:50










  • $begingroup$
    Thank you! is this a general fact? could please give some link to the litterature or a proof that P is irreducible over $mathbb{F}_pRightarrow$ P is irreducible over $mathbb{Z}$?
    $endgroup$
    – PerelMan
    Dec 23 '18 at 4:17








  • 2




    $begingroup$
    @PerelMan A factorisation over $Bbb Z$ induces one over $Bbb F_p$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 23 '18 at 5:19






  • 2




    $begingroup$
    PerelMan, I try and describe the use of modular reduction in irreducibility proofs here. Also observe that Eisenstein's criterion with $p=3$ works here as well (and is the go-to technique for many).
    $endgroup$
    – Jyrki Lahtonen
    Dec 23 '18 at 7:57














0












0








0





$begingroup$


Let $P (X) = X^5 − 6X + 3$.



Prove that it is irreducible over $mathbb{Q}$.



In the solution I have for this exercice, I have litterally:



P is irreducible over $mathbb{F}_5$, therefore, by Gauss Lemma, it is irreducible over $mathbb{Q}$.



I am lacking the necessary knowledge about irreducibility by reducing modulo p. All I know is that there is a bijection between roots of P in $overline{mathbb{Q}}$ and the roots of $overline{P}$ in $overline{mathbb{F}_p}$.



I also know this generalization of Gauss Lemma: in a factorial ring A, with its fraction field K, a primitive $P$ is irreducible in $A[X] Leftrightarrow $ $P$ is irreducible in K[X].



How does this apply to P above?



Thank you for any directions or help.






abstract-algebra







share|cite|improve this question









$endgroup$




Let $P (X) = X^5 − 6X + 3$.



Prove that it is irreducible over $mathbb{Q}$.



In the solution I have for this exercice, I have litterally:



P is irreducible over $mathbb{F}_5$, therefore, by Gauss Lemma, it is irreducible over $mathbb{Q}$.



I am lacking the necessary knowledge about irreducibility by reducing modulo p. All I know is that there is a bijection between roots of P in $overline{mathbb{Q}}$ and the roots of $overline{P}$ in $overline{mathbb{F}_p}$.



I also know this generalization of Gauss Lemma: in a factorial ring A, with its fraction field K, a primitive $P$ is irreducible in $A[X] Leftrightarrow $ $P$ is irreducible in K[X].



How does this apply to P above?



Thank you for any directions or help.






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 23 '18 at 3:40









PerelManPerelMan

538311




538311








  • 3




    $begingroup$
    Because $P$ is irreducible over $mathbb{Z}_5 Rightarrow P$ is irreducible over $mathbb{Z}$. And then $P$ is irreducible over $mathbb{Q}$ by Gauss's Lemma.
    $endgroup$
    – mouthetics
    Dec 23 '18 at 3:50










  • $begingroup$
    Thank you! is this a general fact? could please give some link to the litterature or a proof that P is irreducible over $mathbb{F}_pRightarrow$ P is irreducible over $mathbb{Z}$?
    $endgroup$
    – PerelMan
    Dec 23 '18 at 4:17








  • 2




    $begingroup$
    @PerelMan A factorisation over $Bbb Z$ induces one over $Bbb F_p$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 23 '18 at 5:19






  • 2




    $begingroup$
    PerelMan, I try and describe the use of modular reduction in irreducibility proofs here. Also observe that Eisenstein's criterion with $p=3$ works here as well (and is the go-to technique for many).
    $endgroup$
    – Jyrki Lahtonen
    Dec 23 '18 at 7:57














  • 3




    $begingroup$
    Because $P$ is irreducible over $mathbb{Z}_5 Rightarrow P$ is irreducible over $mathbb{Z}$. And then $P$ is irreducible over $mathbb{Q}$ by Gauss's Lemma.
    $endgroup$
    – mouthetics
    Dec 23 '18 at 3:50










  • $begingroup$
    Thank you! is this a general fact? could please give some link to the litterature or a proof that P is irreducible over $mathbb{F}_pRightarrow$ P is irreducible over $mathbb{Z}$?
    $endgroup$
    – PerelMan
    Dec 23 '18 at 4:17








  • 2




    $begingroup$
    @PerelMan A factorisation over $Bbb Z$ induces one over $Bbb F_p$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 23 '18 at 5:19






  • 2




    $begingroup$
    PerelMan, I try and describe the use of modular reduction in irreducibility proofs here. Also observe that Eisenstein's criterion with $p=3$ works here as well (and is the go-to technique for many).
    $endgroup$
    – Jyrki Lahtonen
    Dec 23 '18 at 7:57








3




3




$begingroup$
Because $P$ is irreducible over $mathbb{Z}_5 Rightarrow P$ is irreducible over $mathbb{Z}$. And then $P$ is irreducible over $mathbb{Q}$ by Gauss's Lemma.
$endgroup$
– mouthetics
Dec 23 '18 at 3:50




$begingroup$
Because $P$ is irreducible over $mathbb{Z}_5 Rightarrow P$ is irreducible over $mathbb{Z}$. And then $P$ is irreducible over $mathbb{Q}$ by Gauss's Lemma.
$endgroup$
– mouthetics
Dec 23 '18 at 3:50












$begingroup$
Thank you! is this a general fact? could please give some link to the litterature or a proof that P is irreducible over $mathbb{F}_pRightarrow$ P is irreducible over $mathbb{Z}$?
$endgroup$
– PerelMan
Dec 23 '18 at 4:17






$begingroup$
Thank you! is this a general fact? could please give some link to the litterature or a proof that P is irreducible over $mathbb{F}_pRightarrow$ P is irreducible over $mathbb{Z}$?
$endgroup$
– PerelMan
Dec 23 '18 at 4:17






2




2




$begingroup$
@PerelMan A factorisation over $Bbb Z$ induces one over $Bbb F_p$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:19




$begingroup$
@PerelMan A factorisation over $Bbb Z$ induces one over $Bbb F_p$.
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:19




2




2




$begingroup$
PerelMan, I try and describe the use of modular reduction in irreducibility proofs here. Also observe that Eisenstein's criterion with $p=3$ works here as well (and is the go-to technique for many).
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 7:57




$begingroup$
PerelMan, I try and describe the use of modular reduction in irreducibility proofs here. Also observe that Eisenstein's criterion with $p=3$ works here as well (and is the go-to technique for many).
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 7:57










1 Answer
1






active

oldest

votes


















1












$begingroup$

There is a general result: Let $R$ be a ring and $I$ be a proper ideal. Fix a nonconstant monic polynomial $p(x) in R[x]$. If $overline{p(x)} in (R/I)[x]$ is irreducible, then $p(x)$ is irreducible in $R[x]$.



You can prove this by considering the contrapositive.






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%2f3050038%2fon-the-irreducibility-of-a-polynomial-and-gauss-lemma%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












    $begingroup$

    There is a general result: Let $R$ be a ring and $I$ be a proper ideal. Fix a nonconstant monic polynomial $p(x) in R[x]$. If $overline{p(x)} in (R/I)[x]$ is irreducible, then $p(x)$ is irreducible in $R[x]$.



    You can prove this by considering the contrapositive.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      There is a general result: Let $R$ be a ring and $I$ be a proper ideal. Fix a nonconstant monic polynomial $p(x) in R[x]$. If $overline{p(x)} in (R/I)[x]$ is irreducible, then $p(x)$ is irreducible in $R[x]$.



      You can prove this by considering the contrapositive.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        There is a general result: Let $R$ be a ring and $I$ be a proper ideal. Fix a nonconstant monic polynomial $p(x) in R[x]$. If $overline{p(x)} in (R/I)[x]$ is irreducible, then $p(x)$ is irreducible in $R[x]$.



        You can prove this by considering the contrapositive.






        share|cite|improve this answer









        $endgroup$



        There is a general result: Let $R$ be a ring and $I$ be a proper ideal. Fix a nonconstant monic polynomial $p(x) in R[x]$. If $overline{p(x)} in (R/I)[x]$ is irreducible, then $p(x)$ is irreducible in $R[x]$.



        You can prove this by considering the contrapositive.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 23 '18 at 5:18









        Aniruddh AgarwalAniruddh Agarwal

        1218




        1218






























            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%2f3050038%2fon-the-irreducibility-of-a-polynomial-and-gauss-lemma%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

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more