Splitting Field of a Polynomial in $mathbb{Q}$












1












$begingroup$


We are studying splitting fields and I just wanted to make sure that I really understood them, so we needed to do the following:



Show that $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a splitting field of the Polynomial $f=X^4 -2X^2 + 2$.



My attempt:



Substituting $z:=X^2$ we get $f'=z^2 - 2z +2$. Computing the roots we get $x_{1,2} = 1pm i$ so $f$ has the roots $y_{1,2,3,4}=pm sqrt{1pm i}$. The splitting field of f is therefore $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$. Now we show that $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ = $mathbb{Q}(sqrt{2}, sqrt{1-i})$.



i) $mathbb{Q}(sqrt{2}, sqrt{1-i})subset mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ is true since $sqrt{1+i}cdot sqrt{1-i} = sqrt{(1+i)cdot(1-i)}=sqrt{2}$



ii)
$
mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i}) subset mathbb{Q}(sqrt{2}, sqrt{1-i})$
Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not 100% sure about $sqrt{1-i}$. Am I allowed to assume that because $sqrt{2}= sqrt{(1+i)cdot(1-i)}= sqrt{1+i}cdot sqrt{1-i}$ then $sqrt{1-i}$ has to be in $mathbb{Q}(sqrt{2}, sqrt{1+i})$ because $sqrt{1+i}$ is and if $sqrt{1-i}$ isn't then $sqrt{2}$ also wouldn't be? I am not sure my argumentation is 100% correct at this point.






abstract-algebra splitting-field







share|cite|improve this question











$endgroup$












  • $begingroup$
    The notation $(mathbb{Q},sqrt{2}, sqrt{1-i})$ is nonstandard, I presume you mean the smallest field containing $mathbb{Q},sqrt{2}$ and $sqrt{1-i}$? That's usually denoted by $mathbb{Q}(sqrt{2}, sqrt{1-i})$
    $endgroup$
    – Wojowu
    Jan 16 at 9:59










  • $begingroup$
    Do you mean $X^4-2X^2+2$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 10:01










  • $begingroup$
    Guys... I'm sorry. I edited both my mistakes. Thank you.
    $endgroup$
    – KingDingeling
    Jan 16 at 10:02
















1












$begingroup$


We are studying splitting fields and I just wanted to make sure that I really understood them, so we needed to do the following:



Show that $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a splitting field of the Polynomial $f=X^4 -2X^2 + 2$.



My attempt:



Substituting $z:=X^2$ we get $f'=z^2 - 2z +2$. Computing the roots we get $x_{1,2} = 1pm i$ so $f$ has the roots $y_{1,2,3,4}=pm sqrt{1pm i}$. The splitting field of f is therefore $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$. Now we show that $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ = $mathbb{Q}(sqrt{2}, sqrt{1-i})$.



i) $mathbb{Q}(sqrt{2}, sqrt{1-i})subset mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ is true since $sqrt{1+i}cdot sqrt{1-i} = sqrt{(1+i)cdot(1-i)}=sqrt{2}$



ii)
$
mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i}) subset mathbb{Q}(sqrt{2}, sqrt{1-i})$
Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not 100% sure about $sqrt{1-i}$. Am I allowed to assume that because $sqrt{2}= sqrt{(1+i)cdot(1-i)}= sqrt{1+i}cdot sqrt{1-i}$ then $sqrt{1-i}$ has to be in $mathbb{Q}(sqrt{2}, sqrt{1+i})$ because $sqrt{1+i}$ is and if $sqrt{1-i}$ isn't then $sqrt{2}$ also wouldn't be? I am not sure my argumentation is 100% correct at this point.






abstract-algebra splitting-field







share|cite|improve this question











$endgroup$












  • $begingroup$
    The notation $(mathbb{Q},sqrt{2}, sqrt{1-i})$ is nonstandard, I presume you mean the smallest field containing $mathbb{Q},sqrt{2}$ and $sqrt{1-i}$? That's usually denoted by $mathbb{Q}(sqrt{2}, sqrt{1-i})$
    $endgroup$
    – Wojowu
    Jan 16 at 9:59










  • $begingroup$
    Do you mean $X^4-2X^2+2$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 10:01










  • $begingroup$
    Guys... I'm sorry. I edited both my mistakes. Thank you.
    $endgroup$
    – KingDingeling
    Jan 16 at 10:02














1












1








1


1



$begingroup$


We are studying splitting fields and I just wanted to make sure that I really understood them, so we needed to do the following:



Show that $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a splitting field of the Polynomial $f=X^4 -2X^2 + 2$.



My attempt:



Substituting $z:=X^2$ we get $f'=z^2 - 2z +2$. Computing the roots we get $x_{1,2} = 1pm i$ so $f$ has the roots $y_{1,2,3,4}=pm sqrt{1pm i}$. The splitting field of f is therefore $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$. Now we show that $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ = $mathbb{Q}(sqrt{2}, sqrt{1-i})$.



i) $mathbb{Q}(sqrt{2}, sqrt{1-i})subset mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ is true since $sqrt{1+i}cdot sqrt{1-i} = sqrt{(1+i)cdot(1-i)}=sqrt{2}$



ii)
$
mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i}) subset mathbb{Q}(sqrt{2}, sqrt{1-i})$
Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not 100% sure about $sqrt{1-i}$. Am I allowed to assume that because $sqrt{2}= sqrt{(1+i)cdot(1-i)}= sqrt{1+i}cdot sqrt{1-i}$ then $sqrt{1-i}$ has to be in $mathbb{Q}(sqrt{2}, sqrt{1+i})$ because $sqrt{1+i}$ is and if $sqrt{1-i}$ isn't then $sqrt{2}$ also wouldn't be? I am not sure my argumentation is 100% correct at this point.






abstract-algebra splitting-field







share|cite|improve this question











$endgroup$




We are studying splitting fields and I just wanted to make sure that I really understood them, so we needed to do the following:



Show that $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a splitting field of the Polynomial $f=X^4 -2X^2 + 2$.



My attempt:



Substituting $z:=X^2$ we get $f'=z^2 - 2z +2$. Computing the roots we get $x_{1,2} = 1pm i$ so $f$ has the roots $y_{1,2,3,4}=pm sqrt{1pm i}$. The splitting field of f is therefore $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$. Now we show that $mathbb{Q}(sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ = $mathbb{Q}(sqrt{2}, sqrt{1-i})$.



i) $mathbb{Q}(sqrt{2}, sqrt{1-i})subset mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i})$ is true since $sqrt{1+i}cdot sqrt{1-i} = sqrt{(1+i)cdot(1-i)}=sqrt{2}$



ii)
$
mathbb{Q}( sqrt{1+i}, -sqrt{1+i}, sqrt{1-i}, -sqrt{1-i}) subset mathbb{Q}(sqrt{2}, sqrt{1-i})$
Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not 100% sure about $sqrt{1-i}$. Am I allowed to assume that because $sqrt{2}= sqrt{(1+i)cdot(1-i)}= sqrt{1+i}cdot sqrt{1-i}$ then $sqrt{1-i}$ has to be in $mathbb{Q}(sqrt{2}, sqrt{1+i})$ because $sqrt{1+i}$ is and if $sqrt{1-i}$ isn't then $sqrt{2}$ also wouldn't be? I am not sure my argumentation is 100% correct at this point.






abstract-algebra splitting-field




abstract-algebra splitting-field






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 10:14









Bernard

120k740115




120k740115










asked Jan 16 at 9:55









KingDingelingKingDingeling

1497




1497












  • $begingroup$
    The notation $(mathbb{Q},sqrt{2}, sqrt{1-i})$ is nonstandard, I presume you mean the smallest field containing $mathbb{Q},sqrt{2}$ and $sqrt{1-i}$? That's usually denoted by $mathbb{Q}(sqrt{2}, sqrt{1-i})$
    $endgroup$
    – Wojowu
    Jan 16 at 9:59










  • $begingroup$
    Do you mean $X^4-2X^2+2$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 10:01










  • $begingroup$
    Guys... I'm sorry. I edited both my mistakes. Thank you.
    $endgroup$
    – KingDingeling
    Jan 16 at 10:02


















  • $begingroup$
    The notation $(mathbb{Q},sqrt{2}, sqrt{1-i})$ is nonstandard, I presume you mean the smallest field containing $mathbb{Q},sqrt{2}$ and $sqrt{1-i}$? That's usually denoted by $mathbb{Q}(sqrt{2}, sqrt{1-i})$
    $endgroup$
    – Wojowu
    Jan 16 at 9:59










  • $begingroup$
    Do you mean $X^4-2X^2+2$?
    $endgroup$
    – Lord Shark the Unknown
    Jan 16 at 10:01










  • $begingroup$
    Guys... I'm sorry. I edited both my mistakes. Thank you.
    $endgroup$
    – KingDingeling
    Jan 16 at 10:02
















$begingroup$
The notation $(mathbb{Q},sqrt{2}, sqrt{1-i})$ is nonstandard, I presume you mean the smallest field containing $mathbb{Q},sqrt{2}$ and $sqrt{1-i}$? That's usually denoted by $mathbb{Q}(sqrt{2}, sqrt{1-i})$
$endgroup$
– Wojowu
Jan 16 at 9:59




$begingroup$
The notation $(mathbb{Q},sqrt{2}, sqrt{1-i})$ is nonstandard, I presume you mean the smallest field containing $mathbb{Q},sqrt{2}$ and $sqrt{1-i}$? That's usually denoted by $mathbb{Q}(sqrt{2}, sqrt{1-i})$
$endgroup$
– Wojowu
Jan 16 at 9:59












$begingroup$
Do you mean $X^4-2X^2+2$?
$endgroup$
– Lord Shark the Unknown
Jan 16 at 10:01




$begingroup$
Do you mean $X^4-2X^2+2$?
$endgroup$
– Lord Shark the Unknown
Jan 16 at 10:01












$begingroup$
Guys... I'm sorry. I edited both my mistakes. Thank you.
$endgroup$
– KingDingeling
Jan 16 at 10:02




$begingroup$
Guys... I'm sorry. I edited both my mistakes. Thank you.
$endgroup$
– KingDingeling
Jan 16 at 10:02










1 Answer
1






active

oldest

votes


















1












$begingroup$


Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not
100% sure about $sqrt{1-i}$.




Actually, the cases $sqrt{1-i}$ and $-sqrt{1-i}$ are trivial since we are considering $mathbb{Q}(sqrt{2}, sqrt{1-i}) $.



Now for $sqrt{1+i}$, your argument is almost correct. Note that $sqrt{1-i}neq 0in mathbb{Q}(sqrt{2}, sqrt{1-i})$. Since $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a field, the multiplicative inverse of $sqrt{1-i} $ is in $mathbb{Q}(sqrt{2}, sqrt{1-i})$. So now we can express $sqrt{1+i}$ as $frac{sqrt{2}}{sqrt{1-i}}$(obtained from the equation you wrote). This implies that $sqrt{1+i}in mathbb{Q}(sqrt{2}, sqrt{1-i})$.






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%2f3075545%2fsplitting-field-of-a-polynomial-in-mathbbq%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












    $begingroup$


    Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not
    100% sure about $sqrt{1-i}$.




    Actually, the cases $sqrt{1-i}$ and $-sqrt{1-i}$ are trivial since we are considering $mathbb{Q}(sqrt{2}, sqrt{1-i}) $.



    Now for $sqrt{1+i}$, your argument is almost correct. Note that $sqrt{1-i}neq 0in mathbb{Q}(sqrt{2}, sqrt{1-i})$. Since $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a field, the multiplicative inverse of $sqrt{1-i} $ is in $mathbb{Q}(sqrt{2}, sqrt{1-i})$. So now we can express $sqrt{1+i}$ as $frac{sqrt{2}}{sqrt{1-i}}$(obtained from the equation you wrote). This implies that $sqrt{1+i}in mathbb{Q}(sqrt{2}, sqrt{1-i})$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$


      Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not
      100% sure about $sqrt{1-i}$.




      Actually, the cases $sqrt{1-i}$ and $-sqrt{1-i}$ are trivial since we are considering $mathbb{Q}(sqrt{2}, sqrt{1-i}) $.



      Now for $sqrt{1+i}$, your argument is almost correct. Note that $sqrt{1-i}neq 0in mathbb{Q}(sqrt{2}, sqrt{1-i})$. Since $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a field, the multiplicative inverse of $sqrt{1-i} $ is in $mathbb{Q}(sqrt{2}, sqrt{1-i})$. So now we can express $sqrt{1+i}$ as $frac{sqrt{2}}{sqrt{1-i}}$(obtained from the equation you wrote). This implies that $sqrt{1+i}in mathbb{Q}(sqrt{2}, sqrt{1-i})$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$


        Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not
        100% sure about $sqrt{1-i}$.




        Actually, the cases $sqrt{1-i}$ and $-sqrt{1-i}$ are trivial since we are considering $mathbb{Q}(sqrt{2}, sqrt{1-i}) $.



        Now for $sqrt{1+i}$, your argument is almost correct. Note that $sqrt{1-i}neq 0in mathbb{Q}(sqrt{2}, sqrt{1-i})$. Since $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a field, the multiplicative inverse of $sqrt{1-i} $ is in $mathbb{Q}(sqrt{2}, sqrt{1-i})$. So now we can express $sqrt{1+i}$ as $frac{sqrt{2}}{sqrt{1-i}}$(obtained from the equation you wrote). This implies that $sqrt{1+i}in mathbb{Q}(sqrt{2}, sqrt{1-i})$.






        share|cite|improve this answer









        $endgroup$




        Now for $sqrt{1+i}$ and $-sqrt{1+i}$ this is trivial. But I am not
        100% sure about $sqrt{1-i}$.




        Actually, the cases $sqrt{1-i}$ and $-sqrt{1-i}$ are trivial since we are considering $mathbb{Q}(sqrt{2}, sqrt{1-i}) $.



        Now for $sqrt{1+i}$, your argument is almost correct. Note that $sqrt{1-i}neq 0in mathbb{Q}(sqrt{2}, sqrt{1-i})$. Since $mathbb{Q}(sqrt{2}, sqrt{1-i})$ is a field, the multiplicative inverse of $sqrt{1-i} $ is in $mathbb{Q}(sqrt{2}, sqrt{1-i})$. So now we can express $sqrt{1+i}$ as $frac{sqrt{2}}{sqrt{1-i}}$(obtained from the equation you wrote). This implies that $sqrt{1+i}in mathbb{Q}(sqrt{2}, sqrt{1-i})$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 10:49









        Thomas ShelbyThomas Shelby

        2,877421




        2,877421






























            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%2f3075545%2fsplitting-field-of-a-polynomial-in-mathbbq%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

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
            Read more

            The Binding of Isaac: Rebirth/Afterbirth

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
            Read more