A concept about normal extension and splitting field











up vote
2
down vote

favorite












Let $K$ be an field extension of $F$. Then I know there is a theorem that says: $K$ is a splitting field of some polynomial over $F$ iff $K$ over $F$ is finite and normal. (Here the normal means every polynomial in $F[x]$ has a root then have all roots.)



Now, $Bbb Q(sqrt[3]{2})$ does not have all roots of $x^3-2$, so it is not normal. Is it possible that $Bbb Q(sqrt[3]{2})$ has all roots of some polynomial $p(x)$, but these roots are all not in $Bbb Q$?






abstract-algebra extension-field







share|cite|improve this question
























  • Is your question about a general $K$, or specifically $K=mathbb Q(sqrt[3]2)$?
    – Y. Forman
    yesterday










  • Specifically $K=Bbb Q(sqrt[3]{2})$. let me edit
    – Eric
    yesterday










  • $Bbb Q(sqrt[3]{2})/Bbb Q(sqrt[3]{2})$ is a (trivial) normal extension, the splitting field of $x in Bbb Q(sqrt[3]{2})[x]$. For extensions of $Bbb Q$ when not specified we mean $K/Bbb Q$ is a normal extension, thus the splitting field of some $p(x) in Bbb Q[x]$.
    – reuns
    yesterday

















up vote
2
down vote

favorite












Let $K$ be an field extension of $F$. Then I know there is a theorem that says: $K$ is a splitting field of some polynomial over $F$ iff $K$ over $F$ is finite and normal. (Here the normal means every polynomial in $F[x]$ has a root then have all roots.)



Now, $Bbb Q(sqrt[3]{2})$ does not have all roots of $x^3-2$, so it is not normal. Is it possible that $Bbb Q(sqrt[3]{2})$ has all roots of some polynomial $p(x)$, but these roots are all not in $Bbb Q$?






abstract-algebra extension-field







share|cite|improve this question
























  • Is your question about a general $K$, or specifically $K=mathbb Q(sqrt[3]2)$?
    – Y. Forman
    yesterday










  • Specifically $K=Bbb Q(sqrt[3]{2})$. let me edit
    – Eric
    yesterday










  • $Bbb Q(sqrt[3]{2})/Bbb Q(sqrt[3]{2})$ is a (trivial) normal extension, the splitting field of $x in Bbb Q(sqrt[3]{2})[x]$. For extensions of $Bbb Q$ when not specified we mean $K/Bbb Q$ is a normal extension, thus the splitting field of some $p(x) in Bbb Q[x]$.
    – reuns
    yesterday















up vote
2
down vote

favorite









up vote
2
down vote

favorite











Let $K$ be an field extension of $F$. Then I know there is a theorem that says: $K$ is a splitting field of some polynomial over $F$ iff $K$ over $F$ is finite and normal. (Here the normal means every polynomial in $F[x]$ has a root then have all roots.)



Now, $Bbb Q(sqrt[3]{2})$ does not have all roots of $x^3-2$, so it is not normal. Is it possible that $Bbb Q(sqrt[3]{2})$ has all roots of some polynomial $p(x)$, but these roots are all not in $Bbb Q$?






abstract-algebra extension-field







share|cite|improve this question















Let $K$ be an field extension of $F$. Then I know there is a theorem that says: $K$ is a splitting field of some polynomial over $F$ iff $K$ over $F$ is finite and normal. (Here the normal means every polynomial in $F[x]$ has a root then have all roots.)



Now, $Bbb Q(sqrt[3]{2})$ does not have all roots of $x^3-2$, so it is not normal. Is it possible that $Bbb Q(sqrt[3]{2})$ has all roots of some polynomial $p(x)$, but these roots are all not in $Bbb Q$?






abstract-algebra extension-field




abstract-algebra extension-field






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday

























asked yesterday









Eric

1,737515




1,737515












  • Is your question about a general $K$, or specifically $K=mathbb Q(sqrt[3]2)$?
    – Y. Forman
    yesterday










  • Specifically $K=Bbb Q(sqrt[3]{2})$. let me edit
    – Eric
    yesterday










  • $Bbb Q(sqrt[3]{2})/Bbb Q(sqrt[3]{2})$ is a (trivial) normal extension, the splitting field of $x in Bbb Q(sqrt[3]{2})[x]$. For extensions of $Bbb Q$ when not specified we mean $K/Bbb Q$ is a normal extension, thus the splitting field of some $p(x) in Bbb Q[x]$.
    – reuns
    yesterday




















  • Is your question about a general $K$, or specifically $K=mathbb Q(sqrt[3]2)$?
    – Y. Forman
    yesterday










  • Specifically $K=Bbb Q(sqrt[3]{2})$. let me edit
    – Eric
    yesterday










  • $Bbb Q(sqrt[3]{2})/Bbb Q(sqrt[3]{2})$ is a (trivial) normal extension, the splitting field of $x in Bbb Q(sqrt[3]{2})[x]$. For extensions of $Bbb Q$ when not specified we mean $K/Bbb Q$ is a normal extension, thus the splitting field of some $p(x) in Bbb Q[x]$.
    – reuns
    yesterday


















Is your question about a general $K$, or specifically $K=mathbb Q(sqrt[3]2)$?
– Y. Forman
yesterday




Is your question about a general $K$, or specifically $K=mathbb Q(sqrt[3]2)$?
– Y. Forman
yesterday












Specifically $K=Bbb Q(sqrt[3]{2})$. let me edit
– Eric
yesterday




Specifically $K=Bbb Q(sqrt[3]{2})$. let me edit
– Eric
yesterday












$Bbb Q(sqrt[3]{2})/Bbb Q(sqrt[3]{2})$ is a (trivial) normal extension, the splitting field of $x in Bbb Q(sqrt[3]{2})[x]$. For extensions of $Bbb Q$ when not specified we mean $K/Bbb Q$ is a normal extension, thus the splitting field of some $p(x) in Bbb Q[x]$.
– reuns
yesterday






$Bbb Q(sqrt[3]{2})/Bbb Q(sqrt[3]{2})$ is a (trivial) normal extension, the splitting field of $x in Bbb Q(sqrt[3]{2})[x]$. For extensions of $Bbb Q$ when not specified we mean $K/Bbb Q$ is a normal extension, thus the splitting field of some $p(x) in Bbb Q[x]$.
– reuns
yesterday












2 Answers
2






active

oldest

votes

















up vote
3
down vote













No.



First, such a polynomial $p(x)$ would have to be irreducible over $mathbb Q$. If it wasn't, it would split into factors at least one of which would be linear. This would yield a root in $mathbb Q$.



If $p(x)$ is such an irreducible polynomial, then we have that $mathbb Q (^3 ! sqrt2)$ contains the splitting field of $p(x)$. But this extension has no intermediate extensions. (By multiplicity of degrees of extensions).



So $mathbb Q (^3 ! sqrt2)$ becomes the splitting field for an irreducible polynomial over $mathbb Q$, which makes it a normal extension. But we know that this isn't the case - The polynomial $x^3 - 2$ has only one root here, out of the three possible.






share|cite|improve this answer






























    up vote
    0
    down vote













    They'd all have to be $notin mathbb Q$, as otherwise $p$ would be reducible.






    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: 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%2f3005001%2fa-concept-about-normal-extension-and-splitting-field%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








      up vote
      3
      down vote













      No.



      First, such a polynomial $p(x)$ would have to be irreducible over $mathbb Q$. If it wasn't, it would split into factors at least one of which would be linear. This would yield a root in $mathbb Q$.



      If $p(x)$ is such an irreducible polynomial, then we have that $mathbb Q (^3 ! sqrt2)$ contains the splitting field of $p(x)$. But this extension has no intermediate extensions. (By multiplicity of degrees of extensions).



      So $mathbb Q (^3 ! sqrt2)$ becomes the splitting field for an irreducible polynomial over $mathbb Q$, which makes it a normal extension. But we know that this isn't the case - The polynomial $x^3 - 2$ has only one root here, out of the three possible.






      share|cite|improve this answer



























        up vote
        3
        down vote













        No.



        First, such a polynomial $p(x)$ would have to be irreducible over $mathbb Q$. If it wasn't, it would split into factors at least one of which would be linear. This would yield a root in $mathbb Q$.



        If $p(x)$ is such an irreducible polynomial, then we have that $mathbb Q (^3 ! sqrt2)$ contains the splitting field of $p(x)$. But this extension has no intermediate extensions. (By multiplicity of degrees of extensions).



        So $mathbb Q (^3 ! sqrt2)$ becomes the splitting field for an irreducible polynomial over $mathbb Q$, which makes it a normal extension. But we know that this isn't the case - The polynomial $x^3 - 2$ has only one root here, out of the three possible.






        share|cite|improve this answer

























          up vote
          3
          down vote










          up vote
          3
          down vote









          No.



          First, such a polynomial $p(x)$ would have to be irreducible over $mathbb Q$. If it wasn't, it would split into factors at least one of which would be linear. This would yield a root in $mathbb Q$.



          If $p(x)$ is such an irreducible polynomial, then we have that $mathbb Q (^3 ! sqrt2)$ contains the splitting field of $p(x)$. But this extension has no intermediate extensions. (By multiplicity of degrees of extensions).



          So $mathbb Q (^3 ! sqrt2)$ becomes the splitting field for an irreducible polynomial over $mathbb Q$, which makes it a normal extension. But we know that this isn't the case - The polynomial $x^3 - 2$ has only one root here, out of the three possible.






          share|cite|improve this answer














          No.



          First, such a polynomial $p(x)$ would have to be irreducible over $mathbb Q$. If it wasn't, it would split into factors at least one of which would be linear. This would yield a root in $mathbb Q$.



          If $p(x)$ is such an irreducible polynomial, then we have that $mathbb Q (^3 ! sqrt2)$ contains the splitting field of $p(x)$. But this extension has no intermediate extensions. (By multiplicity of degrees of extensions).



          So $mathbb Q (^3 ! sqrt2)$ becomes the splitting field for an irreducible polynomial over $mathbb Q$, which makes it a normal extension. But we know that this isn't the case - The polynomial $x^3 - 2$ has only one root here, out of the three possible.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          Praneet Srivastava

          762516




          762516






















              up vote
              0
              down vote













              They'd all have to be $notin mathbb Q$, as otherwise $p$ would be reducible.






              share|cite|improve this answer

























                up vote
                0
                down vote













                They'd all have to be $notin mathbb Q$, as otherwise $p$ would be reducible.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  They'd all have to be $notin mathbb Q$, as otherwise $p$ would be reducible.






                  share|cite|improve this answer












                  They'd all have to be $notin mathbb Q$, as otherwise $p$ would be reducible.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  Richard Martin

                  1,3238




                  1,3238






























                       

                      draft saved


                      draft discarded



















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005001%2fa-concept-about-normal-extension-and-splitting-field%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

                      Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                      Image
                      28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                      Read more

                      Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                      Image
                      up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                      Read more

                      A Topological Invariant for $pi_3(U(n))$

                      Image
                      3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                      Read more