If a field $K$ (of characteristic 0) has no proper extensions of the form $K[sqrt[n]{x}]$, is it...












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$begingroup$


Let $K$ be a field, assume characteristic 0 if this simplifies things. Suppose we know that for any positive integer $n$ and any $x in K$ the polynomial $X^n - x$ has a root in $K$. Is $K$ algebraically closed?



Working on the assumption that the answer is probably "no", how would one construct a counterexample?






abstract-algebra field-theory radicals







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Let $K$ be a field, assume characteristic 0 if this simplifies things. Suppose we know that for any positive integer $n$ and any $x in K$ the polynomial $X^n - x$ has a root in $K$. Is $K$ algebraically closed?



    Working on the assumption that the answer is probably "no", how would one construct a counterexample?






    abstract-algebra field-theory radicals







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Let $K$ be a field, assume characteristic 0 if this simplifies things. Suppose we know that for any positive integer $n$ and any $x in K$ the polynomial $X^n - x$ has a root in $K$. Is $K$ algebraically closed?



      Working on the assumption that the answer is probably "no", how would one construct a counterexample?






      abstract-algebra field-theory radicals







      share|cite|improve this question











      $endgroup$




      Let $K$ be a field, assume characteristic 0 if this simplifies things. Suppose we know that for any positive integer $n$ and any $x in K$ the polynomial $X^n - x$ has a root in $K$. Is $K$ algebraically closed?



      Working on the assumption that the answer is probably "no", how would one construct a counterexample?






      abstract-algebra field-theory radicals




      abstract-algebra field-theory radicals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 21:29









      user26857

      39.3k124183




      39.3k124183










      asked Jan 4 at 9:40









      Bib-lostBib-lost

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      2,040629






















          2 Answers
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          $begingroup$

          The answer in characteristic $0$ is no. For instance, the "root closure"* of $Bbb Q$ still wouldn't let you solve, say, an unsolvable quintic like $x^5-x-1 = 0$, because we know its solutions can't be written using roots in $Bbb Q$, so it can't be written using roots in the root closure either.



          *What I mean by "root closure" is this: Start with $Bbb Q$, and consider it a subfield of $Bbb C$ (or of $overline{Bbb Q}$). To $Bbb Q$, add all $n$th roots of all elements of $Bbb Q$. Then add all $n$'th roots of all elements in that new field, and so on. The root closure is the final result, i.e. the union of these fields. Any element in that field may be written with a finite expression involving only addition, subtraction, multiplication, division, roots and rational numbers.






          share|cite|improve this answer











          $endgroup$





















            4












            $begingroup$

            In positive characteristic the situation is even worse. For instance, the field $K$ with two elements trivially satisfies your property, but the polynomial $x^{2} + x + 1$ has no roots in $K$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @DietrichBurde That's what happens when I only read titles. I need to stop doing that.
              $endgroup$
              – Arthur
              Jan 4 at 10:49








            • 2




              $begingroup$
              Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
              $endgroup$
              – Andreas Caranti
              Jan 4 at 11:07












            • $begingroup$
              @DietrichBurde, thanks. And a Happy New Year to you!
              $endgroup$
              – Andreas Caranti
              Jan 4 at 11:57










            • $begingroup$
              Thank you, you too!!!
              $endgroup$
              – Dietrich Burde
              Jan 4 at 11:59











            Your Answer





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            2 Answers
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            active

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            2 Answers
            2






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            oldest

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            active

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            active

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            4












            $begingroup$

            The answer in characteristic $0$ is no. For instance, the "root closure"* of $Bbb Q$ still wouldn't let you solve, say, an unsolvable quintic like $x^5-x-1 = 0$, because we know its solutions can't be written using roots in $Bbb Q$, so it can't be written using roots in the root closure either.



            *What I mean by "root closure" is this: Start with $Bbb Q$, and consider it a subfield of $Bbb C$ (or of $overline{Bbb Q}$). To $Bbb Q$, add all $n$th roots of all elements of $Bbb Q$. Then add all $n$'th roots of all elements in that new field, and so on. The root closure is the final result, i.e. the union of these fields. Any element in that field may be written with a finite expression involving only addition, subtraction, multiplication, division, roots and rational numbers.






            share|cite|improve this answer











            $endgroup$


















              4












              $begingroup$

              The answer in characteristic $0$ is no. For instance, the "root closure"* of $Bbb Q$ still wouldn't let you solve, say, an unsolvable quintic like $x^5-x-1 = 0$, because we know its solutions can't be written using roots in $Bbb Q$, so it can't be written using roots in the root closure either.



              *What I mean by "root closure" is this: Start with $Bbb Q$, and consider it a subfield of $Bbb C$ (or of $overline{Bbb Q}$). To $Bbb Q$, add all $n$th roots of all elements of $Bbb Q$. Then add all $n$'th roots of all elements in that new field, and so on. The root closure is the final result, i.e. the union of these fields. Any element in that field may be written with a finite expression involving only addition, subtraction, multiplication, division, roots and rational numbers.






              share|cite|improve this answer











              $endgroup$
















                4












                4








                4





                $begingroup$

                The answer in characteristic $0$ is no. For instance, the "root closure"* of $Bbb Q$ still wouldn't let you solve, say, an unsolvable quintic like $x^5-x-1 = 0$, because we know its solutions can't be written using roots in $Bbb Q$, so it can't be written using roots in the root closure either.



                *What I mean by "root closure" is this: Start with $Bbb Q$, and consider it a subfield of $Bbb C$ (or of $overline{Bbb Q}$). To $Bbb Q$, add all $n$th roots of all elements of $Bbb Q$. Then add all $n$'th roots of all elements in that new field, and so on. The root closure is the final result, i.e. the union of these fields. Any element in that field may be written with a finite expression involving only addition, subtraction, multiplication, division, roots and rational numbers.






                share|cite|improve this answer











                $endgroup$



                The answer in characteristic $0$ is no. For instance, the "root closure"* of $Bbb Q$ still wouldn't let you solve, say, an unsolvable quintic like $x^5-x-1 = 0$, because we know its solutions can't be written using roots in $Bbb Q$, so it can't be written using roots in the root closure either.



                *What I mean by "root closure" is this: Start with $Bbb Q$, and consider it a subfield of $Bbb C$ (or of $overline{Bbb Q}$). To $Bbb Q$, add all $n$th roots of all elements of $Bbb Q$. Then add all $n$'th roots of all elements in that new field, and so on. The root closure is the final result, i.e. the union of these fields. Any element in that field may be written with a finite expression involving only addition, subtraction, multiplication, division, roots and rational numbers.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 4 at 11:41

























                answered Jan 4 at 10:12









                ArthurArthur

                116k7116199




                116k7116199























                    4












                    $begingroup$

                    In positive characteristic the situation is even worse. For instance, the field $K$ with two elements trivially satisfies your property, but the polynomial $x^{2} + x + 1$ has no roots in $K$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      @DietrichBurde That's what happens when I only read titles. I need to stop doing that.
                      $endgroup$
                      – Arthur
                      Jan 4 at 10:49








                    • 2




                      $begingroup$
                      Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
                      $endgroup$
                      – Andreas Caranti
                      Jan 4 at 11:07












                    • $begingroup$
                      @DietrichBurde, thanks. And a Happy New Year to you!
                      $endgroup$
                      – Andreas Caranti
                      Jan 4 at 11:57










                    • $begingroup$
                      Thank you, you too!!!
                      $endgroup$
                      – Dietrich Burde
                      Jan 4 at 11:59
















                    4












                    $begingroup$

                    In positive characteristic the situation is even worse. For instance, the field $K$ with two elements trivially satisfies your property, but the polynomial $x^{2} + x + 1$ has no roots in $K$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      @DietrichBurde That's what happens when I only read titles. I need to stop doing that.
                      $endgroup$
                      – Arthur
                      Jan 4 at 10:49








                    • 2




                      $begingroup$
                      Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
                      $endgroup$
                      – Andreas Caranti
                      Jan 4 at 11:07












                    • $begingroup$
                      @DietrichBurde, thanks. And a Happy New Year to you!
                      $endgroup$
                      – Andreas Caranti
                      Jan 4 at 11:57










                    • $begingroup$
                      Thank you, you too!!!
                      $endgroup$
                      – Dietrich Burde
                      Jan 4 at 11:59














                    4












                    4








                    4





                    $begingroup$

                    In positive characteristic the situation is even worse. For instance, the field $K$ with two elements trivially satisfies your property, but the polynomial $x^{2} + x + 1$ has no roots in $K$.






                    share|cite|improve this answer











                    $endgroup$



                    In positive characteristic the situation is even worse. For instance, the field $K$ with two elements trivially satisfies your property, but the polynomial $x^{2} + x + 1$ has no roots in $K$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 4 at 10:50









                    Arthur

                    116k7116199




                    116k7116199










                    answered Jan 4 at 10:15









                    Andreas CarantiAndreas Caranti

                    56.6k34395




                    56.6k34395












                    • $begingroup$
                      @DietrichBurde That's what happens when I only read titles. I need to stop doing that.
                      $endgroup$
                      – Arthur
                      Jan 4 at 10:49








                    • 2




                      $begingroup$
                      Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
                      $endgroup$
                      – Andreas Caranti
                      Jan 4 at 11:07












                    • $begingroup$
                      @DietrichBurde, thanks. And a Happy New Year to you!
                      $endgroup$
                      – Andreas Caranti
                      Jan 4 at 11:57










                    • $begingroup$
                      Thank you, you too!!!
                      $endgroup$
                      – Dietrich Burde
                      Jan 4 at 11:59


















                    • $begingroup$
                      @DietrichBurde That's what happens when I only read titles. I need to stop doing that.
                      $endgroup$
                      – Arthur
                      Jan 4 at 10:49








                    • 2




                      $begingroup$
                      Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
                      $endgroup$
                      – Andreas Caranti
                      Jan 4 at 11:07












                    • $begingroup$
                      @DietrichBurde, thanks. And a Happy New Year to you!
                      $endgroup$
                      – Andreas Caranti
                      Jan 4 at 11:57










                    • $begingroup$
                      Thank you, you too!!!
                      $endgroup$
                      – Dietrich Burde
                      Jan 4 at 11:59
















                    $begingroup$
                    @DietrichBurde That's what happens when I only read titles. I need to stop doing that.
                    $endgroup$
                    – Arthur
                    Jan 4 at 10:49






                    $begingroup$
                    @DietrichBurde That's what happens when I only read titles. I need to stop doing that.
                    $endgroup$
                    – Arthur
                    Jan 4 at 10:49






                    2




                    2




                    $begingroup$
                    Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
                    $endgroup$
                    – Andreas Caranti
                    Jan 4 at 11:07






                    $begingroup$
                    Part of the conversation is gone, but I suppose it is related to the fact that the title appears to restrict the question to characteristic $0$, whereas in the question body it is only stated assume characteristic $0$ if this simplifies things.
                    $endgroup$
                    – Andreas Caranti
                    Jan 4 at 11:07














                    $begingroup$
                    @DietrichBurde, thanks. And a Happy New Year to you!
                    $endgroup$
                    – Andreas Caranti
                    Jan 4 at 11:57




                    $begingroup$
                    @DietrichBurde, thanks. And a Happy New Year to you!
                    $endgroup$
                    – Andreas Caranti
                    Jan 4 at 11:57












                    $begingroup$
                    Thank you, you too!!!
                    $endgroup$
                    – Dietrich Burde
                    Jan 4 at 11:59




                    $begingroup$
                    Thank you, you too!!!
                    $endgroup$
                    – Dietrich Burde
                    Jan 4 at 11:59


















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