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












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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?






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    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



























      1












      1








      1







      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















      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






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      edited Jan 4 at 21:29









      user26857

      39.3k123983




      39.3k123983










      asked Jan 4 at 9:40









      Bib-lostBib-lost

      1,993627




      1,993627






















          2 Answers
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          3














          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























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








          • 2




            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.
            – Andreas Caranti
            Jan 4 at 11:07












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










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



















          3














          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























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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            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























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








            • 2




              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.
              – Andreas Caranti
              Jan 4 at 11:07












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










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
















            3














            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























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








            • 2




              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.
              – Andreas Caranti
              Jan 4 at 11:07












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










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














            3












            3








            3






            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














            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

            111k7105186




            111k7105186










            answered Jan 4 at 10:15









            Andreas CarantiAndreas Caranti

            56.2k34295




            56.2k34295












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








            • 2




              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.
              – Andreas Caranti
              Jan 4 at 11:07












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










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


















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








            • 2




              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.
              – Andreas Caranti
              Jan 4 at 11:07












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










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
















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






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






            2




            2




            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.
            – Andreas Caranti
            Jan 4 at 11:07






            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.
            – Andreas Caranti
            Jan 4 at 11:07














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




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












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




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











            3














            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




























              3














              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


























                3












                3








                3






                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














                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

                111k7105186




                111k7105186






























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