algebraic closure of $mathbb{F}_5$ contains $1$ field of $5^2$ elements












2












$begingroup$


Prove that the algebraic closure of $mathbb{F}_5$ contains only 1 field of $5^2$ elements.



If we denote by $K$ the algebraic closure of $mathbb{F}_5$, and we take $alpha in K$ such that $alpha^2=2$, we know that $[mathbb{F}_5(alpha):mathbb{F}_5]=2$.



Now if we take another element $beta in K$, we must prove $ mathbb{F}_5(beta) = mathbb{F}_5(alpha)$.



I don't know much information about K, maybe if Icompute its cardinal, I can use Sylow theorems to show the unicity of extension of degree 2 over $mathbb{F}_5$?



Thank you.






abstract-algebra extension-field







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  • $begingroup$
    $$K = lim_to mathbb{F}_{5^n}$$ is countably infinite.
    $endgroup$
    – Alex Vong
    Dec 5 '18 at 16:20


















2












$begingroup$


Prove that the algebraic closure of $mathbb{F}_5$ contains only 1 field of $5^2$ elements.



If we denote by $K$ the algebraic closure of $mathbb{F}_5$, and we take $alpha in K$ such that $alpha^2=2$, we know that $[mathbb{F}_5(alpha):mathbb{F}_5]=2$.



Now if we take another element $beta in K$, we must prove $ mathbb{F}_5(beta) = mathbb{F}_5(alpha)$.



I don't know much information about K, maybe if Icompute its cardinal, I can use Sylow theorems to show the unicity of extension of degree 2 over $mathbb{F}_5$?



Thank you.






abstract-algebra extension-field







share|cite|improve this question











$endgroup$












  • $begingroup$
    $$K = lim_to mathbb{F}_{5^n}$$ is countably infinite.
    $endgroup$
    – Alex Vong
    Dec 5 '18 at 16:20
















2












2








2





$begingroup$


Prove that the algebraic closure of $mathbb{F}_5$ contains only 1 field of $5^2$ elements.



If we denote by $K$ the algebraic closure of $mathbb{F}_5$, and we take $alpha in K$ such that $alpha^2=2$, we know that $[mathbb{F}_5(alpha):mathbb{F}_5]=2$.



Now if we take another element $beta in K$, we must prove $ mathbb{F}_5(beta) = mathbb{F}_5(alpha)$.



I don't know much information about K, maybe if Icompute its cardinal, I can use Sylow theorems to show the unicity of extension of degree 2 over $mathbb{F}_5$?



Thank you.






abstract-algebra extension-field







share|cite|improve this question











$endgroup$




Prove that the algebraic closure of $mathbb{F}_5$ contains only 1 field of $5^2$ elements.



If we denote by $K$ the algebraic closure of $mathbb{F}_5$, and we take $alpha in K$ such that $alpha^2=2$, we know that $[mathbb{F}_5(alpha):mathbb{F}_5]=2$.



Now if we take another element $beta in K$, we must prove $ mathbb{F}_5(beta) = mathbb{F}_5(alpha)$.



I don't know much information about K, maybe if Icompute its cardinal, I can use Sylow theorems to show the unicity of extension of degree 2 over $mathbb{F}_5$?



Thank you.






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 Dec 5 '18 at 15:39









Key Flex

7,78461232




7,78461232










asked Dec 5 '18 at 15:36









PerelManPerelMan

531211




531211












  • $begingroup$
    $$K = lim_to mathbb{F}_{5^n}$$ is countably infinite.
    $endgroup$
    – Alex Vong
    Dec 5 '18 at 16:20




















  • $begingroup$
    $$K = lim_to mathbb{F}_{5^n}$$ is countably infinite.
    $endgroup$
    – Alex Vong
    Dec 5 '18 at 16:20


















$begingroup$
$$K = lim_to mathbb{F}_{5^n}$$ is countably infinite.
$endgroup$
– Alex Vong
Dec 5 '18 at 16:20






$begingroup$
$$K = lim_to mathbb{F}_{5^n}$$ is countably infinite.
$endgroup$
– Alex Vong
Dec 5 '18 at 16:20












2 Answers
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$begingroup$

Hint:




  1. Note that $(2alpha)^2=3$, and so every element of $Bbb{F}$ has a square root in $Bbb{F}(alpha)$.


  2. Use the quadratic formula.







share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You can use that for every prime number $p$ and every natural number $n$, a field $mathbb{F}_{p^n}$ of characteristic $p$ and cardinality $p^n$ is a splitting field of the polynomial $f(X)inmathbb{F}_p[X]$ given by $f(X)=X^{p^n}-X$. This follows from the Lagrange's Theorem applied to the group of units $mathbb{F}_{p^n}^{times}$. So if you fix an algebraic closure $overline{mathbb{F}_p}$ the uniqueness of such a subfield follows.






    share|cite|improve this answer









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      2 Answers
      2






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      2 Answers
      2






      active

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      active

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      active

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      0












      $begingroup$

      Hint:




      1. Note that $(2alpha)^2=3$, and so every element of $Bbb{F}$ has a square root in $Bbb{F}(alpha)$.


      2. Use the quadratic formula.







      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Hint:




        1. Note that $(2alpha)^2=3$, and so every element of $Bbb{F}$ has a square root in $Bbb{F}(alpha)$.


        2. Use the quadratic formula.







        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Hint:




          1. Note that $(2alpha)^2=3$, and so every element of $Bbb{F}$ has a square root in $Bbb{F}(alpha)$.


          2. Use the quadratic formula.







          share|cite|improve this answer









          $endgroup$



          Hint:




          1. Note that $(2alpha)^2=3$, and so every element of $Bbb{F}$ has a square root in $Bbb{F}(alpha)$.


          2. Use the quadratic formula.








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 16:27









          MicahMicah

          29.8k1364106




          29.8k1364106























              0












              $begingroup$

              You can use that for every prime number $p$ and every natural number $n$, a field $mathbb{F}_{p^n}$ of characteristic $p$ and cardinality $p^n$ is a splitting field of the polynomial $f(X)inmathbb{F}_p[X]$ given by $f(X)=X^{p^n}-X$. This follows from the Lagrange's Theorem applied to the group of units $mathbb{F}_{p^n}^{times}$. So if you fix an algebraic closure $overline{mathbb{F}_p}$ the uniqueness of such a subfield follows.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You can use that for every prime number $p$ and every natural number $n$, a field $mathbb{F}_{p^n}$ of characteristic $p$ and cardinality $p^n$ is a splitting field of the polynomial $f(X)inmathbb{F}_p[X]$ given by $f(X)=X^{p^n}-X$. This follows from the Lagrange's Theorem applied to the group of units $mathbb{F}_{p^n}^{times}$. So if you fix an algebraic closure $overline{mathbb{F}_p}$ the uniqueness of such a subfield follows.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You can use that for every prime number $p$ and every natural number $n$, a field $mathbb{F}_{p^n}$ of characteristic $p$ and cardinality $p^n$ is a splitting field of the polynomial $f(X)inmathbb{F}_p[X]$ given by $f(X)=X^{p^n}-X$. This follows from the Lagrange's Theorem applied to the group of units $mathbb{F}_{p^n}^{times}$. So if you fix an algebraic closure $overline{mathbb{F}_p}$ the uniqueness of such a subfield follows.






                  share|cite|improve this answer









                  $endgroup$



                  You can use that for every prime number $p$ and every natural number $n$, a field $mathbb{F}_{p^n}$ of characteristic $p$ and cardinality $p^n$ is a splitting field of the polynomial $f(X)inmathbb{F}_p[X]$ given by $f(X)=X^{p^n}-X$. This follows from the Lagrange's Theorem applied to the group of units $mathbb{F}_{p^n}^{times}$. So if you fix an algebraic closure $overline{mathbb{F}_p}$ the uniqueness of such a subfield follows.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 5 '18 at 16:40









                  Dante GrevinoDante Grevino

                  96319




                  96319






























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