The algebraic elements of $K$ extension of $F$ are a subfield of $K$












2












$begingroup$


the question I wanna try to answer is this:




The algebraic elements of $K$ extension of $F$ are a subfield of $K$




I know that:



An element $a in K$ is algebraic on $F$ if $f(a)=0$ for some polynomial $not=0$. Let $S$ be the subfield of algebraic elements of K. So as definition, $S in K$ because $K$ is an extension of S too. But if it's an algebraic extension, must be closed and of course finite.



My approach is that I have two algebraic elements $a,b in K$.
$a+b in K$ and also $ab in K$, and of course they belong to $S$ too.



Maybe I need to write a polynomial with the algebraic elements and show that is finite?



I've some difficult to glue together those pieces for a satisfying proof. Any hints?



Thanks






abstract-algebra







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    the question I wanna try to answer is this:




    The algebraic elements of $K$ extension of $F$ are a subfield of $K$




    I know that:



    An element $a in K$ is algebraic on $F$ if $f(a)=0$ for some polynomial $not=0$. Let $S$ be the subfield of algebraic elements of K. So as definition, $S in K$ because $K$ is an extension of S too. But if it's an algebraic extension, must be closed and of course finite.



    My approach is that I have two algebraic elements $a,b in K$.
    $a+b in K$ and also $ab in K$, and of course they belong to $S$ too.



    Maybe I need to write a polynomial with the algebraic elements and show that is finite?



    I've some difficult to glue together those pieces for a satisfying proof. Any hints?



    Thanks






    abstract-algebra







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      the question I wanna try to answer is this:




      The algebraic elements of $K$ extension of $F$ are a subfield of $K$




      I know that:



      An element $a in K$ is algebraic on $F$ if $f(a)=0$ for some polynomial $not=0$. Let $S$ be the subfield of algebraic elements of K. So as definition, $S in K$ because $K$ is an extension of S too. But if it's an algebraic extension, must be closed and of course finite.



      My approach is that I have two algebraic elements $a,b in K$.
      $a+b in K$ and also $ab in K$, and of course they belong to $S$ too.



      Maybe I need to write a polynomial with the algebraic elements and show that is finite?



      I've some difficult to glue together those pieces for a satisfying proof. Any hints?



      Thanks






      abstract-algebra







      share|cite|improve this question











      $endgroup$




      the question I wanna try to answer is this:




      The algebraic elements of $K$ extension of $F$ are a subfield of $K$




      I know that:



      An element $a in K$ is algebraic on $F$ if $f(a)=0$ for some polynomial $not=0$. Let $S$ be the subfield of algebraic elements of K. So as definition, $S in K$ because $K$ is an extension of S too. But if it's an algebraic extension, must be closed and of course finite.



      My approach is that I have two algebraic elements $a,b in K$.
      $a+b in K$ and also $ab in K$, and of course they belong to $S$ too.



      Maybe I need to write a polynomial with the algebraic elements and show that is finite?



      I've some difficult to glue together those pieces for a satisfying proof. Any hints?



      Thanks






      abstract-algebra




      abstract-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 10:09









      Bernard

      124k741116




      124k741116










      asked Dec 13 '18 at 9:56









      AlessarAlessar

      313115




      313115






















          1 Answer
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          $begingroup$

          Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:22












          • $begingroup$
            I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:27












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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:22












          • $begingroup$
            I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:27
















          2












          $begingroup$

          Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:22












          • $begingroup$
            I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:27














          2












          2








          2





          $begingroup$

          Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.






          share|cite|improve this answer









          $endgroup$



          Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 10:03









          lhflhf

          167k11172404




          167k11172404












          • $begingroup$
            To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:22












          • $begingroup$
            I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:27


















          • $begingroup$
            To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:22












          • $begingroup$
            I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:27
















          $begingroup$
          To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
          $endgroup$
          – Alessar
          Dec 13 '18 at 10:22






          $begingroup$
          To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
          $endgroup$
          – Alessar
          Dec 13 '18 at 10:22














          $begingroup$
          I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
          $endgroup$
          – Alessar
          Dec 13 '18 at 10:27




          $begingroup$
          I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
          $endgroup$
          – Alessar
          Dec 13 '18 at 10:27


















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