Extension with algebraic element is finite












1












$begingroup$


I don't get how to proof this theorem:



Let $E/K$ be a field extension. If $alpha$ is algebraic over K, then $K(alpha ):K<infty$.



I know that we can assume that there exists a nontrivial polynomial $f(X)$ with $f(alpha)=0$. We didn't had the minimal polynomial in class yet. I would very much appreciate your help.



Best, KingDingeling






abstract-algebra field-theory extension-field







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    1












    $begingroup$


    I don't get how to proof this theorem:



    Let $E/K$ be a field extension. If $alpha$ is algebraic over K, then $K(alpha ):K<infty$.



    I know that we can assume that there exists a nontrivial polynomial $f(X)$ with $f(alpha)=0$. We didn't had the minimal polynomial in class yet. I would very much appreciate your help.



    Best, KingDingeling






    abstract-algebra field-theory extension-field







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I don't get how to proof this theorem:



      Let $E/K$ be a field extension. If $alpha$ is algebraic over K, then $K(alpha ):K<infty$.



      I know that we can assume that there exists a nontrivial polynomial $f(X)$ with $f(alpha)=0$. We didn't had the minimal polynomial in class yet. I would very much appreciate your help.



      Best, KingDingeling






      abstract-algebra field-theory extension-field







      share|cite|improve this question











      $endgroup$




      I don't get how to proof this theorem:



      Let $E/K$ be a field extension. If $alpha$ is algebraic over K, then $K(alpha ):K<infty$.



      I know that we can assume that there exists a nontrivial polynomial $f(X)$ with $f(alpha)=0$. We didn't had the minimal polynomial in class yet. I would very much appreciate your help.



      Best, KingDingeling






      abstract-algebra field-theory extension-field




      abstract-algebra field-theory extension-field






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 30 '18 at 14:44









      Jyrki Lahtonen

      108k12166367




      108k12166367










      asked Dec 30 '18 at 11:24









      KingDingelingKingDingeling

      1316




      1316






















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          $begingroup$

          The relation
          $$alpha^n + sum_{i=0}^{n-1} a_i alpha^i = 0$$
          shows that you can write all powers of $alpha$ as linear cobinations of ${ alpha^i}_{0 le i le n-1}$.



          Proof by induction.



          CASE $k=0, dots , n-1$ is obvious. $alpha^k$ is a linear combinations of ${ alpha^i}_{0 le i le n-1}$.



          CASE $k=n$. Follows from
          $alpha^n =- sum_{i=0}^{n-1} a_i alpha^i $



          INDUCTIVE STEP. For $k ge n+1$ you have
          $$alpha^k = alpha cdot alpha^{k-1}$$
          since $alpha^{k-1}$ is a linear combinations of ${ alpha^i}_{0 le i le n-1}$, you can write
          $$alpha^{k-1} = sum_{i=0}^{n-1} b_i alpha^i$$
          Thus
          $$alpha^k = alpha cdot alpha^{k-1} = alpha cdot sum_{i=0}^{n-1} b_i alpha^i = sum_{i=0}^{n-1} b_i alpha^{i+1} = sum_{i=0}^{n-2} b_i alpha^{i+1} + b_{n-1} alpha^n = sum_{i=0}^{n-2} b_i alpha^{i+1} - b_{n-1} sum_{i=0}^{n-1} a_i alpha^i$$
          is a linear combination of ${ alpha^i}_{0 le i le n-1}$.



          This concludes the proof.



          WHAT DOES THIS MEAN: Since
          $$K( alpha) = { sum_j a_j alpha^j : a_j in K }$$
          it shows that $K( alpha)$ is generated by ${ alpha^i}_{0 le i le n-1}$, i.e. it is finitely generated.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you taking the time and help with the problem! :)
            $endgroup$
            – KingDingeling
            Dec 30 '18 at 13:46











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          1 Answer
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          1 Answer
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          $begingroup$

          The relation
          $$alpha^n + sum_{i=0}^{n-1} a_i alpha^i = 0$$
          shows that you can write all powers of $alpha$ as linear cobinations of ${ alpha^i}_{0 le i le n-1}$.



          Proof by induction.



          CASE $k=0, dots , n-1$ is obvious. $alpha^k$ is a linear combinations of ${ alpha^i}_{0 le i le n-1}$.



          CASE $k=n$. Follows from
          $alpha^n =- sum_{i=0}^{n-1} a_i alpha^i $



          INDUCTIVE STEP. For $k ge n+1$ you have
          $$alpha^k = alpha cdot alpha^{k-1}$$
          since $alpha^{k-1}$ is a linear combinations of ${ alpha^i}_{0 le i le n-1}$, you can write
          $$alpha^{k-1} = sum_{i=0}^{n-1} b_i alpha^i$$
          Thus
          $$alpha^k = alpha cdot alpha^{k-1} = alpha cdot sum_{i=0}^{n-1} b_i alpha^i = sum_{i=0}^{n-1} b_i alpha^{i+1} = sum_{i=0}^{n-2} b_i alpha^{i+1} + b_{n-1} alpha^n = sum_{i=0}^{n-2} b_i alpha^{i+1} - b_{n-1} sum_{i=0}^{n-1} a_i alpha^i$$
          is a linear combination of ${ alpha^i}_{0 le i le n-1}$.



          This concludes the proof.



          WHAT DOES THIS MEAN: Since
          $$K( alpha) = { sum_j a_j alpha^j : a_j in K }$$
          it shows that $K( alpha)$ is generated by ${ alpha^i}_{0 le i le n-1}$, i.e. it is finitely generated.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you taking the time and help with the problem! :)
            $endgroup$
            – KingDingeling
            Dec 30 '18 at 13:46
















          2












          $begingroup$

          The relation
          $$alpha^n + sum_{i=0}^{n-1} a_i alpha^i = 0$$
          shows that you can write all powers of $alpha$ as linear cobinations of ${ alpha^i}_{0 le i le n-1}$.



          Proof by induction.



          CASE $k=0, dots , n-1$ is obvious. $alpha^k$ is a linear combinations of ${ alpha^i}_{0 le i le n-1}$.



          CASE $k=n$. Follows from
          $alpha^n =- sum_{i=0}^{n-1} a_i alpha^i $



          INDUCTIVE STEP. For $k ge n+1$ you have
          $$alpha^k = alpha cdot alpha^{k-1}$$
          since $alpha^{k-1}$ is a linear combinations of ${ alpha^i}_{0 le i le n-1}$, you can write
          $$alpha^{k-1} = sum_{i=0}^{n-1} b_i alpha^i$$
          Thus
          $$alpha^k = alpha cdot alpha^{k-1} = alpha cdot sum_{i=0}^{n-1} b_i alpha^i = sum_{i=0}^{n-1} b_i alpha^{i+1} = sum_{i=0}^{n-2} b_i alpha^{i+1} + b_{n-1} alpha^n = sum_{i=0}^{n-2} b_i alpha^{i+1} - b_{n-1} sum_{i=0}^{n-1} a_i alpha^i$$
          is a linear combination of ${ alpha^i}_{0 le i le n-1}$.



          This concludes the proof.



          WHAT DOES THIS MEAN: Since
          $$K( alpha) = { sum_j a_j alpha^j : a_j in K }$$
          it shows that $K( alpha)$ is generated by ${ alpha^i}_{0 le i le n-1}$, i.e. it is finitely generated.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you taking the time and help with the problem! :)
            $endgroup$
            – KingDingeling
            Dec 30 '18 at 13:46














          2












          2








          2





          $begingroup$

          The relation
          $$alpha^n + sum_{i=0}^{n-1} a_i alpha^i = 0$$
          shows that you can write all powers of $alpha$ as linear cobinations of ${ alpha^i}_{0 le i le n-1}$.



          Proof by induction.



          CASE $k=0, dots , n-1$ is obvious. $alpha^k$ is a linear combinations of ${ alpha^i}_{0 le i le n-1}$.



          CASE $k=n$. Follows from
          $alpha^n =- sum_{i=0}^{n-1} a_i alpha^i $



          INDUCTIVE STEP. For $k ge n+1$ you have
          $$alpha^k = alpha cdot alpha^{k-1}$$
          since $alpha^{k-1}$ is a linear combinations of ${ alpha^i}_{0 le i le n-1}$, you can write
          $$alpha^{k-1} = sum_{i=0}^{n-1} b_i alpha^i$$
          Thus
          $$alpha^k = alpha cdot alpha^{k-1} = alpha cdot sum_{i=0}^{n-1} b_i alpha^i = sum_{i=0}^{n-1} b_i alpha^{i+1} = sum_{i=0}^{n-2} b_i alpha^{i+1} + b_{n-1} alpha^n = sum_{i=0}^{n-2} b_i alpha^{i+1} - b_{n-1} sum_{i=0}^{n-1} a_i alpha^i$$
          is a linear combination of ${ alpha^i}_{0 le i le n-1}$.



          This concludes the proof.



          WHAT DOES THIS MEAN: Since
          $$K( alpha) = { sum_j a_j alpha^j : a_j in K }$$
          it shows that $K( alpha)$ is generated by ${ alpha^i}_{0 le i le n-1}$, i.e. it is finitely generated.






          share|cite|improve this answer









          $endgroup$



          The relation
          $$alpha^n + sum_{i=0}^{n-1} a_i alpha^i = 0$$
          shows that you can write all powers of $alpha$ as linear cobinations of ${ alpha^i}_{0 le i le n-1}$.



          Proof by induction.



          CASE $k=0, dots , n-1$ is obvious. $alpha^k$ is a linear combinations of ${ alpha^i}_{0 le i le n-1}$.



          CASE $k=n$. Follows from
          $alpha^n =- sum_{i=0}^{n-1} a_i alpha^i $



          INDUCTIVE STEP. For $k ge n+1$ you have
          $$alpha^k = alpha cdot alpha^{k-1}$$
          since $alpha^{k-1}$ is a linear combinations of ${ alpha^i}_{0 le i le n-1}$, you can write
          $$alpha^{k-1} = sum_{i=0}^{n-1} b_i alpha^i$$
          Thus
          $$alpha^k = alpha cdot alpha^{k-1} = alpha cdot sum_{i=0}^{n-1} b_i alpha^i = sum_{i=0}^{n-1} b_i alpha^{i+1} = sum_{i=0}^{n-2} b_i alpha^{i+1} + b_{n-1} alpha^n = sum_{i=0}^{n-2} b_i alpha^{i+1} - b_{n-1} sum_{i=0}^{n-1} a_i alpha^i$$
          is a linear combination of ${ alpha^i}_{0 le i le n-1}$.



          This concludes the proof.



          WHAT DOES THIS MEAN: Since
          $$K( alpha) = { sum_j a_j alpha^j : a_j in K }$$
          it shows that $K( alpha)$ is generated by ${ alpha^i}_{0 le i le n-1}$, i.e. it is finitely generated.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 11:48









          CrostulCrostul

          27.7k22352




          27.7k22352












          • $begingroup$
            Thank you taking the time and help with the problem! :)
            $endgroup$
            – KingDingeling
            Dec 30 '18 at 13:46


















          • $begingroup$
            Thank you taking the time and help with the problem! :)
            $endgroup$
            – KingDingeling
            Dec 30 '18 at 13:46
















          $begingroup$
          Thank you taking the time and help with the problem! :)
          $endgroup$
          – KingDingeling
          Dec 30 '18 at 13:46




          $begingroup$
          Thank you taking the time and help with the problem! :)
          $endgroup$
          – KingDingeling
          Dec 30 '18 at 13:46


















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