Show that $mathbb{Q}(sqrt{3},sqrt[4]{3}, sqrt[8]{3},…)$ is algebraic over $mathbb{Q}$ but not a finite...












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Show that $mathbb{Q}(sqrt{3},sqrt[4]{3}, sqrt[8]{3},...)$ is algebraic over $mathbb{Q}$ but not a finite extension.




I think for the algebraic part, since for every simple extension, each of those elements can be adjoined and each of these simple extensions has a minimal polynomial that cannot be reduced in $mathbb{Q}$. For example, the simple extension $mathbb{Q}(sqrt{3}, sqrt[4]{3})(sqrt[8]{3})$ has minimal polynomial $x^{8}-3$. And since each simple extension has an increasingly large degree, the degree of the simple extensions over the previous extension gets larger for each attachment. But I am not sure how to express this formally...



For the infinite degree part, I was thinking because the set $left {sqrt{3},sqrt[4]{3}, sqrt[8]{3},...right }$ is linearly independent?






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    $begingroup$



    Show that $mathbb{Q}(sqrt{3},sqrt[4]{3}, sqrt[8]{3},...)$ is algebraic over $mathbb{Q}$ but not a finite extension.




    I think for the algebraic part, since for every simple extension, each of those elements can be adjoined and each of these simple extensions has a minimal polynomial that cannot be reduced in $mathbb{Q}$. For example, the simple extension $mathbb{Q}(sqrt{3}, sqrt[4]{3})(sqrt[8]{3})$ has minimal polynomial $x^{8}-3$. And since each simple extension has an increasingly large degree, the degree of the simple extensions over the previous extension gets larger for each attachment. But I am not sure how to express this formally...



    For the infinite degree part, I was thinking because the set $left {sqrt{3},sqrt[4]{3}, sqrt[8]{3},...right }$ is linearly independent?






    abstract-algebra field-theory extension-field







    share|cite|improve this question









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      3












      3








      3


      0



      $begingroup$



      Show that $mathbb{Q}(sqrt{3},sqrt[4]{3}, sqrt[8]{3},...)$ is algebraic over $mathbb{Q}$ but not a finite extension.




      I think for the algebraic part, since for every simple extension, each of those elements can be adjoined and each of these simple extensions has a minimal polynomial that cannot be reduced in $mathbb{Q}$. For example, the simple extension $mathbb{Q}(sqrt{3}, sqrt[4]{3})(sqrt[8]{3})$ has minimal polynomial $x^{8}-3$. And since each simple extension has an increasingly large degree, the degree of the simple extensions over the previous extension gets larger for each attachment. But I am not sure how to express this formally...



      For the infinite degree part, I was thinking because the set $left {sqrt{3},sqrt[4]{3}, sqrt[8]{3},...right }$ is linearly independent?






      abstract-algebra field-theory extension-field







      share|cite|improve this question









      $endgroup$





      Show that $mathbb{Q}(sqrt{3},sqrt[4]{3}, sqrt[8]{3},...)$ is algebraic over $mathbb{Q}$ but not a finite extension.




      I think for the algebraic part, since for every simple extension, each of those elements can be adjoined and each of these simple extensions has a minimal polynomial that cannot be reduced in $mathbb{Q}$. For example, the simple extension $mathbb{Q}(sqrt{3}, sqrt[4]{3})(sqrt[8]{3})$ has minimal polynomial $x^{8}-3$. And since each simple extension has an increasingly large degree, the degree of the simple extensions over the previous extension gets larger for each attachment. But I am not sure how to express this formally...



      For the infinite degree part, I was thinking because the set $left {sqrt{3},sqrt[4]{3}, sqrt[8]{3},...right }$ is linearly independent?






      abstract-algebra field-theory extension-field




      abstract-algebra field-theory extension-field






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      asked Dec 7 '18 at 6:38









      numericalorangenumericalorange

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          2 Answers
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          $begingroup$

          Any element $alpha$ of $F$ is a rational expression in the numbers adjoined. As such, it can involve only finitely many of the $sqrt[2^k]3$. If in such an expression, $sqrt[2^n]3$ is one with maximal $k$, then all other $sqrt[2^k]3$ are powers of $sqrt[2 k]3$. It follows that $alphainBbb Q(sqrt[2^n]3)$ and $alpha $ is algebraic.



          A different approach for the infinity part: By Eisenstein, the polynomial $X^{2^n}-3$ is irreducible. Hence $[F:Bbb Q]ge[Bbb Q(sqrt[2^n]3):Bbb Q]ge 2^n$, where $n$ is arbitrariy. It follows that $[F:Bbb Q]$ is inifnite.






          share|cite|improve this answer









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          • $begingroup$
            Thanks for the insight! I didn't realize I could use Eisenstein's criterion.
            $endgroup$
            – numericalorange
            Dec 7 '18 at 18:07



















          1












          $begingroup$

          Let $L = mathbb Q(3^{frac 1{2^n}})$.



          Why is $L$ algebraic over $mathbb Q$? It is because the generating set of $L$ is $mathbb Q cup {3^{frac 1{2^n}}}$, which are all contained in $bar{mathbb Q}$, the algebraic closure of $mathbb Q$. Therefore, $L subset bar {mathbb Q}$, which by definition of the algebraic closure implies it is algebraic over $mathbb Q$.



          However, $L$ contains subfields of the form $K_n$, where $K_n = mathbb Q(3^{frac 1{2^n}})$. One can conclude that $[K_n : mathbb Q] = 2^n$, since $x^{2^n} - 3$ is irreducible via the Eisenstein criterion(shifted).



          Then, suppose $L$ were finite, then $[L : mathbb Q] < infty $, but by the tower property it is equal to $[L : K_n][K_n : mathbb Q]$ for each $n$. In particular, $[L : mathbb Q]$ is a multiple of $2^n$ for all $n$, clearly impossible if it is finite.






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            2 Answers
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            2 Answers
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            active

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            5












            $begingroup$

            Any element $alpha$ of $F$ is a rational expression in the numbers adjoined. As such, it can involve only finitely many of the $sqrt[2^k]3$. If in such an expression, $sqrt[2^n]3$ is one with maximal $k$, then all other $sqrt[2^k]3$ are powers of $sqrt[2 k]3$. It follows that $alphainBbb Q(sqrt[2^n]3)$ and $alpha $ is algebraic.



            A different approach for the infinity part: By Eisenstein, the polynomial $X^{2^n}-3$ is irreducible. Hence $[F:Bbb Q]ge[Bbb Q(sqrt[2^n]3):Bbb Q]ge 2^n$, where $n$ is arbitrariy. It follows that $[F:Bbb Q]$ is inifnite.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for the insight! I didn't realize I could use Eisenstein's criterion.
              $endgroup$
              – numericalorange
              Dec 7 '18 at 18:07
















            5












            $begingroup$

            Any element $alpha$ of $F$ is a rational expression in the numbers adjoined. As such, it can involve only finitely many of the $sqrt[2^k]3$. If in such an expression, $sqrt[2^n]3$ is one with maximal $k$, then all other $sqrt[2^k]3$ are powers of $sqrt[2 k]3$. It follows that $alphainBbb Q(sqrt[2^n]3)$ and $alpha $ is algebraic.



            A different approach for the infinity part: By Eisenstein, the polynomial $X^{2^n}-3$ is irreducible. Hence $[F:Bbb Q]ge[Bbb Q(sqrt[2^n]3):Bbb Q]ge 2^n$, where $n$ is arbitrariy. It follows that $[F:Bbb Q]$ is inifnite.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks for the insight! I didn't realize I could use Eisenstein's criterion.
              $endgroup$
              – numericalorange
              Dec 7 '18 at 18:07














            5












            5








            5





            $begingroup$

            Any element $alpha$ of $F$ is a rational expression in the numbers adjoined. As such, it can involve only finitely many of the $sqrt[2^k]3$. If in such an expression, $sqrt[2^n]3$ is one with maximal $k$, then all other $sqrt[2^k]3$ are powers of $sqrt[2 k]3$. It follows that $alphainBbb Q(sqrt[2^n]3)$ and $alpha $ is algebraic.



            A different approach for the infinity part: By Eisenstein, the polynomial $X^{2^n}-3$ is irreducible. Hence $[F:Bbb Q]ge[Bbb Q(sqrt[2^n]3):Bbb Q]ge 2^n$, where $n$ is arbitrariy. It follows that $[F:Bbb Q]$ is inifnite.






            share|cite|improve this answer









            $endgroup$



            Any element $alpha$ of $F$ is a rational expression in the numbers adjoined. As such, it can involve only finitely many of the $sqrt[2^k]3$. If in such an expression, $sqrt[2^n]3$ is one with maximal $k$, then all other $sqrt[2^k]3$ are powers of $sqrt[2 k]3$. It follows that $alphainBbb Q(sqrt[2^n]3)$ and $alpha $ is algebraic.



            A different approach for the infinity part: By Eisenstein, the polynomial $X^{2^n}-3$ is irreducible. Hence $[F:Bbb Q]ge[Bbb Q(sqrt[2^n]3):Bbb Q]ge 2^n$, where $n$ is arbitrariy. It follows that $[F:Bbb Q]$ is inifnite.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 7 '18 at 6:48









            Hagen von EitzenHagen von Eitzen

            278k22269499




            278k22269499












            • $begingroup$
              Thanks for the insight! I didn't realize I could use Eisenstein's criterion.
              $endgroup$
              – numericalorange
              Dec 7 '18 at 18:07


















            • $begingroup$
              Thanks for the insight! I didn't realize I could use Eisenstein's criterion.
              $endgroup$
              – numericalorange
              Dec 7 '18 at 18:07
















            $begingroup$
            Thanks for the insight! I didn't realize I could use Eisenstein's criterion.
            $endgroup$
            – numericalorange
            Dec 7 '18 at 18:07




            $begingroup$
            Thanks for the insight! I didn't realize I could use Eisenstein's criterion.
            $endgroup$
            – numericalorange
            Dec 7 '18 at 18:07











            1












            $begingroup$

            Let $L = mathbb Q(3^{frac 1{2^n}})$.



            Why is $L$ algebraic over $mathbb Q$? It is because the generating set of $L$ is $mathbb Q cup {3^{frac 1{2^n}}}$, which are all contained in $bar{mathbb Q}$, the algebraic closure of $mathbb Q$. Therefore, $L subset bar {mathbb Q}$, which by definition of the algebraic closure implies it is algebraic over $mathbb Q$.



            However, $L$ contains subfields of the form $K_n$, where $K_n = mathbb Q(3^{frac 1{2^n}})$. One can conclude that $[K_n : mathbb Q] = 2^n$, since $x^{2^n} - 3$ is irreducible via the Eisenstein criterion(shifted).



            Then, suppose $L$ were finite, then $[L : mathbb Q] < infty $, but by the tower property it is equal to $[L : K_n][K_n : mathbb Q]$ for each $n$. In particular, $[L : mathbb Q]$ is a multiple of $2^n$ for all $n$, clearly impossible if it is finite.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Let $L = mathbb Q(3^{frac 1{2^n}})$.



              Why is $L$ algebraic over $mathbb Q$? It is because the generating set of $L$ is $mathbb Q cup {3^{frac 1{2^n}}}$, which are all contained in $bar{mathbb Q}$, the algebraic closure of $mathbb Q$. Therefore, $L subset bar {mathbb Q}$, which by definition of the algebraic closure implies it is algebraic over $mathbb Q$.



              However, $L$ contains subfields of the form $K_n$, where $K_n = mathbb Q(3^{frac 1{2^n}})$. One can conclude that $[K_n : mathbb Q] = 2^n$, since $x^{2^n} - 3$ is irreducible via the Eisenstein criterion(shifted).



              Then, suppose $L$ were finite, then $[L : mathbb Q] < infty $, but by the tower property it is equal to $[L : K_n][K_n : mathbb Q]$ for each $n$. In particular, $[L : mathbb Q]$ is a multiple of $2^n$ for all $n$, clearly impossible if it is finite.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Let $L = mathbb Q(3^{frac 1{2^n}})$.



                Why is $L$ algebraic over $mathbb Q$? It is because the generating set of $L$ is $mathbb Q cup {3^{frac 1{2^n}}}$, which are all contained in $bar{mathbb Q}$, the algebraic closure of $mathbb Q$. Therefore, $L subset bar {mathbb Q}$, which by definition of the algebraic closure implies it is algebraic over $mathbb Q$.



                However, $L$ contains subfields of the form $K_n$, where $K_n = mathbb Q(3^{frac 1{2^n}})$. One can conclude that $[K_n : mathbb Q] = 2^n$, since $x^{2^n} - 3$ is irreducible via the Eisenstein criterion(shifted).



                Then, suppose $L$ were finite, then $[L : mathbb Q] < infty $, but by the tower property it is equal to $[L : K_n][K_n : mathbb Q]$ for each $n$. In particular, $[L : mathbb Q]$ is a multiple of $2^n$ for all $n$, clearly impossible if it is finite.






                share|cite|improve this answer









                $endgroup$



                Let $L = mathbb Q(3^{frac 1{2^n}})$.



                Why is $L$ algebraic over $mathbb Q$? It is because the generating set of $L$ is $mathbb Q cup {3^{frac 1{2^n}}}$, which are all contained in $bar{mathbb Q}$, the algebraic closure of $mathbb Q$. Therefore, $L subset bar {mathbb Q}$, which by definition of the algebraic closure implies it is algebraic over $mathbb Q$.



                However, $L$ contains subfields of the form $K_n$, where $K_n = mathbb Q(3^{frac 1{2^n}})$. One can conclude that $[K_n : mathbb Q] = 2^n$, since $x^{2^n} - 3$ is irreducible via the Eisenstein criterion(shifted).



                Then, suppose $L$ were finite, then $[L : mathbb Q] < infty $, but by the tower property it is equal to $[L : K_n][K_n : mathbb Q]$ for each $n$. In particular, $[L : mathbb Q]$ is a multiple of $2^n$ for all $n$, clearly impossible if it is finite.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 7 '18 at 6:55









                астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

                38k33376




                38k33376






























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