If $L_1$, $L_2$ are intermediate fields of $E/K$, then $[L_1 L_2 : K] = [L_1 : K] [L_2 : K] implies L_1 cap...












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Let $E/K$ be a field extension and let $L_1$, $L_2$ be intermediate fields of $E/K$ with $[L_i : K] < infty$, $i = 1,2$. Prove that $[L_1 L_2 : K] = [L_1:K]cdot[L_2:K] implies L_1 cap L_2 = K$.






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    $begingroup$


    Let $E/K$ be a field extension and let $L_1$, $L_2$ be intermediate fields of $E/K$ with $[L_i : K] < infty$, $i = 1,2$. Prove that $[L_1 L_2 : K] = [L_1:K]cdot[L_2:K] implies L_1 cap L_2 = K$.






    field-theory extension-field







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      0












      0








      0





      $begingroup$


      Let $E/K$ be a field extension and let $L_1$, $L_2$ be intermediate fields of $E/K$ with $[L_i : K] < infty$, $i = 1,2$. Prove that $[L_1 L_2 : K] = [L_1:K]cdot[L_2:K] implies L_1 cap L_2 = K$.






      field-theory extension-field







      share|cite|improve this question











      $endgroup$




      Let $E/K$ be a field extension and let $L_1$, $L_2$ be intermediate fields of $E/K$ with $[L_i : K] < infty$, $i = 1,2$. Prove that $[L_1 L_2 : K] = [L_1:K]cdot[L_2:K] implies L_1 cap L_2 = K$.






      field-theory extension-field




      field-theory extension-field






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      edited Dec 14 '18 at 8:14









      Brahadeesh

      6,46942363




      6,46942363










      asked Sep 30 '16 at 14:25









      F.KF.K

      589415




      589415






















          1 Answer
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          $begingroup$

          One can proceed as follows:




          1. If $S$ is a $K$-basis for $L_1$, then show that $S$ is a generating set for $L_1L_2$ as a vector space over $L_2$. In other words,
            $$
            L_1L_2 = text{span}_{L_2}(S)
            $$
            This amounts to showing that the vector space on the right is actually a field (for this, you need that $S$ is a finite set)


          2. Once you have that, it shows that
            $$
            [L_1L_2:K] leq |S| = [L_1:K] qquad (ast)
            $$
            (This is one way to prove that $[L_1L_2:K] leq [L_1:K][L_2:K]$) In your case, since equality holds in $(ast)$, it also shows that $S$ is $L_2$-linearly independent.


          3. Now since $Ksubset L_1cap L_2$, choose a basis $B$ for $L_1cap L_2$ over $K$. Since $B$ is linearly independent over $K$, it may be extended to a basis $S$ of $L_1$ over $K$. Now by part 2, it follows that $B$ is linearly independent over $L_2$. However, $Bsubset L_2$, so it must follows that
            $$
            |B|=1
            $$
            and so $L_1cap L_2 = K$.







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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            One can proceed as follows:




            1. If $S$ is a $K$-basis for $L_1$, then show that $S$ is a generating set for $L_1L_2$ as a vector space over $L_2$. In other words,
              $$
              L_1L_2 = text{span}_{L_2}(S)
              $$
              This amounts to showing that the vector space on the right is actually a field (for this, you need that $S$ is a finite set)


            2. Once you have that, it shows that
              $$
              [L_1L_2:K] leq |S| = [L_1:K] qquad (ast)
              $$
              (This is one way to prove that $[L_1L_2:K] leq [L_1:K][L_2:K]$) In your case, since equality holds in $(ast)$, it also shows that $S$ is $L_2$-linearly independent.


            3. Now since $Ksubset L_1cap L_2$, choose a basis $B$ for $L_1cap L_2$ over $K$. Since $B$ is linearly independent over $K$, it may be extended to a basis $S$ of $L_1$ over $K$. Now by part 2, it follows that $B$ is linearly independent over $L_2$. However, $Bsubset L_2$, so it must follows that
              $$
              |B|=1
              $$
              and so $L_1cap L_2 = K$.







            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              One can proceed as follows:




              1. If $S$ is a $K$-basis for $L_1$, then show that $S$ is a generating set for $L_1L_2$ as a vector space over $L_2$. In other words,
                $$
                L_1L_2 = text{span}_{L_2}(S)
                $$
                This amounts to showing that the vector space on the right is actually a field (for this, you need that $S$ is a finite set)


              2. Once you have that, it shows that
                $$
                [L_1L_2:K] leq |S| = [L_1:K] qquad (ast)
                $$
                (This is one way to prove that $[L_1L_2:K] leq [L_1:K][L_2:K]$) In your case, since equality holds in $(ast)$, it also shows that $S$ is $L_2$-linearly independent.


              3. Now since $Ksubset L_1cap L_2$, choose a basis $B$ for $L_1cap L_2$ over $K$. Since $B$ is linearly independent over $K$, it may be extended to a basis $S$ of $L_1$ over $K$. Now by part 2, it follows that $B$ is linearly independent over $L_2$. However, $Bsubset L_2$, so it must follows that
                $$
                |B|=1
                $$
                and so $L_1cap L_2 = K$.







              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                One can proceed as follows:




                1. If $S$ is a $K$-basis for $L_1$, then show that $S$ is a generating set for $L_1L_2$ as a vector space over $L_2$. In other words,
                  $$
                  L_1L_2 = text{span}_{L_2}(S)
                  $$
                  This amounts to showing that the vector space on the right is actually a field (for this, you need that $S$ is a finite set)


                2. Once you have that, it shows that
                  $$
                  [L_1L_2:K] leq |S| = [L_1:K] qquad (ast)
                  $$
                  (This is one way to prove that $[L_1L_2:K] leq [L_1:K][L_2:K]$) In your case, since equality holds in $(ast)$, it also shows that $S$ is $L_2$-linearly independent.


                3. Now since $Ksubset L_1cap L_2$, choose a basis $B$ for $L_1cap L_2$ over $K$. Since $B$ is linearly independent over $K$, it may be extended to a basis $S$ of $L_1$ over $K$. Now by part 2, it follows that $B$ is linearly independent over $L_2$. However, $Bsubset L_2$, so it must follows that
                  $$
                  |B|=1
                  $$
                  and so $L_1cap L_2 = K$.







                share|cite|improve this answer









                $endgroup$



                One can proceed as follows:




                1. If $S$ is a $K$-basis for $L_1$, then show that $S$ is a generating set for $L_1L_2$ as a vector space over $L_2$. In other words,
                  $$
                  L_1L_2 = text{span}_{L_2}(S)
                  $$
                  This amounts to showing that the vector space on the right is actually a field (for this, you need that $S$ is a finite set)


                2. Once you have that, it shows that
                  $$
                  [L_1L_2:K] leq |S| = [L_1:K] qquad (ast)
                  $$
                  (This is one way to prove that $[L_1L_2:K] leq [L_1:K][L_2:K]$) In your case, since equality holds in $(ast)$, it also shows that $S$ is $L_2$-linearly independent.


                3. Now since $Ksubset L_1cap L_2$, choose a basis $B$ for $L_1cap L_2$ over $K$. Since $B$ is linearly independent over $K$, it may be extended to a basis $S$ of $L_1$ over $K$. Now by part 2, it follows that $B$ is linearly independent over $L_2$. However, $Bsubset L_2$, so it must follows that
                  $$
                  |B|=1
                  $$
                  and so $L_1cap L_2 = K$.








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                answered Oct 5 '16 at 6:27









                Prahlad VaidyanathanPrahlad Vaidyanathan

                26.6k12153




                26.6k12153






























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