Fields extensions over isomorphic fields of different degrees

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What are the simplest examples of situations where in a field $F$ there are two subfields $L_1$ and $L_2$ such that extensions $F/L_1$ and $F/L_2$ are finite, degrees are different
$$
[F:L_1] neq [F:L_2],
$$
but fields $L_1$ and $L_2$ are isomorphic as abstract fields.




abstract-algebra field-theory extension-field




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  • i guess there may not be any such.. I may be wrong...can you tell me what is your motivation in asking this question...
    – user87543
    Oct 9 '14 at 4:35















up vote
6
down vote

favorite
2












What are the simplest examples of situations where in a field $F$ there are two subfields $L_1$ and $L_2$ such that extensions $F/L_1$ and $F/L_2$ are finite, degrees are different
$$
[F:L_1] neq [F:L_2],
$$
but fields $L_1$ and $L_2$ are isomorphic as abstract fields.




abstract-algebra field-theory extension-field




share|cite|improve this question



















  • i guess there may not be any such.. I may be wrong...can you tell me what is your motivation in asking this question...
    – user87543
    Oct 9 '14 at 4:35













up vote
6
down vote

favorite
2









up vote
6
down vote

favorite
2






2





What are the simplest examples of situations where in a field $F$ there are two subfields $L_1$ and $L_2$ such that extensions $F/L_1$ and $F/L_2$ are finite, degrees are different
$$
[F:L_1] neq [F:L_2],
$$
but fields $L_1$ and $L_2$ are isomorphic as abstract fields.




abstract-algebra field-theory extension-field




share|cite|improve this question











What are the simplest examples of situations where in a field $F$ there are two subfields $L_1$ and $L_2$ such that extensions $F/L_1$ and $F/L_2$ are finite, degrees are different
$$
[F:L_1] neq [F:L_2],
$$
but fields $L_1$ and $L_2$ are isomorphic as abstract fields.





abstract-algebra field-theory extension-field





share|cite|improve this question










share|cite|improve this question




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asked Oct 9 '14 at 3:16









Alex

2,7211027




2,7211027











  • i guess there may not be any such.. I may be wrong...can you tell me what is your motivation in asking this question...
    – user87543
    Oct 9 '14 at 4:35

















  • i guess there may not be any such.. I may be wrong...can you tell me what is your motivation in asking this question...
    – user87543
    Oct 9 '14 at 4:35
















i guess there may not be any such.. I may be wrong...can you tell me what is your motivation in asking this question...
– user87543
Oct 9 '14 at 4:35





i guess there may not be any such.. I may be wrong...can you tell me what is your motivation in asking this question...
– user87543
Oct 9 '14 at 4:35











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For any field $k$, and an indeterminate $t$ over $k$, take $F = k(t)$, $L_1 = F$, $L_2 = k(t^2)$. Then $[F : L_1] = 1 ne 2 = [F : L_2]$, but as abstract fields, $L_1 cong L_2$ (as $L_2$ is the field of fractions of $k[t^2]$, which is abstractly isomorphic to a polynomial ring in $1$ variable over $k$).






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    1 Answer
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    active

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    1 Answer
    1






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    active

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    up vote
    2
    down vote



    accepted










    For any field $k$, and an indeterminate $t$ over $k$, take $F = k(t)$, $L_1 = F$, $L_2 = k(t^2)$. Then $[F : L_1] = 1 ne 2 = [F : L_2]$, but as abstract fields, $L_1 cong L_2$ (as $L_2$ is the field of fractions of $k[t^2]$, which is abstractly isomorphic to a polynomial ring in $1$ variable over $k$).






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      up vote
      2
      down vote



      accepted










      For any field $k$, and an indeterminate $t$ over $k$, take $F = k(t)$, $L_1 = F$, $L_2 = k(t^2)$. Then $[F : L_1] = 1 ne 2 = [F : L_2]$, but as abstract fields, $L_1 cong L_2$ (as $L_2$ is the field of fractions of $k[t^2]$, which is abstractly isomorphic to a polynomial ring in $1$ variable over $k$).






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        For any field $k$, and an indeterminate $t$ over $k$, take $F = k(t)$, $L_1 = F$, $L_2 = k(t^2)$. Then $[F : L_1] = 1 ne 2 = [F : L_2]$, but as abstract fields, $L_1 cong L_2$ (as $L_2$ is the field of fractions of $k[t^2]$, which is abstractly isomorphic to a polynomial ring in $1$ variable over $k$).






        share|cite|improve this answer













        For any field $k$, and an indeterminate $t$ over $k$, take $F = k(t)$, $L_1 = F$, $L_2 = k(t^2)$. Then $[F : L_1] = 1 ne 2 = [F : L_2]$, but as abstract fields, $L_1 cong L_2$ (as $L_2$ is the field of fractions of $k[t^2]$, which is abstractly isomorphic to a polynomial ring in $1$ variable over $k$).







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        answered Oct 9 '14 at 4:46









        zcn

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