If $[F(a):F]=5$, find $[F(a^3):F]$.

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so as it is in the question.



Clearly $F(a^3)$ is a subfield of $F(a)$. So we know that $[F(a):F(a^3)][F(a^3):F]=5$ the problem now is, how do we know if $[F(a):F(a^3)]=5$ or $[F(a^3):F]=5$. The trouble is that $F(a^3)$ could possibly be $F(a)$ or $F$ depending on the multiplicative order of $a$. Or so I think, am I making a mistake somewhere? The question then goes on to ask if the argument applies if we change $a^3$ to $a^2$ or $a^4$. Any hints would be appriciated.




abstract-algebra extension-field




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  • 1




    Hint: if $F(a^3)=F$, is it possible $[F(a):F]=5$?
    – Wojowu
    Jul 16 at 18:55










  • @Wojowu ohhh because then the extension is of degree 3!
    – Sorfosh
    Jul 16 at 18:56










  • Or less, but at most 3.
    – Sorfosh
    Jul 16 at 18:58














up vote
1
down vote

favorite












so as it is in the question.



Clearly $F(a^3)$ is a subfield of $F(a)$. So we know that $[F(a):F(a^3)][F(a^3):F]=5$ the problem now is, how do we know if $[F(a):F(a^3)]=5$ or $[F(a^3):F]=5$. The trouble is that $F(a^3)$ could possibly be $F(a)$ or $F$ depending on the multiplicative order of $a$. Or so I think, am I making a mistake somewhere? The question then goes on to ask if the argument applies if we change $a^3$ to $a^2$ or $a^4$. Any hints would be appriciated.




abstract-algebra extension-field




share|cite|improve this question















  • 1




    Hint: if $F(a^3)=F$, is it possible $[F(a):F]=5$?
    – Wojowu
    Jul 16 at 18:55










  • @Wojowu ohhh because then the extension is of degree 3!
    – Sorfosh
    Jul 16 at 18:56










  • Or less, but at most 3.
    – Sorfosh
    Jul 16 at 18:58












up vote
1
down vote

favorite









up vote
1
down vote

favorite











so as it is in the question.



Clearly $F(a^3)$ is a subfield of $F(a)$. So we know that $[F(a):F(a^3)][F(a^3):F]=5$ the problem now is, how do we know if $[F(a):F(a^3)]=5$ or $[F(a^3):F]=5$. The trouble is that $F(a^3)$ could possibly be $F(a)$ or $F$ depending on the multiplicative order of $a$. Or so I think, am I making a mistake somewhere? The question then goes on to ask if the argument applies if we change $a^3$ to $a^2$ or $a^4$. Any hints would be appriciated.




abstract-algebra extension-field




share|cite|improve this question











so as it is in the question.



Clearly $F(a^3)$ is a subfield of $F(a)$. So we know that $[F(a):F(a^3)][F(a^3):F]=5$ the problem now is, how do we know if $[F(a):F(a^3)]=5$ or $[F(a^3):F]=5$. The trouble is that $F(a^3)$ could possibly be $F(a)$ or $F$ depending on the multiplicative order of $a$. Or so I think, am I making a mistake somewhere? The question then goes on to ask if the argument applies if we change $a^3$ to $a^2$ or $a^4$. Any hints would be appriciated.





abstract-algebra extension-field





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 16 at 18:52









Sorfosh

910616




910616







  • 1




    Hint: if $F(a^3)=F$, is it possible $[F(a):F]=5$?
    – Wojowu
    Jul 16 at 18:55










  • @Wojowu ohhh because then the extension is of degree 3!
    – Sorfosh
    Jul 16 at 18:56










  • Or less, but at most 3.
    – Sorfosh
    Jul 16 at 18:58












  • 1




    Hint: if $F(a^3)=F$, is it possible $[F(a):F]=5$?
    – Wojowu
    Jul 16 at 18:55










  • @Wojowu ohhh because then the extension is of degree 3!
    – Sorfosh
    Jul 16 at 18:56










  • Or less, but at most 3.
    – Sorfosh
    Jul 16 at 18:58







1




1




Hint: if $F(a^3)=F$, is it possible $[F(a):F]=5$?
– Wojowu
Jul 16 at 18:55




Hint: if $F(a^3)=F$, is it possible $[F(a):F]=5$?
– Wojowu
Jul 16 at 18:55












@Wojowu ohhh because then the extension is of degree 3!
– Sorfosh
Jul 16 at 18:56




@Wojowu ohhh because then the extension is of degree 3!
– Sorfosh
Jul 16 at 18:56












Or less, but at most 3.
– Sorfosh
Jul 16 at 18:58




Or less, but at most 3.
– Sorfosh
Jul 16 at 18:58










1 Answer
1






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3
down vote



accepted










As you stated, $$5 = [F(a):F] = [F(a):F(a^3)][F(a^3):F].$$ If the second term in the product is 1, then $a^3 in F$, so that $[F(a):F] le 3$. We know that's false! Hence the second value is 5, since it's not 1, and 5 is a prime.






share|cite|improve this answer





















  • So it works for $a^2$ and $a^4$ as well, same argument, right?
    – Sorfosh
    Jul 16 at 18:58










  • Yup! We just get different inequalities, but they're all impossible.
    – Mr. Chip
    Jul 16 at 19:00










  • Thank you! I should have went step further and saw why thats impossible.
    – Sorfosh
    Jul 16 at 19:01










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










As you stated, $$5 = [F(a):F] = [F(a):F(a^3)][F(a^3):F].$$ If the second term in the product is 1, then $a^3 in F$, so that $[F(a):F] le 3$. We know that's false! Hence the second value is 5, since it's not 1, and 5 is a prime.






share|cite|improve this answer





















  • So it works for $a^2$ and $a^4$ as well, same argument, right?
    – Sorfosh
    Jul 16 at 18:58










  • Yup! We just get different inequalities, but they're all impossible.
    – Mr. Chip
    Jul 16 at 19:00










  • Thank you! I should have went step further and saw why thats impossible.
    – Sorfosh
    Jul 16 at 19:01














up vote
3
down vote



accepted










As you stated, $$5 = [F(a):F] = [F(a):F(a^3)][F(a^3):F].$$ If the second term in the product is 1, then $a^3 in F$, so that $[F(a):F] le 3$. We know that's false! Hence the second value is 5, since it's not 1, and 5 is a prime.






share|cite|improve this answer





















  • So it works for $a^2$ and $a^4$ as well, same argument, right?
    – Sorfosh
    Jul 16 at 18:58










  • Yup! We just get different inequalities, but they're all impossible.
    – Mr. Chip
    Jul 16 at 19:00










  • Thank you! I should have went step further and saw why thats impossible.
    – Sorfosh
    Jul 16 at 19:01












up vote
3
down vote



accepted







up vote
3
down vote



accepted






As you stated, $$5 = [F(a):F] = [F(a):F(a^3)][F(a^3):F].$$ If the second term in the product is 1, then $a^3 in F$, so that $[F(a):F] le 3$. We know that's false! Hence the second value is 5, since it's not 1, and 5 is a prime.






share|cite|improve this answer













As you stated, $$5 = [F(a):F] = [F(a):F(a^3)][F(a^3):F].$$ If the second term in the product is 1, then $a^3 in F$, so that $[F(a):F] le 3$. We know that's false! Hence the second value is 5, since it's not 1, and 5 is a prime.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 16 at 18:55









Mr. Chip

3,2031028




3,2031028











  • So it works for $a^2$ and $a^4$ as well, same argument, right?
    – Sorfosh
    Jul 16 at 18:58










  • Yup! We just get different inequalities, but they're all impossible.
    – Mr. Chip
    Jul 16 at 19:00










  • Thank you! I should have went step further and saw why thats impossible.
    – Sorfosh
    Jul 16 at 19:01
















  • So it works for $a^2$ and $a^4$ as well, same argument, right?
    – Sorfosh
    Jul 16 at 18:58










  • Yup! We just get different inequalities, but they're all impossible.
    – Mr. Chip
    Jul 16 at 19:00










  • Thank you! I should have went step further and saw why thats impossible.
    – Sorfosh
    Jul 16 at 19:01















So it works for $a^2$ and $a^4$ as well, same argument, right?
– Sorfosh
Jul 16 at 18:58




So it works for $a^2$ and $a^4$ as well, same argument, right?
– Sorfosh
Jul 16 at 18:58












Yup! We just get different inequalities, but they're all impossible.
– Mr. Chip
Jul 16 at 19:00




Yup! We just get different inequalities, but they're all impossible.
– Mr. Chip
Jul 16 at 19:00












Thank you! I should have went step further and saw why thats impossible.
– Sorfosh
Jul 16 at 19:01




Thank you! I should have went step further and saw why thats impossible.
– Sorfosh
Jul 16 at 19:01












 

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