Relation between finite fields $F_{p^2}$ and $F_p(omega)$.












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$begingroup$


Let $F_p$ be a finite field having $p$ elements(for some prime $p$), not having primitive
$3$rd root of unit say $omega.$ Then can i say that field $F_p(omega)$ and $F_{p^2}$ are isomorphic.



I tried as follows: As both of fields $F_{p^{2}}$ and $F_p(omega)$ are of same dimension $2$ over the field $F_p$ and hence same number of elements. Now any two finite field with same number of elements are isomorphic.



Please suggest me. Thanks in advance.










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$endgroup$








  • 1




    $begingroup$
    Yes, it is correct. Note that $omega$ is the root $x^2+x+1$, thus having order $2$.
    $endgroup$
    – Berci
    Dec 9 '18 at 10:43










  • $begingroup$
    Okh .... thanks .....
    $endgroup$
    – neelkanth
    Dec 9 '18 at 11:02
















0












$begingroup$


Let $F_p$ be a finite field having $p$ elements(for some prime $p$), not having primitive
$3$rd root of unit say $omega.$ Then can i say that field $F_p(omega)$ and $F_{p^2}$ are isomorphic.



I tried as follows: As both of fields $F_{p^{2}}$ and $F_p(omega)$ are of same dimension $2$ over the field $F_p$ and hence same number of elements. Now any two finite field with same number of elements are isomorphic.



Please suggest me. Thanks in advance.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, it is correct. Note that $omega$ is the root $x^2+x+1$, thus having order $2$.
    $endgroup$
    – Berci
    Dec 9 '18 at 10:43










  • $begingroup$
    Okh .... thanks .....
    $endgroup$
    – neelkanth
    Dec 9 '18 at 11:02














0












0








0


1



$begingroup$


Let $F_p$ be a finite field having $p$ elements(for some prime $p$), not having primitive
$3$rd root of unit say $omega.$ Then can i say that field $F_p(omega)$ and $F_{p^2}$ are isomorphic.



I tried as follows: As both of fields $F_{p^{2}}$ and $F_p(omega)$ are of same dimension $2$ over the field $F_p$ and hence same number of elements. Now any two finite field with same number of elements are isomorphic.



Please suggest me. Thanks in advance.










share|cite|improve this question











$endgroup$




Let $F_p$ be a finite field having $p$ elements(for some prime $p$), not having primitive
$3$rd root of unit say $omega.$ Then can i say that field $F_p(omega)$ and $F_{p^2}$ are isomorphic.



I tried as follows: As both of fields $F_{p^{2}}$ and $F_p(omega)$ are of same dimension $2$ over the field $F_p$ and hence same number of elements. Now any two finite field with same number of elements are isomorphic.



Please suggest me. Thanks in advance.







finite-fields






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 9:34







neelkanth

















asked Dec 9 '18 at 8:56









neelkanthneelkanth

2,20921028




2,20921028








  • 1




    $begingroup$
    Yes, it is correct. Note that $omega$ is the root $x^2+x+1$, thus having order $2$.
    $endgroup$
    – Berci
    Dec 9 '18 at 10:43










  • $begingroup$
    Okh .... thanks .....
    $endgroup$
    – neelkanth
    Dec 9 '18 at 11:02














  • 1




    $begingroup$
    Yes, it is correct. Note that $omega$ is the root $x^2+x+1$, thus having order $2$.
    $endgroup$
    – Berci
    Dec 9 '18 at 10:43










  • $begingroup$
    Okh .... thanks .....
    $endgroup$
    – neelkanth
    Dec 9 '18 at 11:02








1




1




$begingroup$
Yes, it is correct. Note that $omega$ is the root $x^2+x+1$, thus having order $2$.
$endgroup$
– Berci
Dec 9 '18 at 10:43




$begingroup$
Yes, it is correct. Note that $omega$ is the root $x^2+x+1$, thus having order $2$.
$endgroup$
– Berci
Dec 9 '18 at 10:43












$begingroup$
Okh .... thanks .....
$endgroup$
– neelkanth
Dec 9 '18 at 11:02




$begingroup$
Okh .... thanks .....
$endgroup$
– neelkanth
Dec 9 '18 at 11:02










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