(New page: \[\#6\]<br>Proof: since$[E:F]=p$, where $p$ is a prime, so we know that $E$ is a Galois extension over $F$, and by Galois theory , the immediate field between $E$ and $F$ can only ha...) |
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Latest revision as of 18:20, 2 July 2013
\[\#6\]
Proof: since$[E:F]=p$, where $p$ is a prime, so we know that $E$ is a Galois extension over $F$, and by Galois theory , the immediate field between $E$ and $F$ can only have two options, either $E$ or $F$, since
$F\subseteq F(a)\subseteq E$, we have $[F(a):F]=p$or $[F(a):F]=1$, so we will get either $F(a)=F$or $F(a)=E$.
