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Let $F(a)$ be Field extension over $F$ such that $[F(a):F]=5$. I know that if $[F(a) : F]$ is odd then $F(a) = F(a^2)$. So how can I show that $F(a)= F(a^2+a+1)$. Could somebody please give me hints.

Thank you!

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It's the same reason as with $F(a)=F(a^2)$, namely that $x^2+x+1$ has degree $2$ (i.e. between $1$ and $5$), so we know that

$$F\subset F(a^2+a+1)\subseteq F(a)$$

The first inclusion is proper because $[F(a^2+a+1):F]\big | [F(a):F]$, and we have the degree to be $1$ or $5$, but if $a^2+a+1=b\in F$, then $x^2+x+(1-b)$ is a smaller degree polynomial which $a$ satisfies, a contradiction.

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  • $\begingroup$ @ Adam Hughhes.. Why is $F(a^2+a+1)\subseteq F(a)$? $\endgroup$ – UserAb Mar 28 '16 at 1:27
  • $\begingroup$ @UserAb $F(a)$ is a field, so it is closed under multiplication and addition. $a\cdot a=a^2\in F(a)$ since it's a multiplication, the rest is addition $\endgroup$ – Adam Hughes Mar 28 '16 at 1:28
  • $\begingroup$ @ Adam Hughhes... wow thanks! $\endgroup$ – UserAb Mar 28 '16 at 1:29
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Hint: If $F\subset K\subset L$ are finite extensions, then $$[L:F]=[K:F]\cdot[L:K]$$ If $[L:F]$ is prime, what can you say for $K$?

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