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Let $E\supseteq F$ be an extension of fields. Show that $\forall u \in E,$ and nonzero $a\in F,$ $F(u)=F(au)$.

My first instinct was to argue with the fact that $F(u)$ is the smallest subfield that contains both $F$ and $u$, so the "$\subseteq$" inclusion is clear. Would the same approach work for the reverse inclusion? I'm not so sure if this method would work this time around, so any other suggestions would be appreciated!

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You need to show $F(u)\subseteq F(au)$ and $F(au)\subseteq F(u)$.

Because $F(au)$ is the smallest subfield of $E$ containing $F$ and $au$, to show $F(au)\subseteq F(u)$ it suffices to show that $au\in F(u)$; but $u\in F(u)$ and $a\in F\subseteq F(u)$, so their product should be in $F(u)$ as well.

Now use similar logic to show $F(u)\subseteq F(au)$, using the fact that $a$ is invertible in $F$.

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note that $a^{-1}\in F\subset F(au)$, so $a^{-1}au=u\in F(au)$, therefore $F(u)=F(au)$

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