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I attempting to solve this proof for my Abstract Algebra II class.

Let $K$ be a finite normal extension of a field $F$, and let $x$ be an element of $G(K/F)$. Prove that if $x(b) = b$ for some element $b \in K$, then $x$ is an element of $G(K/F(b))$.

I attempted to prove by assuming that $x$ is not an element of $G(K/F(b))$ and show that it isn't possible. Though, I am failing to link the field $F$ and $F(b)$ and how the automorphism fixes both $F$ and $F(b)$. Any suggestions would be appreciated.

Thank you.

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    Phosphide, you really should go to your instructor's office hours and get help in person. This question is almost just a matter of definitions, so not understanding it means you are missing a very basic concept.2011-05-05

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By definition, $G(K/F)$ consists of all maps $x\colon K\to K$ such that $x(f)=f$ for all $f\in F$.

By definition, $G(K/F(b))$ consists of all maps $y\colon K\to K$ such that $y(a) = a$ for all $a\in F(b)$.

By definition, $F(b)$ is the smallest subfield of $K$ that contains both $F$ and $b$. Since $K$ is finite over $K$, $F(b)$ is finite over $F$, and you can actually give an expression for what elements of $F(b)$ will "look like", in terms of elements of $F$ and of $b$.

Because $F\subseteq F(b)$, you should be able to note that $G(K/F(b))\subseteq G(K/F)$, because anything that is an automorphism of $K$ and fixes each element of $F(b)$ will also, in particular, fix every element of $F$.

Here you are trying to show that if $x\in G(K/F)$ fixes $b$, in addition to fixing every $f\in F$, then in fact it fixes all of $F(b)$.

In short: the key to this problem is to know the definitions, and to remember what the elements of $F(b)$ "look like."

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    This is actually all that I needed. Thank you for your help.2011-05-05