Field Extensions: Prove $F(a,b)^* = F(a)^* \cap F(b)^*$ Let $F$ be a field,
Let $K$ be a finite extension of $F$,
Let $a, b$ be members of $K$.
Prove $F(a,b)^* = F(a)^* \cap  F(b)^*$
From Pinter's book of abstract algebra chapter 32 problem G2
I'm assuming the asterisk means that zero is not included in the sets.  Not sure on that but it seems to be the usage earlier in this book.
But with or without the identity elements, since $F \subseteq F(a) \subseteq F(a,b)$ and $F \subseteq F(b) \subseteq F(a,b)$, how is "$F(a,b)^* = F(a)^* \cap  F(b)^*$" necessarily true?  Is this a typo, should it be union?  What do the asterisks have to do with it?
I wish I had more to go on, but this is it.
 A: According to the comments, for an intermediate extension $L$, $F\leq L\leq K$, the notation means $$L^* = \mathrm{Gal}(K/L) = \{\sigma\in \mathrm{Aut}(K)\mid \sigma(x)=x\text{ for all }x\in L\}.$$ That is, the automorphisms of $K$ that fix $L$ pointwise.
(The notation often implies that the extension is Galois, but many authors do not require it for the definition; e.g., Lang’s Algebra calls the automorphism group as above “the Galois group”, and has a separate notation to denote the special case when the extension is Galois; in the comments it is stated that there is an assumption the extension will be Galois, so see remark below the horizontal line for a more immediate proof in that situation.)
With that understanding, the equality should be obvious: what are the automorphisms of $K$ that fix $F(a,b)$? Well, they must fix all of $F$, they must fix $a$, and they must fix $b$. Thus, they must fix both $F(a)$ and $F(b)$. Conversely, if an automorphism fixes both $F(a)$ and $F(b)$, then it fixes their compositum, $F(a)F(b) = F(a,b)$, which gives the equality.

If you know the extension is Galois then the equality is even more immediate: the Fundamental Theorem of Galois Theory establishes an inclusion-reversing correspondence between subgroups of $\mathrm{Gal}(K/F)$ and intermediate fields given precisely by the ${}^*$ operator. Since $F(a,b)$ is the smallest field that contains both $F(a)$ and $F(b)$, it follows that $F(a,b)^*$ must be the largest subgroup that is contained in both $F(a)^*$ and $F(b)^*$, i.e., the intersection.
A: I looked up Pinter's book. A couple paragraphs above Theorem 2 (same chapter) he defines $I^* = \operatorname{Gal}(K : I)$ (where $I$ is an intermediate field extension). He calls this the "fixer" of $I$ but I would like to point out that most people call this the "isotropy group" or maybe "stabilizer group" instead.
With this in mind, the exercise should not be too difficult. Basically it just comes down to the facts that $\sigma \in F(a)^*$ if and only if $\sigma(a) = a$ and $\sigma \in F(a,b)^*$ if and only if $\sigma(a) = a, \sigma(b) = b$.
