Prove: \(F(a,b)^*=F(a) \cup F(b)\)

[Kiwi1]

G. Shorter questions relating to automorphisms and Galois Groups

Let F be a field, and K a finite extension of F. Suppose \(a,b \in K\). Prove parts 1-3.

2. \(F(a,b)^*=F(a)^* \cap F(b)^*\)

Surely, they mean the union of F(a) and F(b) and not the intersection? There is no reason to think that \(b \in F(a)\) and therefore no reason to think \(b \in F(a)^* \cap F(b)^*\)?

Also what is the * about? Usualy I would expect \(F(a)^*=F(a)-\{0\}\) but I can see no reason to exclude the zero element. So has the author made typos or am I just confused?

 

[Opalg]

Also what is the * about?

That's the same question that I have. The * notation cannot refer to the nonzero elements. It seems to me that it must indicate some sort of dual structure. I suggest you look bach through whatever text these questions come from, to see where this notation is defined.

 

[Kiwi1]

Could it mean the subset of elements that have an inverse?

 

[steenis]

You could have given us more information about the book you are reading. But on page 327 of that book, I read that $I^*=\mbox{Gal} (K:I)$ is the fixer of $I$. I have no idea what that means.

 

[Kiwi1]

Thanks guys. F(a,b)* is the field of automorphisms of, a larger field containing F(a,b), that don't change each element in F(a,b). With that knowledge solving the problem is straightforward.