# $\sigma$ trivial on $K$ and $F'$ $\implies$ $\sigma$ trivial on $KF'$?

In section 14.4 of Dummit/Foote, there is the following proposition:

Suppose $K/F$ is a Galois extension and $F'/F$ is any extension. Then $KF'/F'$ is a Galois extension, with Galois group

\begin{equation*} \text{Gal}(KF'/F') \cong \text{Gal}(K/K \cap F') \end{equation*}

The proof begins as follows:

• Since $K$ is Galois over $F$, $K$ is the splitting field of some polynomial $f(x) \in F[x]$. Now $F' \supset F$, so $f(x) \in F'[x]$. It follows that $KF'/F'$ is the splitting field for $f(x) \in F'[x]$. Hence $KF'/F'$ is a Galois extension.

• Since $K/F$ is Galois, every embedding of $K$ fixing $F$ is an automorphism of $K$, so the map \begin{equation*} \phi: \text{Gal}(KF'/F') \to \text{Gal}(K/F) \end{equation*} defined by $\sigma \mapsto \sigma|_K$ is well defined.

• $\phi$ is clearly a homomorphism, and \begin{equation*} \text{ker}(\phi) = \{ \sigma \in \text{Gal}(KF'/F'): \sigma|_K = 1\} \end{equation*}

• Now any $\sigma \in \text{Gal}(KF'/F')$ is trivial on $F'$. If $\sigma$ is also in the kernel of $\phi$, then $\sigma$ is trivial on $K$.

• It follows that any $\sigma \in \text{ker}(\phi)$ is trivial on the composite $KF'$.

It is this last bullet point that I am confused about. If $F'/F$ is a finite extension, then I can see how $\sigma\in \ker(\phi)$ gives $\sigma$ trivial on $KF'$, since any element of $KF'$ can be written out explicitly in terms of the basis elements of $K$ and $F'$.

So my question is:

If $F'$ is infinite-dimensional over $F$, how does one argue that $\sigma$ remains trivial on the composite $KF'$?

• One does not need finiteness of the extension to write out the elements explicitly. – Starfall Apr 7 '17 at 20:30
• @Starfall could you be a bit more specific...? How would you write a generic element of $KF'$? – Sam Y. Apr 8 '17 at 0:03
• math.stackexchange.com/questions/1226078/… – Ben West Apr 11 '17 at 1:25

Presumably $K$ and $F'$ here are both subfields of some big field $L$. By definition $KF'$ is the smallest subfield of $L$ containing both $K$ and $F'$. Now note that $L^\sigma=\{x\in L:\sigma(x)=x\}$ is a subfield of $L$, and it contains both $K$ and $F'$. Therefore $KF'\subseteq L^\sigma$.
So you don't actually need to know what elements of $KF'$ look like to prove this. As for what they do look like, every element of $KF'$ can be written as a quotient $$\frac{a_1b_1+\dots+a_nb_n}{c_1d_1+\dots+c_md_m}$$ where $a_1,\dots,a_n,c_1,\dots,c_m\in K$ and $b_1,\dots,b_n,d_1,\dots,d_m\in F'$. Indeed, clearly any such expression must be an element of $KF'$. Conversely, the set of such expressions is a subfield of $L$ (if you add, subtract, multiply or divide two such expressions you get another one of the same form) and contains $K$ and $F'$, and therefore every element of $KF'$ must be of that form.
• Great answer. One question though: in the quotient you wrote above, we would need the restriction that there exists some $i$ for which $c_id_i \neq 0_L$, right? (I know this is probably obvious, but just wanted to make sure we are on the same page.) – Sam Y. Apr 11 '17 at 16:32
• You need more than that: you need the restriction that the entire sum $c_1d_1+\dots+c_md_m$ is nonzero. (It's possible for the sum to be zero even if the individual terms are nonzero.) – Eric Wofsey Apr 11 '17 at 16:46