$L^H$ is a subfield of L

Question

Let $L$ be a field and $H < \mathrm{Aut}(L)$. Then: $L^H:=\lbrace a\in L:\phi(a)=a \ \forall \phi \in H \rbrace$ is a subfield of $L$.

My idea was to show it like that

$0 \in L^H$ since $\phi(0)=0$, $1\in L^H$ since $\phi(1)=1$. Now it has to be shown that it's closed under field operations. I thought about using a homomorphism.

In which way should it continue from here?

Answer 1

To complete the problem, use the fact that $\phi$ is a homomorphism of fields for all $\phi\in H.$ As an example, let $a,b\in L^H,$ and let $\phi\in H.$ Then \begin{align*} \phi(a + b) &= \phi(a) + \phi(b)\\ &= a + b, \end{align*} so $L^H$ is closed under addition. Multiplication is similar, and proving that $L^H$ has additive and multiplicative inverses will also crucially use the fact that $\phi$ is a homomorphism.

Below is the rest of the proof in all the gory detail; don't look unless you want to be spoiled!

Let $a,b\in L^H,$ and let $\phi\in H.$ Then $$\phi(ab) = \phi(a)\phi(b) = ab,$$ so $ab\in L^H.$ Thus $L^H$ is closed under multiplication. Now let's show that $L^H$ has additive inverses. Let $a\in L^H$ and $\phi\in H.$ Then $$0 = \phi(0) = \phi(a - a) = \phi(a) + \phi(-a) = a + \phi(-a).$$ Thus, $\phi(-a)$ is an additive inverse of $a.$ But, as additive inverses are unique, we must have $\phi(-a) = -a,$ so that $L^H$ has additive inverses. Finally, we show that $L^H$ has multiplicative inverses. Let $a\in L^H\setminus\{0\},$ and let $\phi\in H.$ Then $$1= \phi(1) = \phi(a\cdot a^{-1}) = \phi(a)\phi(a^{-1}) = a\cdot\phi(a^{-1}). $$ This implies that $\phi(a^{-1})$ is a multiplicative inverse of $a,$ but multiplicative inverses are unique! So, $\phi(a^{-1}) = a^{-1},$ and thus $L^H$ has multiplicative inverses as well. This completes the proof that $L^H$ is a field.