I am working on the following problem:
Let $K/F$ be a finite, normal field extension and $L/K$ be any field extension. Furthermore, let $\varphi:K\longrightarrow L$ be an $F$-homomorphism. Show that $\varphi(K)\subseteq K$.
I have seen similar versions of this question that leverage the primitive element theorem to obtain an $n$ element basis where $n=[K:F]$. However, if my understanding is correct, you can only use the primitive element theorem for finite extensions with a finite number of intermediate fields. This applies to finite separable extensions, but here I only know my extension is finite and normal. I know that since my extension is normal, it is the splitting field of some polynomial over $F$, but after that I am getting stuck. Am I approaching this problem correctly? Do I even need the primitive element theorem here?
Best Answer
I was thinking too hard. We should be able to do it like this. $\newcommand{\ph}{\varphi}$
Since $K/F$ is normal, it is the splitting field of some polynomial $f(x)\in F[x]$. Denote by $\alpha_1,\ldots,\alpha_n$ the roots of $f(x)$. Since $K/F$ is finite (do we need this here?) we may write $K=F(\alpha_1,\ldots,\alpha_n)$. Now let $$ v=\sum_{i_1,\ldots,i_n}\lambda_{i_1\ldots i_n}\alpha_1^{i_1}\cdots\alpha_n^{i_n}\in K $$ where each $\lambda_{i_1\ldots i_n}\in F$. We have $$ \begin{align*} \ph(v)&=\sum_{i_1,\ldots,i_n}\ph(\lambda_{i_1\ldots i_n})\ph(\alpha_1^{i_1}\cdots\alpha_n^{i_n})\\ &=\sum_{i_1,\ldots,i_n}\lambda_{i_1\ldots i_n}\ph(\alpha_1)^{i_1}\cdots\ph(\alpha_n)^{i_n}\in K. \end{align*} $$ This is because $\ph$ is an $F$-homomorphism, so it fixes $F$ pointwise and permutes the roots of $f(x)$. Hence $\ph(K)\subseteq K$.