1. ## Confusion about Normal Field Extension

Hello I was a little bit confused by something and would appreciate any help.

Let $\mathbb{F}$ be a field, and let $\mathbb{K}:\mathbb{F}$ be a finite, normal field extension. So it is the splitting field of some polynomial $f\in\mathbb{F}[x]$.

Let $\theta:K \rightarrow \overline{\mathbb{F}}$ be an $\mathbb{F}$-homomorphism (i.e. it fixes the ground field) into the algebraic closure of $\mathbb{F}$.

I am trying to show that $\theta(\mathbb{K}) = \mathbb{K}$.

So $\mathbb{K}=\mathbb{F}(\alpha_1,\ldots,\alpha_s)$ where $\alpha_i$ are the roots of the polynomial $f$. It follows then that $\theta$ permutes these root since it is an injective field homomorphism, but why does this then automatically mean $\theta (\mathbb{K}) = \mathbb{K}$ ? Is it true that $\mathbb{F}(\theta(\alpha_1),\ldots,\theta(\alpha_s )\)=\theta(\mathbb{F}(\alpha_1,\ldots,\alpha_s))?$ Thanks for any help

2. what does a typical element of $\mathbb{F}(\alpha_1,\dots ,\alpha_s)$ look like? what happens when you apply $\theta$ to such an element?

3. I think it looks like some multivariate polynomial function of $\alpha_1,\ldots,\alpha_n$,

so if i apply theta i just replace each $\alpha_i$ by $\theta(\alpha_i)$)?

But why is $\theta(\mathbb{K})$ generated by $\{ \theta(\alpha_1),\ldots\theta(\alpha_n)\}$?

4. because θ just permutes the roots, and fixes F. a permutation is a bijection, and θ maps the set { $\alpha_1,\dots ,\alpha_s$} to itself.

5. ok thanks