Same extension degree, then there is a field isomorphism

Question

I want to show the following theorem

Let $E/F$ and $K/F$ be field extensions with $[E:F]=n=[K:F]$. Then $\sigma : E\rightarrow K$ is an isomorphism, if $\sigma$ is an embedding such that $\sigma|_{F} = id_F$.

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So my starting point is $\sigma$ is injective field homomorphism. To show that is an isomorphism I need to show $\sigma$ is surjective. For the element in $F$ it seems no problem but how about the elements in $K$ but not in $F$?

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Trial from the comment : @Servaes

Let $\{w_1, \cdots, w_n\}$ be a basis of $E$ over $F$ and $\{v_1, \cdots, v_n\}$ be the basis of $K$ over $F$, any element $x \in E$ can be written as $x = \sum_{i=1}^n a_i w_i$, and applying $\sigma$, I have \begin{align} \sigma(x) = \sigma \left(\sum_{i=1}^n a_i w_i \right) = \sum_{i=1}^n \sigma(a_i) \sigma(w_i) = \sum_{i=1}^n a_i \sigma(w_i) \end{align} I now $\sigma(0)=0$, and thus I notices that $\{\sigma(w_i)\}$ is linearly independent set. I am trying to find surjectiveness out of definition. i.e., $\forall y \in K$, $\exists x \in E$, such that $y=\sigma(x)$, so in this sense I need to prove $\{\sigma(w_i)\}$ spans $K$.... But having trouble of showing this.

Another way : I know for the injective map between finite sets implies surjectiveness. Since $E, K$ are finite $n$ dimensional vector space over $F$, Is it safe to say surjective is naturally obtained?

Answer 1

Hint: Both $E$ and $K$ are $n$-dimensional $F$-vector spaces.

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Details:

The fact that $\sigma\vert_F=\operatorname{id}_F$ implies that $\sigma$ is a homomorphism of $F$-vector spaces. Because $\sigma$ is injective, its image is also $n$-dimensional, and hence it is surjective.