Suppose I have a field $k,$ an algebraically closed field $L,$ and an embedding $k \hookrightarrow L.$ Then we know that for any algebraic extension of $k, E$ we can extend the embedding to $E \hookrightarrow L.$ However, if we take $L$ to be the algebraic closure of $k$ and $E$ to be a nontrivial finite extension of $L$ then $E$ is algebraic over $k$ and we have an embedding of $E \hookrightarrow L.$ So this would mean that $L$ and $E$ are infinite dimensional, otherwise, if $L$ and $E$ were finite, $E$ would have strictly greater dimension and such an embedding would not make sense. However, we have something like $\mathbb{C}$ that is a finite dimensional algebraic closure of $\mathbb{R}.$ So would this mean that there are no nontrivial finite extensions of algebraically closed fields in the first place?
For finite fields, we can always take its polynomial rings and modulo out by some irreducible. For example, $\mathbb{R}[X]/X^2 + 1.$ However, this does not work for algebraically closed fields. But I feel that we should be able to make nontrivial extensions for any field. Is this not the case?
Best Answer
Your idea is almost a complete proof. You start off well:
Let $k$ be a field and $L$ an algebraic closure of $k$. Let $K$ be a finite extension of $L$. Then $K/L$ is algebraic, hence $K/k$ is algebraic, hence there exists an embedding $\iota:\ K\ \longrightarrow L$.
Now let $\alpha\in K$ and let $f\in L[X]$ be its minimal polynomial over $L$. Then $f$ is irreducible over $L$ and $\iota(\alpha)\in L$ is a root of $f$, so $f$ is linear. Hence $\alpha\in L$ and so $K=L$.
As for the question in your comment, the answer is no. If $k$ is a field then $k[x]/\mathfrak{m}$ is a finite and simple extension of $k$, but not every extension of fields is finite and simple.