Corollary of Lifting Lemma for Algebraic Closures

Question

If $L$ is algebraic extension of a field $K$ and $F$ is an algebraically closed field. Then for any nonzero homomorphism $j: K \to F$, there exist $i: L \to F$ such that $i|_K = j$

(note $i|_K$ is the restriction of $i$ to K)

Now apparently the following is an immediate corollary:

For any nonzero homomorphisms $j:K \to L$, $i:K \to F$ with L algebraic over $j(K)$ and F algebraically closed, there exists $h: L \to F$ so that $h(j(a)) = i(a)$ for all $a\in K$.

If my notes are correct then the hint for this exercise is use Zorn, I did not see it, so I decided to think a little for myself. So in short I am wondering if this proof is correct or am I missing anything.

Proof:

Using the setup we can get a homomorphism $h': j(K) \to F$, where $h' = i \circ j'^{-1}$ and $j': K \to j(K)$, $j'(a) = j(a)$ for all $j\in K$.

So applying the theorem to h' we get our h.

$\forall a \in K$, $h(j(a)) = h'(j(a)) = i(a)$

Any feedback is greatly appreciated

Answer 1

Yeah, if I'm not mistaken, the proof is right. Zorn's lemma is supposed to be used only for the main theorem above, not the corollary.

I think the proof should have a few more clarifications at some points, so I'd write

Since $j$ is injective, the map $j':K\rightarrow j(K)$ defined by $j'(a)=j(a)$ is an ismorphism, so then we form the map $h'=i\circ (j')^{-1}: j(K)\rightarrow F$. Also, $L/j(K)$ is algebraic. Applying the theorem to $h'$ we get our $h$. For all $a\in K$, $h(j(a)) = h'(j(a)) = i(a)$.