abstract algebra - Algebraic Closure & Algebraic Elements

I learn that the algebraic closure of rational numbers are algebraic numbers. Is this true in general, namely

For all field $F$ and its algebraic closure $\bar F$, does $\alpha\in \bar F$ imply $\alpha$ is algebraic over $F$?

My idea:

The set of algebraic elements in the algebraic closure $F$ forms a field, since for every algebraic elements $\alpha, \beta \in \bar F$, $\alpha+\beta, -\alpha, \alpha \beta, \alpha^{-1} \in F(\alpha, \beta)$. But $[F(\alpha, \beta)]= [F(\alpha)(\beta): F(\alpha)][F(\alpha):F]$ is finite, so all the four of them is again an algebraic element.

I think that what I conjectured is not true. Actually, in the famous proof of the existence of algebraic closure over a field (due to E. Artin), the algebraic closure of $F$ is constructed as follows:

1. Construct a field $F_1$ over $F_0:=F$ by taking quotient ring over a huge polynomial ring and its maximal ideal so that every polynomial $p(x)\in F[x]$ has a root.


2. Construct $F_2$, $F_3$, ... similarly.


3. Define $E=\bigcup_{k=0}^{\infty} F_k$. Then $E$ is the algebraic closure of $F$ that we want.

If the field of all algebraic numbers over any field is algebraic closed, then we do not have to construct $F_2$, $F_3$, ... sequentially. It seems that if $F$ is perfect, then every algebraic element over $F$ is contained in $F_1$, so $F_1$ is already sufficient. We don’t have to bother to write down the rest of the proof.

Of course, this is not a proof, but I can’t find any counterexample. Any supporting materials can be helpful. Thanks in advance.

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As Don Thousand said in the comments, this is true since it is impossible to add transcendental elements in the process I mentioned. However, I cannot come up with a proof that explains why elements in $F_2$, $F_3$, ... remains to be algebraic, nor did I find any material that support that claim. Is that true? Can you please provide a prove?