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.

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?

• By definition this is true. It's impossible to introduce transcendental elements via the process you mentioned as they can't be roots of polynomials in $F[x]$. Commented May 2, 2019 at 13:56
• @DonThousand But why can’t transcendental elements be roots of some polynomials in $F_1[x]$, or $F_{100}[x]$? Commented May 2, 2019 at 14:06
• Because the closure contains those sub-fields? Commented May 2, 2019 at 14:10
• @DonThousand I’m sorry but this looks likes a circular explanation. I mean, how do you prove that elements in $F_2,$F_3, \ldots$are still algebraic in$F\$? Commented May 2, 2019 at 14:23
• What textbook are you following? I can guarantee that a proof of this is included. Commented May 2, 2019 at 14:30

Suppose $$E/F$$ is a field extension such that $$F$$ is algebraic closed in $$E$$, and $$K$$ is the set of all algebraic elements (of $$F$$) in $$E$$.$$K$$ forms a field. $$K$$ is also algebraic closed by the following arguments:
Suppose $$\alpha\in F$$, and $$K(\alpha)$$ is an algebraic extension of $$K$$. Then $$K(\alpha)/F$$ is algebraic since both $$K(\alpha)/K$$ and $$K/F$$ is algebraic. By now, we know that $$\alpha$$ satisfies some polynomial $$f(x)\in F[x]$$, and $$\alpha$$ is an algebraic element in $$F$$. So $$\alpha\in K$$, and $$K(\alpha)$$ cannot be a proper extension of $$K$$. This leads to the conclusion.