Quadratic closure of $Z_2$

Question

Find or disprove the existence of a minimal quadratically closed field extension of $Z_2$
I.e. $$\forall b,c \in S, \exists x\in S, x^2+ bx+c = 0$$ Note: Because $S$ is a field we can reduce every polynomial to a monic one by multiplying by $a^{-1}$.

Attempt 1
In $Z_2$, $x^2+x+1$ is irreducible, by introducing a new element, $b$, such that $b^2 + b + 1$(Equivalent to $Z_2[b]/(b^2+b+1)$).
I thought the field was quadratically closed.
but, $x^2+x+b$ was irreducible.
I tried to add more elements but it became hard to test all the quadratics.
Attempt 2
I tried proving $$\forall a, a +a=0 \Rightarrow \exists b,c \in S, \forall x\in S, x^2+ bx+c \neq 0$$ By constructing $b$ and $c$, I didn’t manage to.

Answer 1

Let $F$ be an algebraic closure of $\Bbb Z_2$. Now consider the following sequence of sets of subfields of $F$: $\mathcal X_1 = \{\Bbb Z_2\}$, and for any $n>1$ we set $\mathcal X_n$ to contain all elements of $\mathcal X_{n-1}$, as well as all quadratic extensions of all of the fields in $\mathcal X_{n-1}$ (still as subfields of $F$).

Now take the union $\mathcal X$ of all these families. It contains $\Bbb Z_2$. For any field $E\in \mathcal X$, any quadratic extension of $E$ is also in $\mathcal X$. Each element of $\mathcal X$ can be reached from $\Bbb Z_2$ doing some finite number of quadratic extensions. And finally, any two fields of $\mathcal X$ are both subfields of some other common superfield $\mathcal X$.

The union of all of the fields in $\mathcal X$ (as subfields / subsets of $F$) gives you the smallest quadratically closed subfield of $F$.

Answer 2

All finite extensions of $\mathbb{F}_q$ are of the form $\mathbb{F}_{q^n}$ for some positive $n$. Moreover, $\mathbb{F}_{q^a} \subseteq \mathbb{F}_{q^b}$ if and only if $a \mid b$.

Also, finite fields of all prime power orders exist and are unique up to isomorphism. This means that any degree two extension of $\mathbb{F}_2$ will give you $\mathbb{F}_4$ and any degree two extension of $\mathbb{F}_4$ will give you $\mathbb{F}_{16}$ and so on.

Note that $\mathbb{F}_{2^{2n}}$ is always a degree two extension of $\mathbb{F}_{2^n}$ (and it exists). So clearly no finite field extension will do. What about an infinite one?

If you need another hint: review the construction of $\mathbb{F}_p^{\rm alg}$.