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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.

razivo
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2 Answers2

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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 $F$.

Arthur
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  • And $\mathcal F \neq F$? – razivo Aug 06 '20 at 14:55
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    @razivo $\mathcal F$ is a set of subfields of $F$. They are, as such, not equal. Yes, they are two different symbols. That may have been careless of me. I'll fix. – Arthur Aug 06 '20 at 14:55
  • I’m not sure I understand you, for example, in the natural numbers the sequence $F_n = {n}$’s union is the entire set. – razivo Aug 06 '20 at 15:02
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    @razivo It is, a priori, possible that we construct all of $F$ this way, yes. This is not a problem or an issue in any way. We construct the smallest quadratically closed subfield of $F$, and that's what matters. – Arthur Aug 06 '20 at 15:08
  • It is a different question, but one that doesn’t deserve a full post. – razivo Aug 06 '20 at 15:09
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    @razivo Also note that it is entirely possible to use known facts about finite fields to simplify this construction ($\mathcal X_n$ is just ${\Bbb F_2, \Bbb F_4, \ldots, \Bbb F_{2^{2^{n-1}}}}$). But with this construction you can quadratically close any field. In partucular $\Bbb Q$. – Arthur Aug 06 '20 at 15:11
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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 $\mathbb{F}_p^{\rm alg}$.

Sera Gunn
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  • This isn’t an answer, I was suspecting so, thanks for the confirmation. – razivo Aug 06 '20 at 14:42
  • How can we prove there doesn’t exist some field $S$ s.t. $\forall n, F_{2^n} \subsetneq S \subsetneq F_{2^\infty}$? As it is weird if the minimal quadratic closure is also an algebraic closure. – razivo Aug 06 '20 at 14:49
  • @razivo Show that any quadratically closed extension of $\mathbb{F}2$ must contain $\mathbb{F}_4$ and $\mathbb{F}{16}$ and, in general, $\mathbb{F}{2^{2^n}}$ for all $n$. By the way, I realize now I made some small mistakes in my answer: 1. it should be $\mathbb{F}{2^a} \subseteq \mathbb{F}{2^b}$ iff $a \mid b$ and 2. a degree two extension of $\mathbb{F}{2^n}$ would be $\mathbb{F}_{2^{2n}}$ not $2^{n + 1}$. Sorry about that. – Sera Gunn Aug 06 '20 at 14:57