Find or disprove the existence of a minimal quadratically closed field extension of Z2 I.e.∀b,c∈S,∃x∈S,x2+bx+c=0Note: Because S is a field we can reduce every polynomial to a monic one by multiplying by a−1.

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Kendall Clark · Accepted Answer

Let F be an algebraic closure of Z2. Now consider the following sequence of sets of subfields of F:X1={Z2}, and for any n&amp;gt;1 we set Xn to contain all elements of Xn−1, as well as all quadratic extensions of all of the fields in Xn−1 (still as subfields of F).Now take the union X of all these families. It contains Z2. For any field E∈X, any quadratic extension of E is also in X. Each element of X can be reached from Z2 doing some finite number of quadratic extensions. And finally, any two fields of X are both subfields of some other common superfield X.The union of all of the fields in X (as subfields / subsets of F) gives you the smallest quadratically closed subfield of F.

Find or disprove the existence of a minimal

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