I have to proof that is not possible with compass-and-straightedge to construct a square which has surface equal to a disk.
Let $M\subset \mathbb{C}$ with $\{0, 1\} \subset M$ and let $\cal{M}$ the set of numbers that can be constructed from $M$.
Let $ c(M) = \{ \overline{z} : z \in M \}$ and let $Q(M \cup c(M)):= \{ \cap \, K : K \, \ \text{subfield}, \, \,M \cup c(M) \subset K \}$.
I have some trouble to proof that the field-extension $Q(M \cup c(M)) \subset \cal{M}$ is algebraic (is the essential step in the squaring a circle problem).
Any suggestions? Thanks in advance!