Can abstract algebra be used to prove points are constructible? I am doing a self-study of abstract algebra, and have a basic question:
Using field extensions, abstract algebra can clearly be used to prove certain points are not constructible (with compass and straight edge).  But, can algebra be used to prove points are constructible with the same instruments? 
 A: Sort of. Once you've proven that a point is constructible only if its coordinates can be obtained by sequences of quadratic extensions of $\mathbb Q$, then you need to prove that any element of any such extension can actually be constructed, turning your "only if" into an "if and only if". In order to do this, we need to provide explicit steps to construct such a point, thus we do need to step outside of algebra and use classical geometry, but that's not really surprising.
Once one has proven that $(x, y)$ is constructible iff $(x, 0)$ and $(y, 0)$ are, the usual way is to find explicit geometric constructions to construct:


*

*$(a + b, 0)$ from $(a, 0)$ and $(b, 0)$.

*$(a - b, 0)$ from $(a, 0)$ and $(b, 0)$.

*$(ab, 0)$ from $(a, 0)$ and $(b, 0)$.

*$(a/b, 0)$ from $(a, 0)$ and $(b, 0)$.

*$(\sqrt a, 0)$ from $(a, 0)$.


Points 1 through 4 show that if we can construct any element of a field, we can construct the whole field. Since quadratic extensions are always obtained by adjoining square roots, point 5 shows that we can obtain any quadratic extension we like. Thus all points covered by our "only if" criterion are actually constructible.
Points 1 and 2 are easy using a few compass strokes. 3 and 4 are more complex, making use of the concept of similar triangles. Point 5 is an application of the Pythagorean theorem.
Thus, for example, I can conclude that the point $(1+\sqrt {2 - \sqrt 6}, \frac {1 + \sqrt 5} 2)$ is constructible.
