1.7. Algebraic reformulation 23 2 . Again we can apply Proposition 1.6.5 to get a quadratic extension F of Q such that both 1 and 2 are F -lines. Similarly, if (a, b) is the point of intersection of a line from Ln and a circle from Cn or of two circles from Cn, there is a quadratic extension F of Q such that these lines and circles are F -lines and F -circles. But then Lemma 1.6.9 implies that a and b belong to a quadratic extension of F , which is a quadratic extension of Q. This completes the proof. Exercises 1. Let F = Q( 2) show that F ( 7) = {w + x 2 + y 7 + z 14 : w, x, y, z Q}. 2. If F = Q( 2), show that F ( 3) = F ( 6). 3. Show that 3 / Q( 2). 4. Is there a rational number a such that Q( 6)( 7 + 2 6) = Q( a)? 5. If F = {Fα : α A} is a collection of fields contained in R, show that α is a field. 6. If an F -line passes through a point (x0,y0) and x0 F , show that y0 F . If an F -circle passes through a point (x0,y0) and x0 F , does it follow that y0 F ? 7. Show that a circle is an F -circle if and only if it passes through three F -points. 8. If a circle passes through an F -point and the tangent line to the circle at that point is an F -line, must the circle be an F -circle? 9. This exercise requires a knowledge of linear algebra over arbitrary fields. (a) If the field F is a quadratic extension of Q, show that F is a vector space over Q. (b) If K is a simple quadratic extension of the field F , show that K is a vector space over F of dimension 2. What is a basis? (c) if F is a quadratic extension of Q of order n, show that the dimension of F as a vector space over Q is also n. What is a basis? 1.7. An algebraic reformulation of the trisection problem In this section we take a crucial step in solving the trisection problem by reformulating the problem in terms of algebra. With any given angle an equation is presented and trisecting the angle is equivalent to finding a solution of the equation in the field K of constructible numbers. Say that an angle is constructible if the two lines that form the angle are constructible. The first step is to show that every constructible angle can
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